Zachary Scherr
1
Math 370 HW 4
Solutions
Book Problems
1. 2.3.1
Solution:
(a) This is not a group. Sure the integers are closed under subtraction, but subtraction is not even
associative. For example,
1 − (2 − 3) = 2
while
(1 − 2) − 3 = −4.
There is also no identity element for subtraction.
(b) This is not a group. The set of positive integers is closed under × and we do have associativity
and a multiplicative identity. There are not, however, multiplicative inverses.
(c) This is a group. Notice that we can summarize the two rules for multiplication in one by saying
ai · aj = ai+j
(mod 7) .
• G is closed under the operation given. This is because when 0 ≤ i, j ≤ 6 we have 0 ≤ i + j
(mod 7) ≤ 6.
• The operation is associative follows from the fact that modular arithmetic addition is
associative. We have
ai · (aj · ak )
=
ai · (aj+k
=
a(i+j)+k
(mod 7) )
(mod 7)
= ai+(j+k)
= ai+j
(mod 7)
(mod 7)
· ak = (ai · aj ) · ak .
• a0 is an identity element for this operation since ai · a0 = a0 · ai = ai for all i.
• The inverse of a0 is a0 and for 1 ≤ i ≤ 6, the inverse of ai is a7−i since
ai · a7−i = a7−i · ai = a0 .
Note that this group is secretly the cyclic group e, a, a2 , a3 , a4 , a5 , a6 discussed in class.
(d) This is a group assuming that 0 ∈ G, which is a reasonable assumption since we should consider
integers as n = n1 . Let’s check
• G is closed under the operation. If
a
b
and
c
d
are fractions with b, d odd then
a
c
ad + bc
+ =
b
d
bd
and bd is odd.
• Addition of rational number is associative so addition in G is associative as well. Let’s
just check
c
e
a cf + ed
a(df ) + b(cf + ed)
a
+
+
=
+
=
b
d f
b
df
b(df )
a
(ad + bc)f + bde
ad + bc
e
c e
=
=
+ =
+
+ .
(bd)f
bd
f
b
d
f
• G has an identity element, namely 0 = 01 .
• G has inverses since if
a
b
∈ G then b is odd so
−a
b
∈ G is the inverse.
Zachary Scherr
Math 370 HW 4
Solutions
2. Herstein 2.3.2
Solution: We have to be a little bit careful here. We are asked to prove this statement for all
integers, not just positive integers. We can first prove the statement for all non-negative integers
using induction and then use that result to extend to negative integers.
So let’s prove that (ab)n = an bn for n ≥ 0. We prove this by induction on n. The base case n = 0 is
trivial. Suppose for some k ≥ 0 we know that (ab)k = ak bk for every a, b ∈ G. Then for any a, b ∈ G
we have
(ab)k+1
=
(ab)k (ab) = (ak bk )(ab)
=
(ak (bk a))b = (ak (abk ))b
=
(ak+1 bk )b = ak+1 bk+1 .
Now suppose that n ≥ 1. For any a, b ∈ G we have
(ab)−n = (b−1 a−1 )n = (a−1 b−1 )n = a−n b−n
by our above result.
3. Herstein 2.3.3
Solution: Suppose that (ab)2 = a2 b2 for all a, b ∈ G. Then for any a, b ∈ G we get
a−1 (ab)2 b−1 = a−1 a2 b2 b−1
which simplifies as
ba = ab.
4. Herstein 2.3.8
Solution: Let G be a finite group. First we show that for each a ∈ G there exists a non-negative
integer na so that ana = e. To do so, first observe that since G is finite, there exist positive integers
i < j so that
ai = aj .
Multiplying both sides by a−i gives
e = aj−i
so we may as well take na = j − i. Now let
N=
Y
na .
a∈G
Then
since G is finite we have that N is a well-defined positive integer. For each a ∈ G we have
na N so there is a positive integer ka such that
na ka = N.
Thus for any a ∈ G we get
aN = ana ka = (ana )ka = eka = e.
Page 2
Zachary Scherr
Math 370 HW 4
Solutions
5. Herstein 2.3.9
Solution:
(a) Let G = {e, a, b} where e is the identity element of G. Notice that we cannot have ab = a
since that would imply b = e. Similarly we cannot have ab = b so we must have ab = e. Thus
b = a−1 and since every group element commutes with its inverse we see that a and b commute
and hence G is abelian.
(b) Let G be a group of 4 elements and suppose a and b are distinct non-identity elements of G.
If ab = e then the argument above shows that a and b commute. Thus we may assume that
ab 6= e in which case the four elements
e, a, b, ab ∈ G
are distinct. But then G, having only four elements, must equal
G = {e, a, b, ab}.
Since ba ∈ G we must have ba equal to one of the four elements listed. But ba 6= e since ab 6= e
and ba 6= a and ba 6= b since a and b are non-identity. Thus we must have ba = ab and so we
see that a and b commute.
(c) Let G have order 5. Assume for the sake of contradiction that there exist elements a, b ∈ G
which don’t commute. Then neither a nor b is the identity element, and a quick check as above
shows that the five elements
e, a, b, ab, ba ∈ G
are distinct. Thus since G has five elements we get
G = {e, a, b, ab, ba}.
