SOLUTIONS TO MATH 3110 HOMEWORK
From Lang’s Undergraduate Algebra, 3rd edition
Chapter I, §2, # 1,2.
1.
If 1 ≤ m ≤ n are integers, define the binomial coefficients
µ
¶
n!
n
=
,
m
m!(n − m)!
using the convention 0! = 1. Prove that
µ
¶ µ
¶ µ
¶
n
n
n+1
+
=
.
m−1
m
m
Proof.
To prove the above equation, we write the left-hand side of the equation in terms of factorials
and simplify:
µ
¶ µ
¶
n!
n!
n
n
+
=
+
m−1
m
(m − 1)!(n − (m − 1))! m!(n − m)!
n!
n!
=
+
(m − 1)!(n − m + 1))! m!(n − m)!
(n!)m + (n!)(n − m + 1)
=
m!(n − m + 1)!
n!(n + 1)
=
m!(n + 1 − m)!
µ
¶
(n + 1)!
n+1
=
.
=
m
m!(n + 1 − m)!
2.
Prove that for all integers x and y, we have
¶
n µ
X
n
(x + y) =
xi y n−i .
i
n
i=0
Proof. We will prove the above equation using weak induction (1st form) on n ≥ 1. In the case
n
0, both sides
of the equation are equal to 1 (provided x 6= −y). In the case, n = 1, we have
µ=¶
µ ¶
1
1
=1=
so that the sum on the right simplifies to x + y. Now assume that
1
0
¶
k µ
X
k
(x + y) =
xi y k−i
i
k
i=0
1
(this is our inductive hypothesis) and we will prove that the equation holds for n = k + 1. We have
(x + y)k+1 = (x + y)(x + y)k
à k µ
!
X k ¶
i k−i
= (x + y)
xy
i
i=0
à k µ
!
à k µ
!
X k ¶
X k ¶
i k−i
i k−i
=x
xy
+y
xy
i
i
i=0
i=0
! Ã k µ
!
à k µ
X k ¶
X k ¶
xi+1 y k−i +
xi y k+1−i
=
i
i
i=0
i=0
Ãk−1 µ
! Ã k µ
!
X k ¶
X k ¶
k+1
i+1 k−i
i k+1−i
=x
+
x y
+
xy
+ y k+1 .
i
i
i=0
i=1
Reindexing the first summation (replacing i by i − 1), we obtain
à k µ
! Ã k µ
!
¶
X
X k ¶
k
k+1
k+1
i k−i+1
i k+1−i
(x + y)
=x
+
xy
+
xy
+ y k+1
i−1
i
i=1
i=1
à k µµ
!
¶
µ
¶¶
X
k
k
k+1
i k+1−i
=x
+
+
xy
+ y k+1
i−1
i
i=1
!
à k µ
X k+1 ¶
= xk+1 +
xi y k+1−i + y k+1
(by Problem 1).
i
i=1
µ
¶
µ
¶
k+1
k+1
Since
=1=
, we conclude that
0
k+1
(x + y)k+1 =
k+1
Xµ
i=0
k+1
i
¶
xi y (k+1)−i .
Chapter I, §5, # 2,4,5,8,9.
2.
Let m and n be nonzero integers written in the form
m = pi11 pi22 · · · pirr
and
n = pj11 pj22 · · · pjrr ,
where iv and jv are nonnegative integers and p1 , p2 , . . . , pr are distinct prime numbers.
(a)
pk11 pk22
Show that the greatest common divisor (m, n) of m and n can be expressed as a product
· · · pkr r where k1 , k2 , . . . , kr are nonnegative integers. Express kv in terms of iv and jv .
Claim: (m, n) = pk11 pk22 · · · pkr r with kv = min(iv , jv ).
Proof. First, we show that if d is a common divisor of m and n such that
¡
¢
n
Given m
d , d = 1, there exists a, b ∈ Z such that
³n´
³m´
+b
= 1.
a
d
d
2
¡m
d
¢
, nd = 1, then (m, n) = d.
Multiplying by d, we obtain the linear combination
am + bn = d.
Since (m, n) is the smallest positive integer that can be written as a linear combination of m and n
and since (m, n) is the largest of all common divisors, we have that (m, n) = d. Now we use the
factorizations
m = pi11 pi22 · · · pirr and n = pj11 pj22 · · · pjrr
and note that
min(i1 ,j1 ) min(i2 ,j2 )
(ir ,jr )
d = p1
p2
· · · pmin
r
¡
¢
n
is a common divisor of m and n. We only need to show that m
d , d = 1 where
m
i −min(i1 ,j1 ) i2 −min(i2 ,j2 )
= p11
p2
· · · pirr −min(ir ,jr )
d
and
n
j −min(i1 ,j1 ) j2 −min(i2 ,j2 )
= p11
p2
· · · pjrr −min(ir ,jr ) .
d
Note that for each v, at most one of
iv − min(iv , jv )
and
jv − min(iv , jv )
is nonzero. If there exists an integer e > 1 that is a common divisor of
m
d
and
n
d,
then it can be uniquely
factored into a product primes and any prime p in its factorization is also a common divisor of m
d and
n
contradiction since only one of the powers of pv in m
d . So, p = pv for some 1 ≤ v ≤ r, which gives a
d
¡
¢
n
and nd can be nonzero. Thus, we conclude that m
d , d = 1, which gives (m, n) = d.
(b) Define the notion of least common multiple, and express the least common multiple of m
and n as a product p`11 p`22 · · · p`rr with nonnegative integers `v . Express kv in terms of iv and jv .
Let S be the set of all positive common multiples of m and n:
S = {s ≥ 1 | s = ma = nb for some a, b ∈ Z} .
Since mn ∈ S (so that S is nonempty) and S is a subset of the nonnegative integers, S must have a
least element by the Well-Ordering Property. The least element of S is called the least common multiple
of m and n and will be denoted [m, n]. While it is not necessary to provide a proof when working this
problem, we note that
4.
max(i1 ,j1 ) max(i2 ,j2 )
(ir ,jr )
[m, n] = p1
p2
· · · pmax
.
r
Let n be an integer ≥ 2.
(a)
Show that any integer x is congruent modulo n to a unique integer m such that 0 ≤ m < n.
3
Proof. By Theorem 2.1 (the Division Algorithm), there exist unique integers q and m such that
x = qn + m with 0 ≤ m < n. Reducing modulo n, we have
x ≡ qn + m ≡ m (mod n).
Note that if
x ≡ m ≡ m0
(mod n)
for
0 ≤ m ≤ m0 < n,
then n|(m0 − m) and −n < m0 − m < n. The only integer within this range that is divisible by n is
m0 − m = 0 so that m0 = m. Thus, m must be unique.
(b) Show that any integer x 6= 0, relatively prime to n, is congruent to a unique integer m relatively
prime to n, such that 0 < m < n.
Proof.
