MATH 113-6: INTRODUCTION TO ABSTRACT ALGEBRA
SOLUTION 2
(1) P22, 22.
First suppose that the order of x is finite, then |x| = n for some n ∈ Z+ . This
means xn = 1 and none of x, x2 , · · · , xn−1 is equal to the identity element 1. For
g −1 xg, we have (g −1 xg)2 = g −1 xgg −1 xg = g −1 x2 g. Then by induction, we get
(g −1 xg)k = g −1 xk g of any k ∈ Z+ . So (g −1 xg)n = g −1 xn g = g −1 1g = 1. Moreover,
we have (g −1 xg)k = g −1 xk g 6= 1 for any k = 1, 2, · · · , n − 1. Otherwise xk =
(gg −1 )xk (gg −1 ) = g(g −1 xk g)g −1 = g1g −1 = 1, which contradicts with the fact that
none of x, x2 , · · · , xn−1 is equal to 1. So the order of g −1 xg is n.
Then we suppose that the order of x is infinite, i.e. none of x, x2 , · · · , xn , · · · is
equal to 1. By the same argument as above, none of g −1 xg, (g −1 xg)2 , · · · , (g −1 xg)n , · · ·
is equal to 1. So the order of g −1 xg is ∞.
Since ba = (a−1 a)ba = a−1 (ab)a, the fact that |x| = |g −1 xg| implies |ab| =
−1
|a (ab)a| = |ba|.
(2) P22, 25.
For any two elements a, b ∈ G, we need to prove ab = ba. Since x2 = 1 for any
x ∈ G, we have (ab)2 = a2 = b2 = 1.
Since (ab)2 = 1 and a2 = b2 = 1, we have abab = (ab)2 = 1 = a2 b2 = aabb. By the
cancellation law in groups, we cancel the first a and the last b in the above equality,
then we get ba = ab. So G is an abelian group.
(3) P23, 32.
Suppose that xi = xj for some i, j ∈ {0, 1, · · · , n − 1}. Without loss of generality,
we suppose that i ≥ j, then xi−j = 1. Since i, j ∈ {0, 1, · · · , n − 1} and i ≥ j, we
have 0 ≤ i − j ≤ n − 1. It contradicts with |x| = n. So 1, x, x2 , · · · , xn−1 are distinct
from each other. These are n distinct elements in G, so we have |G| ≥ |x|.
(4) P27, 2.
For any element in x ∈ D2n , if x is not a power of r, we have x = srk for some
k ∈ {0, 1, · · · , n − 1}.
Then we have
rx = r(srk ) = (rs)rk = (sr−1 )rk = srk−1 = (srk )r−1 = xr−1 .
So the proof is done.
(5) P28, 4.
For any element x ∈ D2n , either x = rl or x = srl for some l ∈ {0, 1, · · · , n − 1}.
If x = rl , then rk x = rk rl = rl rk = xrk .
If x = srl , then
rk x = rk (srl ) = (rk s)rl = (sr−k )rl = s(r−k rl ) = s(rk rl ) = (srl )rk = xrk .
1
Here we have that r−k = rk since 1 = rn = r2k = rk rk . So rk commutes with all
elements in D2n .
We need also to show that for any x ∈ D2n with x 6= 1, rk , there exists y ∈ D2n
such that xy 6= yx.
If x = rl , we take y = s. Then xy = rl s = sr−l and yx = srl . Since rl 6= 1 and
rl 6= rk , we have r2l 6= 1 and rl 6= r−l . So xy = sr−1 6= srl = yx.
If x = srl , we take y = r. Then xy = (srl )r and yx = r(srl ) = (rs)rl =
(sr−1 )rl = (srl )r−1 . Since n ≥ 3, r 6= r−1 holds. So xy = (srl )r 6= (srl )r−1 = yx.
(6) P32, 1.
We have that σ = (135)(24) and τ = (15)(23). Then we can do the standard
computations and get
σ 2 = (153),
στ = (2534),
τ σ = (1243)
and
τ 2 σ = ((15)(23))2 σ = σ = (135)(24).
(7) P33, 5.
Since all the four cycles are disjoint from each other, we have
((1 12 8 10 4)(2 13)(5 11 7)(6 9))k = (1 12 8 10 4)k (2 13)k (5 11 7)k (6 9)k .
If this power is the identity, since these four cycles are disjoint from each other,
we need each of (1 12 8 10 4)k , (2 13)k , (5 11 7)k , (6 9)k is the identity. For a cycle
with n elements, its kth power is the identity if and only if n | k. So we need that
5, 2, 3, 2 are all factors of k, i.e. k is a common multiple of 5, 2, 3, 2.
The smallest such k is 30, and the order of this element is 30.
(8) P33, 14.
Let σ ∈ Sn be an element with order p, with p being a prime number. We write
down the cycle decomposition of σ = σ1 σ2 · · · σn , such that σ is a product of disjoint
nontrivial cycles σ1 , · · · , σn .
Then σ k = σ1k σ2k · · · σnk . Since σi s are disjoint from each other, σ p is the identity
if and only if all of σ1p , · · · , σnp are the identity. So for each σi , its cycle length is a
factor of p. Since each σi is not a trivial cycle (length at least 2) and p is a prime
number, the length is exactly p. So σ is a product of disjoint p-cycles, while disjoint
cycles commute with each other.
Suppose that the cycle decomposition of σ is a product of commuting p-cycles,
then it is straight forward to check that the order of σ is p.
If p is not a prime number, |σ| = p does not imply σ is a product of commuting
p-cycles. For example, we can take p = 4 and σ = (12)(3456).
(9) P35, 1.
We
count
all
candirectly
theorder
ofGL
2 (F2 ),
by listing
the elements, which
1 0
1 0
1 1
1 1
0 1
0 1
are
,
,
,
,
and
. For simplicity, here
0 1
1 1
0 1
1 0
1 0
1 1
we do not write bars on 0 and 1.
Here we also give a proof which proves P35 problem 7, and use that 24 − 23 −
2
2 + 2 = 6.
a b
For a matrix M =
∈ GL2 (Fp ), we first count the possibilities of (a, b).
c d
The pair (a, b) lies Fp × Fp , so there might be p2 many choices. As long as (a, b) 6=
(0, 0), we can get a pair (c, d) such that M ∈ GL2 (Fp ). So there are actually p2 − 1
many choices of (a, b).
a b
Now we fix a nonzero pair (a, b). To make sure
lies in GL2 (Fp ), we need
c d
(a, b) and (c, d) are linearly independent. This is equivalent to that (c, d) does not
lie in the line passing through the nonzero vector (a, b) ∈ Fp × Fp . Since there are
p vectors in the line, there are p2 − p many choices of (c, d).
By considering the pairs (a, b) and (c, d), there are (p2 −1)(p2 −p) = p4 −p3 −p2 +p
elements in GL2 (Fp ).
(10) P35, 8.
We only prove the case for n = 2. Suppose that there
are A, B ∈ GL2 (F )
A
0
with AB 6= BA, then for any n > 2, we consider An =
, Bn =
0 I(n−2)×(n−2)
B
0
. Then AB 6= BA implies An Bn 6= Bn An .
0 I(n−2)×(n−2)
1 a
For the n = 2 case, for any nonzero element a ∈ F , consider A =
and
0 1
1
a
1 0
1 + a2 a
B=
. Then AB =
and BA =
. Since a 6= 0 and
a 1
a
1
a 1 + a2
F is a field, a2 6= 0. So 1 + a2 6= 1 and AB 6= BA.