Now what is a2 ? We can’t have a2 = ab since a 6= b and similarly we cannot have a2 = ba. We
also cannot have a2 = b since we are assuming that a and b do not commute, yet a commutes
with a2 . Obviously a2 6= a as well so by process of elimination we must get a2 = e. But then
a·e =
a
a·a
e
=
a·b =
a · ab =
ab
a2 b = b
so by process of elimination we get a · ba = ba. This is absurd since a 6= e and we have our
contradiction.
6. Herstein 2.3.10
Solution: Suppose that every element of G is its own inverse. Then for any a, b ∈ G we have
(ab)−1 = ab. Of course we also know that (ab)−1 = b−1 a−1 and so putting these facts together gives
ab = (ab)−1 = b−1 a−1 = ba.
7. Herstein 2.3.11
Page 3
Zachary Scherr
Math 370 HW 4
Solutions
Solution: Let G be a group of even order and let S = G − {e}. Then S is a subset of G of odd
order. For every a ∈ S we have that a 6= e so a−1 6= e which means a−1 ∈ S. Thus for every a ∈ S
we can attempt to pair off a with a−1 in S. Since inverses are unique, the fact that S has odd order
means that there is some element s ∈ S which pairs off with itself. This element satisfies s = s−1 ,
which is equivalent to s2 = e.
8. Herstein 2.3.18
Solution: This is a great (and hard) question and I want to try to get you to think about it in
several ways. The group S3 has 6 elements and is non-abelian so it is worthwhile to understand it
and possibly generalize the construction. Using my notation from class, we saw that there are two
elements σ, τ ∈ S3 with the properties:
• σ2 = τ 3 = e
• τ σ = στ −1 = στ 2
The consequence of these rules is that all elements of S3 look like
{e, τ, τ 2 , σ, στ, στ 2 }.
We can try to generalize this idea as follows. Our first pass will only use symbols, but then we will
try to recognize the resulting group in a more natural fashion.
Consider symbols s and t, and let Dn be the set
{e, t, t2 , . . . , tn−1 , s, st, st2 , . . . , stn−1 }.
As a set, Dn has 2n elements. We can turn Dn into a non-abelian group by specifying a particular
binary operation. We will define an associative multiplication by specifying that multiplication
satisfies the following rules:
• s2 = tn = e, and n and 2 are the smallest positive integers with this property.
• ts = st−1 = stn−1
• e is the multiplicative identity element.
Our task now is to check that we’ve actually defined a group. This means that we must check
that Dn is closed under our multiplication rule and that Dn has inverses (existence of the identity
element and associativity of multiplication are built into the definition).
To prove closure, we first prove by induction that for any integer k in the range 0 ≤ k < n we have
tk s = stn−k . The base case k = 0 is clear because
s = t0 s = stn
since tn = t0 = e. Thus assume we know that tk s = stn−k for some 0 ≤ k < n − 1. Then
tk+1 s = t · (tk s) = t · (stn−k ) = (ts) · tn−k = (st−1 ) · tn−k = stn−k−1 = stn−(k+1) .
We can use this result to check that Dn is closed under multiplication. Every element of Dn is of
the form si tj where 0 ≤ i < 2 and 0 ≤ j < n. Conisder the product
0
0
(si tj )(si tj ).
Page 4
Zachary Scherr
Math 370 HW 4
Solutions
0
If i0 = 0 then this product is just si tj+j . If j + j 0 < n then we immediately get an element of Dn ,
0
0
while if j + j 0 ≥ n then si tj+j = si tj+j −n is in Dn . If i0 = 1 then our above result says
0
0
0
0
(si tj )(stj ) = si (tj s)tj = si (stn−j )tj = si+1 tn−j+j .
and if 0 ≤ a < 2 and 0 ≤ b < n are chosen so that a ≡ i + 1 (mod 2) and b ≡ n − j 0 + j (mod n)
then this element is actually
sa tb ∈ Dn .
Thus we see that Dn is closed under multiplication. To check for inverses, note that ti · tn−i = e
while
sti · sti = s(ti · s) · ti = s(s · tn−i ) · ti = s2 tn = e.
The fact that Dn is non-abelian is built into the definition. We know that ts = stn−1 . It cannot be
the case that ts = st for otherwise
stn−1 = ts = st
which implies (by cancellation) tn−2 = e which is absurd as n > 2.
Solution: The above solution works perfectly well to construct a non-abelian group of size 2n,
but it doesn’t really give us intuition about where this group comes from. We will illustrate the
geometric intuition in the example D8 . Consider a regular octagon sitting in the plane. What are
all the ways to pick the octagon up, move it in space, and return it to its original position? If you
think about this for a while, you might recognize that there are 16 possible orientations of the moved
octagon. Illustrating this with a stop sign we have:
The first stop sign on the second line corresponds to flipping along the horizontal axis. Let me call
this flipping move s. The second stop sign in the first line corresponds to rotating the stop sign by
45◦ counterclockwise. Call this rotation t. Then it’s clear that if we perform t eight times or if we
perform s two times then we haven’t moved the stop sign at all. In group language this says
s2 = t8 = e.
