Hungerford: Algebra
I.2. Homomorphisms and Subgroups
2. A group G is abelian if and only if the map G 7→ G by x 7→ x−1 is an automorphism.
Sketch of Proof Suppose that G is abelian. Exercise (I.11)(iii) shows that x 7→ x−1 is a
group homomorphism. Since the identity is the only element whose inverse is the identity,
this map is injective. Since (x−1 )−1 = x, this map is surjective, and so an automorphism.
Conversely, if x 7→ x−1 is a group homomorphism, then Exercise (I.11)(iii) must hold,
and so by Exercise (I.11), G must be abelian.
Question: What happens if we replace ”automorphism” by ”homomorphism”?
3. Let Q8 be the
(under ordinary
 group 
 matrix
 multiplication) generated by th complex
0 1
0 i
 and B = 
, where i2 = −1. Show that Q8 is a nonmatrices A = 
−1 0
i 0
abelian group of order 8.

0
Proof: Let E = AB = 
1
−1 0

. Note that Q8 = {I, A, B, E, −I, −A, −B, −E}.
4. Let
H be thegroup (under
 ordinary
 matrix multiplication) generated by th real matrices
C=
0
1
 and D = 
0 1
. Show that H is a nonabelian group of order 8, which
−1 0
1 0
is not isomorphic to Q8 , but isomorphic to D4∗ .
Proof: Direct computation yields

C 2 = −I, C 3 = −C and C 4 ; D2 = I, CD = 
1
0
0 −1

 , C 2 D = −D, C 3 D = −CD = DC.
Therefore, H = {I, C, C 2 , C 3 , D, CD, −D, −CD} is a group. As CD 6= DC, H is not
abelian. Using the notation in Exercise I-1.4, the map

I C C2 C3

I
CD −D −CD
D
R R 2 R 3 Tx
T13
is a group isomorphism between D4∗ and H.
1
Ty
T24

.
Suppose that there if an isomorphism f : Q8 7→ H. Then f (I) = I. Since f is one-toone, one of f (A), f (B) and f (E) (f (A), say) must be in {C 2 , D, CD, −D, −CD}. Note that
for any X ∈ {C 2 , D, CD, −D, −CD}, X 2 = I. Therefore, f (I) = I = (f (A))2 = f (A2 ) =
f (−I), contrary to the assumption that f is a bijection. Thus such an isomorphism does
not exist.
11. If G is a group, then C = {a ∈ G|ax = xa∀x ∈ G} is an abelian subgroup of G.
Proof Since the identity e ∈ C, C 6= ∅. Suppose that a, b ∈ C. For any x ∈ G, since
a, b ∈ C, (ab)x = a(bx) = (ax)b = x(ab), and so ab ∈ C. Moreover, from ax = xa, we have
xa−1 = a−1 x, and so a−1 ∈ C. Therefore, C ≤ G. Since a ∈ C, for any b ∈ C ⊆ G, ab = ba
and so C is abelian.
15. Let G be a group and Aut(G) the set of all isomorphisms of G.
(a) Aut(G) is a group with composition of functions as binary operation.
(b) Aut(Z) ∼
= Z2 × Z2 , Aut(Ap ) ∼
= Zp−1 (p prime).
= Z2 ∼
= Aut(Z6 ), Aut(Z8 ) ∼
(c) What is Aut(Zn ) for arbitrary n ∈ N∗ ?
Sketch of Solution: (a) Since the composition of bijections is also a bijection, and the
composition of homomorphisms is also a homomorphism, composition is a binary operation
in Aut(G), with the identity map 1G serving as the identity of Aut(G). For any f ∈ Aut(G),
since f is a bijection, f −1 : G 7→ G exists as a bijective map. For any a, b ∈ G, since f is a
bijection, ∃a0 , b0 ∈ G such that a = f (a0 ) and b = f (b0 ). Hence f −1 (ab) = f −1 (f (a0 )f (b0 )) =
f −1 (f (a0 b0 )) = a0 b0 = f −1 (a)f −1 (b), and so f −1 ∈ Aut(G).
(b) Note that in a cyclic group H = hai, if f ∈ Aut(H), then hf (a)i = H, and so a
generator must be mapped to another generator. In the cyclic group Z, there are only two
generators 1 and −1. Therefore, Aut(Z) = {1Z , f }, where 1Z (1) = 1 and f (1) = −1. As
f 2 = 1bf Z , Aut(Z) ∼
= Z2 .
Useful Fact Recall that in a cyclic group H = hai, if |a| = n < ∞, then for any m,
|am | =
n
(n,m) .
Therefore, in Zn = h1i, for any element m ∈ Zn , |m| =
n
(n,m) .
If follows that
if m ∈ Zn , then h1i = Zn if and only if (n, m) = 1.
In particular, Z6 has only two generators, 1 and 5. Let fi denote the map in Aut(Z6 ) such
that fi (1) = i, where i ∈ {1, 5}. Then f1 is the identity and f 2 = f1 , and so Aut(Z6 ) ∼
= Z2 .
5
Apply the Useful Fact to Z8 . Then the generators of Z8 are 1, 3, 5, 7. Let fi ∈ Aut(Z8 )
2
be the map such that fi (1) = i, where i ∈ {1, 3, 5, 7}. Then Aut(Z8 = {f1 , f3 , f5 , f7 } with
f1 being the identity. As 32 ≡ 52 ≡ 72 ≡ 1 (mod 8), every element of Aut(Z8 has order 2.
Thus Aut(Z8 ) ∼
= Z2 ⊕ Z2 .
Apply the Useful Fact to Zp for a prime p, we have Aut(Zp ) = {fi : 1 ≤ i ≤ p − 1},
where fi (1) = i for each i ∈ {1, 2, · · · p − 1}. Since p is a prime, (p − 1, p) = 1, and so there
exists an integer a ∈ {2, 3, · · · , p − 1} such that a(p − 1) ≡ 1 (mod p). This means that
Aut(Zp ) ∼
= hfa i ∼
= Zp−1 .
(c) The examples in (b) suggest the following. Let Z∗n = {a ∈ Zn − {0} : (a, n) = 1}.
With mod n multiplication, (Z∗n , ·) forms an abelian group. By the Useful Fact, ∀a ∈ Z∗n ,
that fa x = ax defines an automorphism fa ∈ Aut(Zn ). As fa (1) = a, wdirect computation
of composition function yields fa fb = fab , and so φ(fa ) = a defines a homomorphism from
Aut(Zn onto Z∗n . Moreover, if φ(fa ) = φ(fb ), then a = fa (1) = fb (1) = b, and so φ must be
a bijection. Hence Aut(Zn ) ∼
= Z∗ .
n
To see the structure of Z∗n , we can proceed these steps:
(Step 1) Show that if (m, n) = 1, then Aut(Zmn ) ∼
= Aut(Zm ⊕ Aut(Zn ).
(Step 2) Show that if (m, n) = 1, then Zmn ∼
= Zm ⊕ Zn .
m−1
∼
(Step 3) Show that Aut(Z2m ) = ⊕
Z2 .
i=1
(Step 4) Show that if p > 2 is a prime, then Zpm ∼
= Zp−1 ⊕ ⊕m−1
i=1 Zp .
Combine these steps to conclude: if k ≥ 0 is an integer, and if, (when k > 0), p1 , p2 , ..., pk are
mk
1
distinct odd primes , and m0 , m1 , ..., mk are nonnegative integers such that n = 2m0 pm
1 · · · pk ,
then
Aut(Zn ) ∼
=
 mj −1
0 −1