From part (a), we know that there exists a unique integer m such that
x ≡ m (mod n)
with
0 ≤ m < n.
We just need to show that when (x, n) = 1, it is not possible for m = 0 and that (m, n) = 1. If m = 0,
then x = nq for some q ∈ Z and (x, n) = n, contradicting the assumption that n and x are relatively
prime. Also, if (m, n) = d > 1, then de = m and df = n for some e, f ∈ Z. From the Division
Algorithm, we have
x = qn + m = qdf + de = d(qf + e)
so that d|x. Of course, if d|x and d|n, then (x, n) ≥ d by definition, contradicting our assumption.
Thus, m and n must also be relatively prime.
(c) Let ϕ(n) be the number of integers m relatively prime to n, such that 0 < m < n. We call ϕ
the Euler phi function. We also define ϕ(1) = 1. If n = p is a prime number, what is varphi(p)?
Note that any divisor of any of 1, 2, . . . , p − 1 must also be less than p. Since the only divisors of p are
1 and p, we have (i, p) = 1 for all 1 ≤ i ≤ p − 1. Thus, ϕ(p) = p − 1.
(d)
Determine ϕ(n) for each integer n with 1 ≤ n ≤ 10.
ϕ(1) = 1
ϕ(2) = 1
ϕ(3) = 2
ϕ(4) = 2
5.
ϕ(5) = 4
ϕ(6) = 2
ϕ(7) = 6
ϕ(8) = 4
ϕ(9) = 6
ϕ(10) = 4
Chinese Remainder Theorem. Let n, n0 be relatively prime positive integers and let a, b ∈ Z.
Show that the congruences
x ≡ a (mod n)
x ≡ b (mod n0 )
can be solved simultaneously with some x ∈ Z.
4
Proof. Assuming that (n, n0 ) = 1, there exist x1 , x2 ∈ Z such that x1 n + x2 n0 = 1. Then we have
solutions to the congruences
x1 n ≡ 1 − x2 n0 ≡ 1 (mod n0 ),
x2 n0 ≡ 1 − x1 n ≡ 1 (mod n).
Setting
x = ax2 n0 + bx1 n,
we obtain an integer that satisfies the congruences
x ≡ a (mod n)
x ≡ b (mod n0 ),
completing the proof of the claim.
Now we generalize the theorem stated above. Let n1 , n2 , . . . nr be positive integers that are pairwise
relatively prime (ie., (ni , nj ) = 1 for i 6= j). Then the system of congruences
x ≡ a1
(mod n1 )
x ≡ a2
..
.
(mod n2 )
x ≡ ar
(mod nr )
can be solved with some x ∈ Z.
Proof.
Let N = n1 n2 · · · nr and for each 1 ≤ i ≤ r define
Ni =
N
= n1 · · · ni−1 ni+1 · · · nr .
ni
Since (Ni , ni ) = 1, it is possible to find an integer xi such that
Ni xi ≡ 1
(mod ni )
for each i. Consider the integer
x = a1 N1 x1 + a2 N2 x2 + · · · ar Nr xr
and observe that Ni ≡ 0 (mod nj ) whenever i 6= j. Thus, for each i, we have
x ≡ a1 N1 x1 + a2 N2 x2 + · · · ar Nr xr ≡ ai Ni xi ≡ ai
(mod ni ),
providing a solution to the system of congruences.
8.
µ
p
n
Let p be a prime number and n and integer, 1 ≤ n ≤ p − 1. Show that the binomial coefficient
¶
is divisible by p.
5
Proof.
By definition we have
µ
p
n
¶
=
p!
,
n!(p − n)!
which gives the number of n-element subsets of a p-element set. In particular,
integer. We obtain the integer equation
µ
p
n
µ
p
n
¶
is a positive
¶
n!(p − n)! = p(p − 1)!
so that p divides the left-hand side of the equation. However, all of the prime factors of n! and (p − n)!
are less than p and must therefore be relatively prime to p. Applying Euclid’s Lemma to
¯µ
¶
¯ p
n!(p − n)! where (p, n!(p − n)!) = 1,
p ¯¯
n
µ
we conclude that p divides
p
n
¶
.
6
9.
For all integers x, y and primes p show that
(x + y)p ≡ xp + y p
Proof.
(mod p).
From Chapter I, §2, # 2, we have
µ
¶
µ
¶
p
p
p−1
p
p
(x + y) ≡ x +
x y + ··· +
xy p−1 + y p .
1
p−1
From Problem 8, we know that all of the binomial coefficients
µ
¶ µ
¶
µ
¶
p
p
p
≡
≡ ··· ≡
≡0
1
2
p−1
(mod p),
resulting in the above claim.
Exercise 1.
Let x, y, z ∈ Z with z 6= 0 such that
xz ≡ yz
Prove that
x≡y
Proof.
(mod n).
µ
mod
n
(z, n)
¶
.
From the definition of
xz ≡ yz
(mod n),
we have that n|(xz − yz). So, for some m ∈ Z, we have
nm = (x − y)z
n
z
(z, n)m = (x − y)
(z, n)
(z, n)
n
z
m = (x − y)
(z, n)
(z, n)
=⇒
=⇒
so that
¯
n ¯¯
z
,
(x − y)
(z, n) ¯
(z, n)
³
´
n
z
with (z,n)
, (z,n)
= 1 (this was a consequence of the first part of the proof of Chapter I, §5, # 2(a)).
We can write
n
z
1=a
+b
for some a, b ∈ Z,
(z, n)
(z, n)
which gives
(x − y) = a(x − y)
The integer
n
(z,n)
n
z
+ b(x − y)
.
(z, n)
(z, n)
divides both of the expressions on the right side of this equation, and must therefore
divide (x − y) resulting in the congruence
x≡y
µ
mod
as desired.
7
n
(z, n)
¶
,
Chapter II, §1, # 1-4, 6, 9, 11, 13, 15-17, Exercise 2, Exercise 3.
1.
Let G be a group and a, b, c ∈ G. If ab = ac, show that b = c.
Proof.
So,
Since G is a group, every element must have an inverse. In particular, a−1 ∈ G since a ∈ G.
ab = ac
=⇒
a−1 ab = a−1 ac
=⇒
b = c.
This property is usually referred to as the “cancellation property” of groups.
2.
Let G, G0 be finite groups of orders m, n, respectively. What is the order of G × G0 ?
By definition,
G × G0 = {(g, g 0 ) | g ∈ G and g 0 ∈ G0 } .
If x is a fixed element in G, there are n ordered pairs in G × G0 of the form (x, ·). The element x can
in be chosen from m possibilities, creating a total of mn elements in the direct product.
3.
Let x1 , x2 , . . . , xn be elements of a group G. Show (by induction) that
−1
(x1 · · · xn )−1 = x−1
n · · · x1 .