What happens if we perform ts? That is, we first flip horizontally then we rotate counter clockwise.
This gives us the second stop sign on the second line. Check that this gives the same stop sign as
first rotating clockwise by 45◦ and then flipping horizontally! In group language this says
ts = st−1 .
This group of order 16 is exactly the group we constructed above! Moreover, this construction works
for any regular n-gon. If we take any flip s, and a rotation t which is counter-clockwise by 2π/n,
then we will see that s2 = tn = e and ts = st−1 . Thus the group generated by these “symmetries”
will be Dn . We’ve recognized our group as “symmetries” of a geometric shape.
Page 5
Zachary Scherr
Math 370 HW 4
Solutions
Solution: Here is one other idea motivated by the second solution. Again let’s think about
the octagon. The octagon has 8 vertices so we can label them, starting from the top left, as
x1 , x2 , x3 , . . . , x8 . What do our “symmetries” from the second solution do to the vertices? Check
that s has the net effect
x1
7→
x6
x2
7→
x5
x3
7→
x4
x4
7→
x3
x5
7→
x2
x6
7→ x1
x7
7→ x8
x1
7→
x8
x2
7→
x1
x3
7→
x2
x4
7→
x3
x5
7→
x4
x6
7→
x5
x7
7→
x6
x8
7→
x7 .
whereas t has the net effect
In this form, one can see that s and t are really one-to-one correspondences of a set with 8 elements
to itself. Again in group language this says that really
s, t ∈ S8 .
What is the subgroup generated by s and t in S8 ? One can compose these maps in all possible ways
and check that there are 16 distinct functions generated by s and t. You can check this by hand,
but we don’t need to since we already know that s and t should generate a group of size 16. We’ve
seen it happen several times already! For fun, you may want to try to come up with formulas for
elements s and t in Sn which generate Dn for various choices of n.
2
For Fun
1. 2.3.4
Solution: This problem is easier than it looks at first glance. We are told that for some integer i
we have
(ab)i
= ai bi
(ab)i+1
= ai+1 bi+1
(ab)i+2
= ai+2 bi+2
Page 6
Zachary Scherr
Math 370 HW 4
Solutions
We can combine the first two equations to get:
ai+1 bi+1
=
(ab)i+1
=
(ab)i (ab)
=
ai bi ab
and so cancelling from both sides gives abi = bi a. Similarly, we can combine the second two equations
to get
ai+2 bi+2
=
(ab)i+2
=
(ab)i+1 (ab)
= ai+1 bi+1 ab
or equivalently abi+1 = bi+1 a. Thus
bi+1 a
= abi+1
= abi · b
= bi a · b
and we can cancel a bi from both sides to get
ba = ab.
Since a and b were arbitrary we see that G is abelian.
2. 2.3.5
Solution: It is true that for any a, b ∈ S3 we have
(ab)6 = a6 b6
and
(ab)7 = a7 b7 .
Since S3 is non-abelian, we have our example. The easiest way to prove this is to use a theorem
I will present on Wednesday which says that if G is a finite group, then for any a ∈ G we have
a#G = e. Regardless, one can just check the for any g ∈ S3 we have g 6 = e. Thus it follows easily
that
e = (ab)6 = a6 b6
and
ab = (ab)7 = a7 b7 .
3. Define a group G as follows. As a set, G will consist of all strings made out of the letters s and t,
including the empty string, and define a product on G via string concatenation. That is, for example,
ssttts · tstt = sstttststt and if e denotes the empty string then sstts · e = sstts. Let us impose on G the
relations that ss = ttt = stststst = e the empty string. Prove that G is non-abelian and that #G = 24.
Show that this group G is really the symmetric group S4 is disguise.
Page 7
Zachary Scherr
Math 370 HW 4
Solutions
Solution: This is a very difficult question given how little we know about groups. Instead of
answering this in full, I will show why it is reasonable to expect the group defined to actually be S4 .
The group G is generated by two symbols s and t with the relations s2 = t3 = (st)4 = e. Consider
the two maps, f and g, in S4 . The map f is defined by
x1
7→
x4
x2
7→
x2
x3
7→
x3
x4
7→
x1
x1
7→ x2
x2
7→ x3
x3
7→ x1
x4
7→ x4 .
and the map g is defined by
then it is easy to see that f 2 = g 3 = e. Moreover, the map f ◦ g sends
x1
7→
x2
x2
7→
x3
x3
7→
x4
x4
7→
x1
and so it is apparent that (f g)4 = e. One can check, and at this point in the course it’s painful to
do so, that f and g generate all of S4 . We don’t yet know how to write down meaninful functions
between groups, but once we do, you will realize that our description of S4 in this way will lead to
a one-to-one correspondence between S4 and G. The idea, which is easy to describe yet challenging
to make rigorous, is to try to define a function from S4 to G which maps f to s and g to t. You
may want to think about this, and come back to it when we have learned about homomorphisms.
If anyone has a clever solution to this problem which doesn’t use much brute force computation, I’d
love to hear about it.
Page 8