⊕m
Z2 ⊕kj=1 Zpj ⊕i=1
Zpj

i=1

if k > 0 and mi > 0 (0 ≤ i ≤ k)




if k = 0 and m0 > 0
⊕kj=1 Zpj
0 −1
⊕m
i=1 Z2
mj −1
⊕i=1
Zpj
if k > 0, mi > 0 (1 ≤ i ≤ k) and m0 = 0
16. For each prime p, the subgroup Z(p∞ ) is benerated by the set {1/pn |n ∈ N∗ }.
Proof Since {1/pn |n ∈ N∗ } ⊆ Z(p∞ ), it suffices to show that Z(p∞ ) ⊆ h{1/pn |n ∈ N∗ }i,
which amounts to show any element in Z(p∞ ) can be obtained by adding elements in
{1/pn |n ∈ N∗ }. Since any element in Z(p∞ ) can be written as n/pi = n(1/pi ),
Z(p∞ ) ⊆ h{1/pn |n ∈ N∗ }i,
as desired.
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18. (a) Let G be a group and {Hi |i ∈ I} a (non empty) family of subgroups. State and
prove a condition that will imply that
S
i∈I
Hi is a subgroup of G.
(b) Give an example of a group G and a family of subgroups {Hi |i ∈ I} such that
h
S
i∈I
S
i∈I
Hi 6=
Hi i.
Proof (a) Consider the statement: for a family of subgroups {Hi |i ∈ I} of a group G,
S
S
i∈I
Hi = h
i∈I
Hi (that is,
S
i∈I
Hi i if and only if the binary operation of G is also a binary operation of
S
i∈I
Hi is closed under multiplication).
To prove this statement, it suffices to prove the sufficiency. Let H =
S
i∈I
Hi . Since
e ∈ H, N 6= ∅. For any a, b ∈ H, since H is closed under multiplication, ab ∈ H. As a ∈ H,
there is an Hi such that a ∈ Hi . But Hi ≤ G, and so a−1 ∈ Hi ⊆ H. Therefore, H ≤ G.
(b) Take G = Z, and H1 = h2i and H2 = h3i. Then 2 ∈ H1 and 3 ∈ H2 , but
5 = 2 + 3 ∈ hH1 ∪ H2 i − H1 ∪ H2 , and so H1 ∪ H2 6≤ G.
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