We begin with the n = 2 can and use weak induction. Let x1 , x2 ∈ G. Then
−1
−1 −1
−1
(x1 x2 )(x−1
2 x1 ) = x1 (x2 x2 )x1 = x1 x1 = e
−1
shows that (x1 x2 )−1 = x−1
2 x1 . Now assume that
−1
(x1 · · · xk )−1 = x−1
k · · · x1 .
Then
−1
−1
−1
(x1 · · · xk+1 )−1 = ((x1 · · · xk )xk+1 )−1 = x−1
= x−1
k+1 (x1 · · · xk )
k+1 xk · · · x1 ,
using both the n = 2 and n = k cases.
Additively, the above statement can be written as
−(x1 + · · · + xn ) = (−xn ) + · · · + (−x1 ).
4.
(a)
Let G be a group and x ∈ G. Suppose that there is an integer n ≥ 1 such that xn = e.
Show that there is an integer m ≥ 1 such that x−1 = xm .
8
Proof. Assuming that xn = e for n ≥ 1, we see that the set of all positive integers k such that xk = e
is nonempty. By the Well-Ordering Property, it must have a least element, which we denote by d and
call the order or period of x. By Theorem 1.1,
hxi = {e, x, x2 , . . . xd−1 }.
In particular, the element xd−1 satisfies xxd−1 = xd = e so that x−1 = xd−1 . Of course, we only have
the inequality d − 1 ≥ 0 (not d − 1 ≥ 1 as required). However, in the case d − 1 = 0, we have d = 1 so
that x = e. Since x−1 = e in this case, the integer m = 1 satisfies the claim.
(b)
x = e.
Let G be a finite group. Show that given x ∈ G, there exists an integer n ≥ 1 such that
n
Proof. We consider the subgroup hxi, which must be finite since it is contained in G. Also, x−1 ∈ hxi
which implies that there exists a nonnegative integer m such that x−1 = xm . It follows that n = m + 1
satisfies the claim:
xm+1 = xxm = xx−1 = e.
6.
Let G be a group such that x2 = e for all x ∈ G. Prove that G is abelian.
Proof. Note that the property we are assuming of G is equivalent to saying that every element in G
is its own inverse. Assume that x and y are two arbitrary elements of G. We know from Problem 3 of
this section (the n = 2 case), that (xy)−1 = y −1 x−1 . Since x and y are their own inverses, we have
(xy)−1 = yx.
On the other hand, xy ∈ G is its own inverse ((xy)−1 = xy) so that we obtain xy = yx.
9.
There exists a group G of order 8 having generators denoted by i, j, k such that
ij = k,
jk = i,
ki = j,
and i2 = j 2 = k 2 ,
where we denote i2 by m.
(a)
Show that every element of G can be written in the form
e, i, j, k, m, mi, mj, mk,
and hence that these are precisely the distinct elements of G.
9
Proof.
We must show that hi, j, ki = {e, i, j, k, m, mi, mj, mk}. First, note that the properties
ij = k,
jk = i,
ki = j
yield
mj = i2 j = ik,
mk = j 2 k = ji,
mi = k 2 i = kj,
showing that G is nonabelian. Next, we show that m2 = e by multiplying it times each of the generators:
m2 i = mkj = jij = jk = i
im2 = imi2 = ikji = mjji = m2 i = i,
m2 j = mik = kjk = ki = j
jm2 = jmj 2 = jikj = mkkj = m2 j = j,
m2 k = mji = iki = ij = k
km2 = kmk 2 = kjik = miik = m2 k = k.
To complete the proof, we refer to the multiplication table given in part (b), noting that both the closure
and inverse properties are satisfied.
(b)
Make up a multiplication table.
Table 1: Multiplication table for the quaternion group.
e
i
j
k
m
mi
mj
mk
e
e
i
j
k
m
mi
mj
mk
i
i
m
k
mj
mi
e
mk
j
j
j
mk
m
i
mj
k
e
mi
k
k
j
mi
m
mk
mj
i
e
m
m
mi
mj
mk
e
i
j
k
mi
mi
e
mk
j
i
m
k
mj
mj
mj
k
e
mi
j
mk
m
i
mk
mk
mj
i
e
k
j
mi
m
11. (*Modified*) Let r denote a 90◦ rotation of a square in the counterclockwise direction and let s
denote a flip about line of reflection. In class, we defined the dihedral group D4 to be the group
{e, r, r2 , r3 , s, sr, sr2 , sr3 | s2 = e = r4 and rs = sr3 }.
Construct a multiplication table for D4 .
13.
Let G be a group and H a subgroup. Let x ∈ G (which we fix) and define
xHx−1 := {xyx−1 | y ∈ H} ⊆ G.
Show that xHx−1 is a subgroup of G.
10
Table 2: Multiplication table for the dihedral group D4 .
e
r
r2
r3
s
sr
sr2
sr3
e
e
r
r2
r3
s
sr
sr2
sr3
r
r
r2
r3
e
sr3
s
sr
sr2
r2
r2
r3
e
r
sr2
sr3
s
sr
r3
r3
e
r
r2
sr
sr2
sr3
s
s
s
sr
sr2
sr3
e
r
r2
r3
sr
sr
sr2
sr3
s
r3
e
r
r2
sr2
sr2
sr3
s
sr
r2
r3
e
r
sr3
sr3
s
sr
sr2
r
r2
r3
e
Proof. Since H is a subgroup of G, we have that e ∈ H. Thus, e = xx−1 = xex−1 ∈ xHx−1 , from
which we see that xHx−1 6= ∅. So, it remains to be shown that xHx−1 satisfies the closure and inverse
properties. Let z, z 0 ∈ xHx−1 with z = xyx−1 and z 0 = xy 0 x−1 . Then
zz 0 = (xyx−1 )(xy 0 x−1 ) = xy(x−1 x)y 0 x−1 = x(yy 0 )x−1 ∈ xHx−1
since yy 0 ∈ H. We also have that y −1 ∈ H so that xy −1 x−1 ∈ xHx−1 and
(xyx−1 )(xy −1 x−1 ) = xy(x−1 x)y −1 x−1 = x(yy −1 )x−1 = xx−1 = e.
It follows that (xyx−1 )−1 = xy −1 x−1 , and we conclude that xHx−1 is a subgroup of G.
15. A root of unity in C is a number ζ such that ζ n = 1 for some positive integer n. We say that ζ
is an nth root of unity. Describe the set of nth roots of unity in C. Show that this set is a cyclic group
of order n.
Proof.
Recall from class that to extend the R-valued function x 7→ ex to C, we defined
eiθ := cos(θ) + i sin(θ).
This extension seemed to be the natural choice after investigating the MacLaurin Series expansions for
ex , sin x, and cos x. Note that
e2πi = cos(2π) + i sin(2π) = 1
and that solving the equation
cos(θ) + i sin(θ) = 1
yields only the solutions θ = 2πm, where m ∈ Z. Thus, eiθ is periodic with period 2π and we can define
the element ζn = e2πi/n , which is clearly an nth root of unity. In fact, we see that whenever 0 ≤ k < n,
the element ζnk = e2πik/n is an nth root of unity and the periodicity of eiθ implies that
¯
n
o
¯
µn := hζn i = e2πik/n ¯ 0 ≤ k < n
11
is a subgroup of C having order n.
16. Let G be a finite cyclic group of order n. Show that for each positive integer d dividing n, there
exists a subgroup of order d.
We have that G = hai = {e, a, a2 , . . . , an−1 } for some a ∈ G that satisfies an = e (with n
being the order of a). Let d be a positive integer that divides n and consider the subgroup han/d i. Since
(an/d )d = an = e, it follows that the elements
Proof.
e, an/d , (an/d )2 , . . . , (an/d )d−1
are all in han/d i. We only need to show that the are distinct to complete the proof. Suppose that
(an/d )i = (an/d )j ,
for
0≤i≤j<d
(without loss of generality, we are assuming i ≤ j). Then we have
(an/d )i (an/d )−i = (an/d )j (an/d )−i
implies (an/d )j−i = e. However, the inequalities
0≤i≤j<d
=⇒
0≤j−i<d
=⇒ 0 ≤ (n/d)(j − i) < (n/d)d = n
contradict the fact that a has order n. Thus, the powers
e, an/d , (an/d )2 , . . . , (an/d )d−1
are distinct and |han/d i| = d.
17.
Let G be a finite cyclic group of order n and let a be a generator. Let r be a nonzero integer
that is relatively prime to n (note that G is nontrivial in order for (r, n) to be defined - ie., n > 1).
(a)
Proof.
Show that ar is also a generator for G.
Consider the subgroup har i ⊆ hai = G. To show that the subgroup is equal to G, we must
prove that the elements
e, ar , (ar )2 , . . . , (ar )n−1
are distinct. Suppose that
(ar )i = (ar )j
for some
0 ≤ i ≤ j < n.
Then
(ar )i (ar )−i = (ar )j (ar )−i
12
=⇒
(ar )j−i = e,
where 0 ≤ j − i < n. However, the only powers of a that result in e are multiples of the order. Thus,
n|r(j − i). From (r, n) = 1, there exists a, b ∈ Z such that ar + bn = 1. Multiplying by j − i, we have
the equation.
(j − i) = a(j − i)r + b(j − i)n.
The integer n divides both of the terms on the right and must therefore divide j − i. The only such
integer within the range 0 ≤ j − i < n is j − i = 0. We conclude that i = j and har i = hai.
(b)
Show that every generator of G can be written in this form.
Proof. We are trying to prove that if har i = hai, then (r, n) = 1. To accomplish this, we will prove
that har i = ha(r,n) i. There exists x, y ∈ Z such that (r, n) = rx + ny. Since
a(r,n) = arx+ny = (ar )x (an )y = (ar )x ∈ har i,
it follows that ha(r,n) i ⊆ har i. From the definition of a greatest common divisor, we have r = (r, n)s for
some s ∈ Z. Thus,
ar = (a(r,n) )s ∈ ha(r,n) i,
which implies har i ⊆ ha(r,n) i. Hence, we conclude that ha(r,n) i = har i.
(c)
Let p be a prime number, and G a cyclic group of order p. How many generators does G have?
The integers 1, 2, . . . , p − 1 are all relatively prime to p. From part (a) of this problem, we see that the
p − 1 elements a, a2 , . . . , ap−1 are all generators.
Exercise 2.
Find all subgroups of D4 and identify which ones are cyclic.
First, we list all of the cyclic subgroups:
hei = {e} (the trivial subgroup)
hri = {e, r, r2 , r3 } = hr3 i
hr2 i = {e, r2 }
hsi = {e, s}
hsri = {e, sr}
hsr2 i = {e, sr2 }
hsr3 i = {e, sr3 }
Note that D4 itself is not a cyclic subgroup and the remaining noncyclic subgroups all have order 4.
Since no element of these groups can have an element with order 4, every nonidentity element will have
order 2. After checking a few possibilities, the remaining subgroups are found to be
{e, s, r2 , sr2 }
and
13
{e, r2 , sr, sr3 }.
Exercise 3.
Proof.
Prove that if m, n ∈ Z are nonzero, then hm, ni = h(m, n)i.
By definition, the subgroup
hm, ni = {am + bn | a, b ∈ Z}.
The variation of Theorem 3.1/3.2 from Chapter I that we stated in class asserted that (m, n) can be
written as a linear combination of m and n:
(m, n) = xm + yn
for some
x, y ∈ Z.
Then for every integer c, we have
c(m, n) = c(xm + yn) = (cx)m + (cy)n ∈ hm, ni,
implying that h(m, n)i ⊆ hm, ni. On the other hand, since (m, n) is a divisor of m and n, we can write
m = d(m, n)
and
n = e(m, n),
for some d, e ∈ Z. Thus, we have
am + bn = ad(m, n) + be(m, n) = (ad + be)(m, n) ∈ h(m, n)i,
implying that hm, ni ⊆ h(m, n)i. Putting together the two set containments, we obtain
hm, ni = h(m, n)i.
Chapter II, §3, # 1, 4, 5, 10, 11, Exercise 4.
1. Let R× be the multiplicative group of nonzero real numbers. Describe explicitly the kernel of the
absolute value homomorphism x 7→ |x| of R× into itself. What is the image of this homomorphism?
We are given that the absolute value defines a homomorphism f : R× −→ R× (writing f (a) = |a|).
This follows from the familiar property |ab| = |a| · |b|. The kernel consists of all elements that map to
the identity 1. Namely,
Ker(f ) = {±1}.
Since |a| > 0 for all a ∈ R× , the image is given by
f (R× ) = {b ∈ R× | b > 0}.
4.
Let G be a group, a ∈ G, and define the map ca : G −→ G by
ca (x) = axa−1 .
(a)
Show that ca : G −→ G is an automorphism of G.
14
Proof.
Let x, y ∈ G. Then
ca (x)ca (y) = (axa−1 )(aya−1 ) = ax(a−1 a)ya−1 = axya−1 = ca (xy),
showing that ca preserves the group operation. To see that the map is injective, note that
ca (x) = ca (y)
=⇒
axa−1 = aya−1
=⇒
a−1 axa−1 a = a−1 aya−1 a
=⇒
x = y.
Finally, for any x ∈ G, the element a−1 xa 7→ x, proving that the map is surjective.
(b)
Show that the set of all such maps ca with a ∈ G is a subgroup of Aut(G).
Proof. By part (a), the set of all such maps is a subset of Aut(G) and since there is at least one
element in G, namely eG , the subset is nonempty. We just need to show that it is closed under the
operation in Aut(G), composition, and that inverses are also in the set. Let a, b ∈ G and consider
(ca ◦ cb )(x) = ca (cb (x)) = ca (bxb−1 ) = abxb−1 a−1 = (ab)x(ab)−1 = cab ,
for all
x ∈ G.
Now, note that (ca )−1 = ca−1 since
(ca ◦ ca−1 )(x) = ca (ca−1 (x)) = ca (a−1 xa) = aa−1 xaa−1 = x
and
(ca−1 ◦ (ca )(x) = ca−1 (ca (x)) = ca−1 (axa−1 ) = a−1 axa−1 a = x
are the identity automorphisms. The resulting group is given a name in the next problem.
5. Let the notation be as in the previous problem. Show that the association a 7→ ca is a homomorphism of G into Aut(G). The image of this homomorphism is called the group of inner automorphisms
of G and we denote it by Inn(G). Thus , an inner automorphism of G is one which is equal to ca for
some a ∈ G.
Proof. In the previous problem, we proved that a 7→ ca is a homomorphism when we showed that
ca ◦ cb = cab .
10.
Let f : G −→ G0 be a group isomorphism and a ∈ G. Show that the order of a is the same as
the order of f (a).
Proof.
If n ≥ 1 satisfies an = eG , then from the basic properties of homomorphisms, we have that
eG0 = f (eG ) = f (an ) = (f (a))n .
Thus, |f (a)| ≤ |a|. On the other hand, since f is an isomorphism, f −1 : G0 −→ G exists and also
satisfies the properties of homomorphisms. Let a0 = f (a) so that f −1 (a0 ) = a and note that for any
m ≥ 1 such that a0m = eG0 , we have
eG = f −1 (eG0 ) = f −1 (a0m ) = (f −1 (a0 ))m = am .
15
Thus, |a| ≤ |f (a)|. Hence, we conclude that |a| = |f (a)|, and note that these arguments hold even if the
order is infinite.
11.
Let G be a cyclic group and f : G −→ G0 a homomorphism. Show that the image of G is cyclic.
Proof. Suppose that G = hai, then we claim that G0 = hf (a)i. Let a0 ∈ f (G), then by definition,
a0 = f (b) for some b ∈ G. However, b = am for some m ∈ Z so that a0 = f (am ) = (f (a))m , proving
our claim.
Exercise 4.
abelian.
Let f : G −→ G0 be a group isomorphism. Prove that G is abelian if and only if G0 is
Proof. We begin by proving that f −1 : G −→ G (which necessarily exists since f is bijective) is a
homomorphism. Let a0 , b0 ∈ G0 such that f (a) = a0 and f (b) = b0 and consider
f −1 (a0 )f −1 (b0 ) = f −1 (f (a))f −1 (f (b)) = ab = f −1 (f (ab)) = f −1 (f (a)f (b)) = f −1 (a0 b0 ).
Thus, f −1 must also be an isomorphism. Now assume that G is abelian: ab = ba for all a, b ∈ G. Then
a0 b0 = f (a)f (b) = f (ab) = f (ba) = f (b)f (a) = b0 a0 ,
proving that G0 must also be abelian. On the other hand, if we assume that G0 is abelian, then
ab = f −1 (a0 )f −1 (b0 ) = f −1 (a0 b0 ) = f −1 (b0 a0 ) = f −1 (b0 )f −1 (a0 ) = ba
proves that G must also be abelian.
Chapter II, §4, # 1, 6, 7, 10, 11, 21, Exercise 5.
1. Let G be a group and H a subgroup. If x, y ∈ G, define x to be equivalent to y if x is in the coset
yH. Prove that this is an equivalence relation.
Proof. We must show that the reflexive, symmetric, and transitive properties are all satisfied. First,
note that eG ∈ H implies that x = xeG ∈ xH so that x is equivalent to itself (the reflexive property).
Next, assume that that x ∈ yH. Then x = yh for some h ∈ H. Since H is a group, h−1 ∈ H and we
have y = xh−1 ∈ xH. Thus, if x is equivalent to y, then y is equivalent to x (the symmetric property).
Finally, assume that x ∈ yH and y ∈ zH. Then x = yh and y = zh0 for some h, h0 ∈ H. We have that
x = yh = zh0 h ∈ zH
since H is closed. So, if x is equivalent to y and y is equivalent to z, we have shown that x is equivalent
to z (the transitive property).
16
6.
Let H1 and H2 be two normal subgroups of a group G. Show that H1 ∩ H2 is normal.
Proof.
First, we prove that H1 ∩ H2 is a subgroup. It must be nonempty since eG ∈ H1 and eG ∈ H2
implies that eG ∈ H1 ∩ H2 . Let a, b ∈ H1 ∩ H2 . Since H1 and H2 are both closed, we have ab ∈ H1 and
ab ∈ H2 so that ab ∈ H1 ∩ H2 . Also, a−1 ∈ H1 and a−1 ∈ H2 implies a−1 ∈ H1 ∩ H2 . So far, we have
only assumed that H1 and H2 are subgroups in order to prove that H1 ∩ H2 is a subgroup. Now we use
the assumption that they are normal subgroups:
xH1 x−1 ⊆ H1
and
xH2 x−1 ⊆ H2
for all x ∈ G.
Let h ∈ H1 ∩ H2 and consider xhx−1 . Since h ∈ H1 , we have xhx−1 ∈ H1 and since h ∈ H2 , we have
xhx−1 ∈ H2 . Thus, xhx−1 ∈ H1 ∩ H2 for all h in H1 ∩ H2 . Hence, x(H1 ∩ H2 )x−1 ⊆ H1 ∩ H2 .
7.
Let f : G −→ G0 be a group homomorphism and let H 0 be a normal subgroup of G0 . Let
H = f −1 (H 0 ) (the preimage of H 0 ).
(a)
Prove that H is normal in G.
Proof. We have already shown that H is a subgroup of G. To see that H is normal, fix and element
x ∈ G and let h ∈ H. Then f (h) = h0 for some h0 ∈ H 0 . We have
f (xhx−1 ) = f (x)f (h)f (x−1 ) = f (x)h0 f (x)−1 ∈ H 0
since f (x) ∈ G0 and H 0 is normal in G0 . Thus, by the definition of H, we have that xhx−1 ∈ H implying
that xHx−1 ⊆ H.
(b)
Prove Theorem 4.8.
Theorem 4.8 Let f : G −→ G0 be a surjective homomorphism. Let H 0 be a normal subgroup of
G0 and let H = f −1 (H 0 ). Then H is normal and the map x 7→ f (x)H 0 is a homomorphism of G onto
G0 /H 0 whose kernel is H. Therefore, we obtain an isomorphism
G/H ∼
= G0 /H 0 .
Proof. In part (a), we showed that H is normal. Since H and H 0 are normal, it makes sense to
define the quotient groups G/H and G0 /H 0 . We must first show that the map φ : G −→ G0 /H 0 given
by φ(x) = f (x)H 0 is a homomorphism. If x, y ∈ G, then
φ(x)φ(y) = f (x)H 0 f (y)H 0 = f (x)f (y)H 0 = f (xy)H 0 = φ(xy).
Applying the First Isomorphism Theorem for Groups to φ, we obtain the isomorphism
G/(Kerφ) ∼
= φ(G).
17
(1)
It is assumed in Theorem 4.8 that f is surjective, so that for any x0 ∈ G0 , there exists an x ∈ G such
that f (x) = x0 . If we consider the coset x0 H 0 in G0 /H 0 , we see that φ(x) = x0 H 0 . Hence, φ is also
surjective and the isomorphism (1) becomes
G/(Kerφ) ∼
= G0 /H 0 .
It remains to be shown that Kerφ = H. Let h ∈ H, in which case we have f (h) ∈ H 0 . Then
φ(h) = f (h)H 0 = H 0 (the identity coset in G0 /H 0 ), proving that H ⊆ Kerφ. Now suppose that
k ∈ Kerφ. Then φ(k) = f (k)H 0 = H 0 , implying that f (k) ∈ H 0 . By the definition of H, we have k ∈ H
so that Kerφ ⊆ H. We conclude that H = Kerφ, giving the desired isomorphism.
10.
Let G be a group. Define the center of G, usually denoted Z(G), to be the set of all elements
a ∈ G such that ax = xa for all x ∈ G. Show that the center is a subgroup, and that it is a normal
subgroup. Show that it is the kernel of the conjugation homomorphism x 7→ γx in Exercise 5, §3.
Proof.
First, we show that Z(G) is a group. Note that ex = x = xe for all x ∈ G implies that
e ∈ Z(G) so that Z(G) 6= ∅. Now let a, b ∈ Z(G). By definition, ax = xa and bx = xb for all x ∈ G so
that
(ab)x = a(bx) = a(xb) = (ax)b = (xa)b = x(ab)
for all x ∈ G.
Thus, Z(G) is closed. Multiplying both sides of ax = xa on the right and left by a−1 yields
a−1 axa−1 = a−1 xaa−1
xa−1 = a−1 x
=⇒
for all x ∈ G.
So, Z(G) contains inverses and must be a subgroup of G. Now, we show that Z(G) is the kernel of the
homomorphism f : G −→ Inn(G) given by a 7→ ca . (Note: in the statement of Problem 10, Lang uses
γ in place of c to distinguish between the homomorphisms G −→ Inn(G) and G −→ Aut(G). Since it
seems more natural to denote the conjugation map by c, we will not adopt Lang’s new notation.) First,
let a ∈ Z(G) and notice that the map
ca (x) = axa−1 = (xa)a−1 = x(aa−1 ) = x = ce (x)
is the identity map. This proves that Z(G) ⊆ Kerf . Now let b ∈ Kerf so that cb (x) = ce (x) for all
x ∈ G. We have
bxb−1 = x
=⇒
bx = xb for all x ∈ G,
proving that b ∈ Z(G) and Kerf ⊆ Z(G). We conclude that Kerf = Z(G) and note that we have
previously shown that the kernel of a homomorphism is normal.
11.
Let G be a group and H a subgroup. Let NH be the set of all x ∈ G such that xHx−1 = H.
Show that NH is a group containing H and that H is normal in NH . The group NH is called the
normalizer of H.
18
Proof. First, we prove that NH is a subgroup of G (in which case, it will necessarily be a group).
Since eHe−1 = H, we have that e ∈ NH and NH 6= ∅. Suppose that a, b ∈ NH so that aHa−1 = H and
bHb−1 = H. Then
(ab)H(ab)−1 = abHb−1 a−1 = aHa−1 = H,
so that ab ∈ NH . Also,
H = aHa−1
=⇒
a−1 Ha = a−1 aHa−1 a = H
implies that a−1 ∈ NH . Thus, NH is a subgroup of G. If h ∈ H, we have hHh−1 = hH = H, so that
H ⊆ NH . To see that H is normal in NH , we observe that by the definition of NH , whenever a ∈ NH ,
aHa−1 = H.
21.
(a)
Let G be an abelian group and H a subgroup. Show that G/H is abelian.
Proof.
Assume that G is abelian and let xH and yH be two left cosets of H in G. Note also that
since xh = hx for every x ∈ G and h ∈ H, xH = Hx, H is necessarily normal, and G/H is defined.
Then
(xH)(yH) = (xy)H = (yx)H = (yH)(xH)
shows that G/H must also be abelian.
(b) Let G be a group and H a normal subgroup. Show that G/H is abelian if and only if H
contains all the elements xyx−1 y −1 for all x, y ∈ G.
Define the commutator subgroup Gc to be the subgroup generated by all elements xyx−1 y −1 with
x, y ∈ G. Such elements are called commutators.
Proof.
Let x, y ∈ G and assume that G/H is abelian so that xy = yx. Then
(xH)(yH) = (yH)(xH)
⇐⇒
(xH)(yH)((xH)(yH))−1 = (eH) = H
⇐⇒
(xH)(yH)(xH)−1 (yH)−1 = H
⇐⇒
(xH)(yH)(x−1 H)(y −1 H) = H
⇐⇒
(xyx−1 y −1 )H = H
⇐⇒
xyx−1 y −1 ∈ H,
proving both implications.
Exercise 5.
Prove that SL2 (R) is a normal subgroup of GL2 (R) and that GL2 (R)/SL2 (R) ∼
= R× .
Proof. Let G = GL2 (R) and H = SL2 (R) and define the map f : G −→ R× to be the determinant
map f (A) = det(A). We have already noted that f is a homomorphism since
det(AB) = det(A)det(B).
19
The identity in R× is 1 so that
Kerf := {A ∈ G | f (A) = 1} = H,
by the definition of H. We have already shown that the kernel of a homomorphism is always normal.
Applying the First Isomorphism Theorem for Groups, we have
G/H ∼
= f (G).
To complete the proof, we only need to show that f is surjective. Pick x ∈ R× and note that
µµ
¶¶
x 0
f
= x,
0 1
proving that f (G) = R× .
??
An interesting consequence of this result is the observation that for A ∈ GL2 (R), the left coset
ASL2 (R) = {B ∈ GL2 (R) | det(B) = det(A)}.
Although this consequence can be shown directly, it follows immediately from Exercise 5.
Chapter II, §6, # 1-3, 6, 7.
1.
Determine the sign (equivalently, even or odd) of the following permutations.
µ
¶
1 2 3
(a)
= (1 2 3) = (1 3)(1 2) =⇒ even.
2 3 1
µ
(b)
µ
(c)
µ
(d)
µ
(e)
µ
(f )
2.
¶
1 2
3 1
3
2
1
3
2
2
3
1
1 2
2 3
3
1
4
4
1
2
2
1
3
4
4
3
1
3
2
2
3
4
4
1
= (1 3 2) = (1 2)(1 3)
=⇒
even.
¶
= (1 3)
=⇒
odd.
¶
= (1 2 3) = (1 3)(1 2)
=⇒
even.
¶
= (1 2)(3 4)
=⇒
even.
¶
= (1 3 4) = (1 4)(1 3)
=⇒
even.
In each one of the cases of Exercise 1, write the inverse of the permutation.
(a)
(1 3 2).
(b)
(1 2 3).
(c)
(1 3).
20
(d)
(1 3 2).
(e)
(1 2)(3 4).
(f )
(1 4 3).
3.
Show that the number of odd permutations of {1, 2, . . . , n} for n ≥ 2 is equal to the number of
even permutations.
Proof. For any odd permutation ρ, note that (1 2)ρ is even and (1 2)ρ 6= (1 2)σ whenever ρ 6= σ. So,
there are at least as many even permutations as there are odd permutations. On the other hand, if ρ
even, then (1 2)ρ is odd and (1 2)ρ 6= (1 2)σ whenever ρ 6= σ shows that there are at least as many odd
permutations as there are even permutations. One outcome of this observation is that |An | = (n!)/2 for
n ≥ 2.
6.
Two cycles (i1 , . . . , ir ) and (j1 , . . . , js ) are said to be disjoint if no integer iν is equal to and
integer jµ . Prove that a permutation is equal to a product of disjoint cycles.
Proof. Let σ be a permutation of Jn = {1, 2, . . . , n}. In order to write σ as a product of disjoint
cycles, we begin by picking any a1 ∈ Jn . Define
a2 = σ(a1 ),
a3 = σ(a2 ) = σ 2 (a1 ),
....
The infinite sequence a1 , σ(a1 ), σ 2 (a1 ), . . . must have some elements being repeated (since Jn is finite).
Suppose that σ i (a1 ) = σ j (a1 ) for some j > i. Then a1 = σ m (a1 ) where m = j − i. Thus,
σ = (a1 a2 . . . am ) . . . .
Of course, σ may not consist of only one cycle. Next, we pick any element b1 from Jn that did not
appear in the cycle (a1 a2 . . . am ). Proceeding as before, we obtain a cycle (b1 b2 . . . bk ). Note that
this this new cycle has no elements in common with (a1 a2 . . . am ) because if σ i (a1 ) = σ j (b1 ), then
σ i−j (a1 ) = b1 , contradicting our choice of b1 . This process may be continued until we exhaust all of the
elements in Jn , yielding
σ = (a1 a2 . . . am )(b1 b2 . . . bk ) · · · (c1 c2 . . . cs ),
a product of disjoint cycles.
Chapter III, §1, # 3-5, 7, 10-13, Exercise 6.
3.
Let R be an integral domain. If a, b, c ∈ R, a 6= 0, and ab = ac, then prove that b = c.
Proof. Given that ab = ac, we can write ab − ac = 0. Since R is a ring, we have a(b − c) = 0. Now,
a 6= 0 and R integral implies that b − c = 0. Thus, b = c.
21
4.
Let R be an integral domain, and a ∈ R, a 6= 0. Show that the map x 7→ ax is an injective
mapping of R into itself.
Proof. Assume that ax = ay for some x, y ∈ R. Then by the previous problem x = y, proving that
x 7→ ax is injective.
5.
Let R be a finite integral domain. Show that R is a field.
Proof. Let a be any nonzero element of R. We must show that a is a unit. If a = 1, then a−1 = 1 ad
we are done. So, assume a 6= 1 and consider the sequence a, a2 , a3 , . . . . Since R is finite, there exists
i < j such that ai = aj , implying aj−i = 1. Since a 6= 1, we have that j − i > 1 and aj−i−1 is the
inverse of a.
7.
Let r be a ring and x ∈ R. We define x to be nilpotent if there exists a positive integer n such
that xn = 0. If x is nilpotent, prove that 1 + x is a unit, and so is 1 − x.
Proof.
Since x is nilpotent, we have xn = 0 for some positive integer n. Thus, 1 − xn = 1 gives
1 = 1 − xn = (1 − x)(1 + x + x2 + · · · + xn−1 ).
Also, 1 + xn = 1 implies
1 = 1 + xn = (1 + x)(1 − x + x2 − · · · + (−1)n−1 xn−1 ).
Hence, 1 − x and 1 + x are units.
√
10. Let R be the set of numbers of type a + b 2 where a, b ∈ Q. Show that R is a ring, and in fact
that R is a field.
Proof.
and
√
√
First, we observe that R is a nonempty subset of R. If a + b 2, c + d 2 ∈ R, then
√
√
√
(a + b 2) + (c + d 2) = (a + c) + (b + d) 2 ∈ R
√
√
√
(a + b 2)(c + d 2) = (ac + 2bd) + (ad + bc) 2 ∈ R
show that R is closed under both addition and multiplication. Also,
√
√
√
−(a + b 2) = −a + (−b) 2 and 1 = 1 + 0 2
show that additive inverses and 1 are contained in R. Thus, R is a subring of R which is clearly
√
commutative since R is commutative. To see that R is field, let a + b 2 be a nonzero element in R.
Then
µ
¶
√
√
a
b
2 = 1,
− 2
(a + b 2)
a2 − 2b2
a − 2b2
√
√
proving that a + b 2 is a unit. Hence, R is s field. This field is usually denoted by Q( 2).
22
√
11. Let R be the set of numbers of type a + b 2 where a, b are integers. Show that R is a ring, but
not a field.
Proof.
The arguments used in the previous proof to prove that the given set is a subring hold here as
√
well. However, it remains to be seen that R is not a field. Consider the element 1 + 2 2. If there exists
√
a + b 2 ∈ R such that
√
√
(1 + 2 2)(a + b 2) = 1,
then a and b must satisfy the system of equations
a + 4b = 1
2a + b = 0.
√
The solutions are a = − 17 and b = 27 , which are not integers. Thus, 1 + 2 2 is not a unit and R is not
√
a field. This ring is usually denoted by Z[ 2].
12.
Let R be the set of numbers of type a + bi where a, b are integers and i =
√
−1. Show that R is
a ring. List all of its units.
Proof.
Let a + bi, c + di ∈ R, which is a nonempty subset of the ring C. Then
(a + bi) + (c + di) = (a + c) + (b + d)i ∈ R
and
(a + bi)(c + di) = (ac − bd) + (ad + bc)i ∈ R
show that R is closed under addition and multiplication. Also,
−(a + bi) = −a + (−b)i
and
1 = 1 + 0i
prove that R is a subring of C. The ring R described here is called the Gaussian Integers and is
usually denoted by Z[i]. In order to find the units in Z[i], consider the norm map N : Z[i] −→ Z given
by
N (a + bi) = a2 + b2 .
This map preserves the operation of multiplication (you must check this!) and hence, maps units to
units. The only units in Z are ±1. However, N (Z[i]) is the set of nonnegative integers, implying that
the units in Z[i] are exactly the elements that N maps to 1. Solving
a2 + b2 = 1
gives the four solutions
a = 1, b = 0,
a = −1, b = 0,
a = 0, b = 1,
corresponding to Z[i]× = {±1, ±i}.
23
a = 0, b = −1,
13.
Let R be the set of numbers of type a + bi where a, b ∈ Q. Show that R is a field.
Proof.
We may follow the same steps as those in the previous proof to show that R is a (commutative)
subring of C. To set that R forms a field, let a + bi be a nonzero element in R and note that
µ
¶
b
a
(a + bi)
−
i
= 1.
a2 + b2
a2 + b2
Hence, every nonzero element in R is a unit. The resulting field is usually denoted by Q(i).
Exercise 6.
Prove that for any prime p, Z/pZ is a field.
Proof.
On the first exam, we proved that an element a+nZ in Z/nZ is a unit if and only if (a, n) = 1.
In the case of a prime p, every nonzero left coset has a representative that is relatively prime to p. Thus,
Z/pZ is a field.
Chapter III, §2, # 1, 4, 8.
1.
Show that a field has no ideal other than the zero ideal and the field itself.
Proof. Let F be a field and I a nonzero ideal of F . We will show that 1F ∈ I as we proved in class
that any ideal containing 1 must be the entire ring that it is contained in. Since I is not the zero ideal,
it contains some nonzero element x. Because x is contained in F , it is a unit and there exists x−1 ∈ F .
Ideals “absorb” products from F , so that we have 1F = x−1 x ∈ I.
4.
Let R be a ring and J1 , J2 ideals. Show that J1 ∩ J2 is an ideal.
Proof.
First, note that 0R ∈ J1 and 0R ∈ J2 so that J1 ∩ J2 is nonempty. Now let x, y ∈ J1 ∩ J2 , then
by definition, x, y ∈ J1 and x, y ∈ J2 . So, −x, x+y ∈ J1 and −x, x+y ∈ J2 , implying −x, x+y ∈ J1 ∩J2 .
Also, if r ∈ R, then rx, xr ∈ J1 and rx, xr ∈ J2 , implying rx, xr ∈ J1 ∩ J2 . Thus, J1 ∩ J2 must be an
ideal of R.
8.
The following example will be of interest in calculus. Let R be the ring of infinitely differentiable
functions defined, say, on the open interval −1 < t < 1. Let Jn be the set of functions f ∈ R such
that Dk f (0) = 0 for all integers k with 0 ≤ k ≤ n. Here D denotes the derivative, so Jn is the set of
functions all of whose derivatives up to order n vanish at 0. Show that Jn is an ideal in R.
Proof. The proof of this problem requires use of several properties you encountered in calculus.
Namely, that the derivative of a sum is the sum of the derivatives and that constants can be pulled
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out in front of derivatives. These properties, along with several inductive arguments can be used to
complete the proof. The details are left to you.
Chapter III, §3, # 1, 3, 6, 17-21, Exercise 7.
1.
Let f : R −→ R0 be a ring-homomorphism. Show that the image of f is a subring of R0 .
Proof. Note that any ring-homomorphism f : R −→ R0 is a group-homomorphism from the additive
group of R into the additive group of R0 and we have already proved that f (R) is an additive subgroup
of the additive group of R0 . Let x0 , y 0 ∈ f (R) such that f (x) = x0 and f (y) = y 0 . Then xy ∈ R and
x0 y 0 = f (x)f (y) = f (xy)
shows that x0 y 0 ∈ f (R). Finally, we proved in class that f (1R ) = 1R0 , implying that 1R0 ∈ f (R).
3. (a) Let n be a positive integer, and let Zn = Z/nZ be the factor ring of Z modulo n. Show that
the units of Zn are precisely those residue classes x having a representative integer x 6= 0 and relatively
prime to n.
Proof.
For a proof of this claim, see your first exam.
(b) Let x be an integer relatively prime to n. Let ϕ be the Euler function. Show that xϕ(n) ≡ 1
(mod n).
Proof.
Consider the set
S = {a ∈ Z | 0 < a < n and (a, n) = 1}.
Then by definition, S has cardinality ϕ(n). Such a set is called a reduced residue system modulo
n and we may write it in the form S = {a1 , a2 , . . . , aϕ(n) }. The elements in S are exactly the representatives that yield units in Z/nZ. If x ∈ S, then the set {xa1 , xa2 , . . . , xaϕ(n) } is equal to the set S (you
should check this). Thus, multiplying together all of the elements in each of the two sets results in the
same product modulo n:
a1 a2 · · · aϕ(n) ≡ xa1 xa2 · · · xaϕ(n)
ϕ(n)
≡x
a1 a2 · · · aϕ(n)
(mod n)
(mod n)
Now (ai , n) = 1 for all i so that the number a1 a2 · · · aϕ(n) is also relatively prime to n. Canceling this
factor from both sides of the above congruence completes the proof.
6.
Let F be a finite field having q elements. Prove that xq−1 = 1 for every nonzero element x ∈ F .
Show that xq = x for every element x ∈ F .
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Proof. If F is a finite field with q elements, then the group of units of F , denoted F × , has order
q − 1. If x ∈ F × (ie., x is a nonzero element of F ), then a corollary of Lagrange’s Theorem implies
that xq−1 = 1F . This proves the first part of the problem. To prove the second part note that if x ∈ F × ,
then we can multiply both sides of the equation xq−1 = 1F by x to obtain xq = x. If x 6∈ F × is in F ,
then x = 0F and we have 0q = 0, completing the proof.
??
Problems 17-19 were proved in class, so their solutions will not be given here. ??
20.
Let K be a field of characteristic p. Show that (x + y)p = xp + y p for all x, y ∈ K.
Proof. This problem is a direct application of Chapter 1, §5, Problem 9. The details of the proof are
left to you.
21.
Proof.
Let K be a finite field of characteristic p. Show that the map x 7→ xp is an automorphism of K.
Let f : K −→ K denote the map x 7→ xp . Then
f (xy) = (xy)p = xp y p = f (x)f (y),
and by the previous problem,
f (x + y) = (x + y)p = xp + y p = f (x) + f (y).
Since K is finite, it is sufficient to show that f is injective. Suppose that f (x) = f (y), which is equivalent
to xp = y p . From problem 6 of this section, we have x = y and f is one-to-one.
Exercise 7.
Let R be a ring and M an ideal. Prove the following properties.
(a)
Congruence modulo M is an equivalence relation.
(b)
If x ≡ y (mod M ) and z ∈ R, then
xz ≡ yz
(c)
(mod M )
and
zx ≡ zy
(mod M ).
If x ≡ y (mod M ) and x0 ≡ y 0 (mod M ), then
xx0 ≡ yy 0
(mod M )
x + x0 ≡ y + y 0
and
(mod M ).
We proved similar properties for congruence modulo n in Z. The proofs for this general case follow the
same approach and are up to you to complete.
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