Hungerford: Algebra
II.4. The Action of a Group on a Set
1. Let G be a group and A a normal subgroup of G.
(a) Show that G/A operates on A by conjugation.
(b) Let CG (A) denote the centralizer of A in G. Show that G/CG (A) is isomorphic to a subgroup
of Aut(A).
(c) Suppose that A is abelian. Obtain a homomorphism from G/A 7→ Aut(A).
Proof: (a) Let Ω = A and G acts on Ω by conjugation: ∀g ∈ G and ∀x ∈ Ω, g(x) = gxg −1 . Since
Ω = A G, g(x) ∈ Ω. As eG (x) = x and g2 g1 (x) = g2 (g1 xg1−1 ) = g2 (g1 xg1−1 )g −1 = (g2 g1 )(x), this
defines A as a G-set.
(b) Moreover, each g : A 7→ A defined by g(x) = gxg −1 is a bijection and a homomorphism (see
(c) below). Therefore, we have a map φ : G 7→ Aut(A) by φ(g)(x) = g(x), for any x ∈ Ω = A. As
A is a G-set, this map φ is a group homomorphism. Note that Ker(φ) = CG (A), and so by the first
isomorphism theorem, G/CG (A) is isomorphic to a subgroup of Aut(A).
∼ (G/CG (A))/(CG (A)/A).
(c) If A is abelian, then A ⊆ CG (A). By the 3rd isomorphism theorem, G/A =
This suggests the following map G/A 7→ Aut(A): ∀g ∈ G/A, and ∀x ∈ A, f (g)(x) = gxg −1 . To see
that this map is well defined, we pick g, g 0 ∈ g ∈ G/A. Then, there is an a ∈ A such that g 0 = ga.
As A is abelian, axa−1 = x. it follows that
f (g 0 )(x) = (g 0 )x(g 0 )−1 = (ga)x(a−1 g −1 ) = gxg −1 = f (g)(x),
and so this map f is well defined. As ∀x, x0 ∈ A, and ∀g ∈ G/A, f (g)(xx0 ) = g(xx0 )g −1 =
gxg −1 gx0 g −1 = f (g)(x)f (g)(x0 ), f (g) is a homomorphism from A into A. To see that f (g is an
epimorphism, we pick x ∈ A. Then x0 = g −1 (x) ∈ A. Thus x = f (e)(x) = f (g)(x0 ), and so f (g) is
onto. The same argument also indicates that f (g) is one to one. Hence f (g) is in aut(A).
2. If H, K are subgroups of G such that H K, show that K ≤ NG (H).
Proof: Note that NG (H) = {g ∈ G : gH = Hg}. Since H K, ∀k ∈ K, kH = Hk, and so
k ∈ NG (H).
3. If a group G contains an element a having exactly two conjugates, then G has a proper normal
subgroup N 6= {e}.
Proof: Let G acts on Ω = G by conjugation. Then the orbit aG has exactly two elements. Therefore
[G : Ga ] = |aG | = 2, and so N = Ga G. If N = {e}, then |G| = 2. It follows that G ∼
= Z2 , and
so G is abelian. But if G is abelian, every conjugate class has exactly one element, contrary to the
assumption that G has an element a with exactly two conjugates.
4. Let H be a subgroup of G. The centralizer of H in G is
CG (H) = {g ∈ G : ∀h ∈ H, gh = hg}.
Show that CG (H) ≤ NG (H).
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Proof: Recall that NG (H) = {g ∈ G : gH = Hg}. Thus ∀c ∈ CG (H), and ∀h ∈ H, ch = hc, and so
cH = Hc. Thus c ∈ NG (H).
5. Let H be a subgroup of G. The factor group NG (H)/CG (N ) is isomorphic to a subgroup of
Aut(H).
Proof: Let NG (H) act on H by conjugation. As shown in the proof of Exercise 1 in this section,
this action induces a homomorphism from NG (H) 7→ Aut(H) with kernel CG (H). By the first
isomorphism theorem, NG (H)/CG (N ) is isomorphic to a subgroup of Aut(H).
7. Let G be a group, and let In(G) be the set of all inner automorphisms of G. Show that
In(G) Aut(G).
Proof: Let φ ∈ Aut(G). For each ginG, let σg ∈ In(G) be such that ∀x ∈ G, σg (x) = gxg −1 . Then
∀x ∈ G,
φ(σg )φ−1 (x) = φ(gφ−1 (x)g −1 ) = φ(g)x(φ(g))−1 = σφ(g) (x).
Thus φσφ−1 ∈ In(G).
8. Exhibit an automorphism of Z6 that is not an inner automorphism.
Proof: As Z6 is abelian, and as if G is abelian, then In(G) = {e}, it suffices to find a φ ∈
Aut(Z6 ) − {e}. Write Z6 = {0, 1, 2, 3, 4, 5}. Define φ(i) = 5i. Then φ ∈ Aut(Z6 ) − {e}.
9. If G/C(G) is cyclic, then G is abelian.
Proof: It suffices to show that G = C(G). Suppose that G − C(G) 6= ∅. Since G/C(G) is cyclic,
there exists an element g ∈ G − C(G) such that G/C(G) = hgi and |g| > 1. We shall show that
g ∈ C(G) also, contrary to the assumption that g ∈ G − C(G).
Arbitrarily pick g 0 ∈ G − C(G). Then g 0 ∈ G/C(G). As G/C(G) = hgi, there exists an integer k
such that g 0 = g k . Therefore, for some element c ∈ C(G), we have g 0 = cg k . It follows by c ∈ C(G)
that gg 0 = gcg k = cg k g = g 0 g, and so g ∈ C(G).
11. Let G be a group containing an element a not of order 1 or 2. Show that G has a nonidentity
automorphism.
Proof: Consider the automorphism φ(x) = x−1 . We can verify that this is an automorphism
(Exercise I-2.2). As a is not of order 1 or 2, φ(a) 6= a, and so φ is not the identity map.
12. Any finite group is isomorphic to a subgroup of An .
Proof: Let G be a group with |G| = m < ∞. By Cayley’s Theorem, G is isomorphic to a subgroup
of Sm . Therefore, it suffices to show that Sm is isomorphic to a subgroup of An , where n = 2m.
This is the same to construct a monomorphism from Sm into An .
To construct a monomorphism from Sm into An , we assume that Sm is the symmetric group
0
on {1, 2, ..., m}. Let {10 , 20 , ..., m0 } be a set disjoint from {1, 2, ..., m}, and Sm
be the symmetric
0 0
0
0
0
group on the set {1 , 2 , ..., m }, and let σ 7→ σ denote the isomorphism from Sm onto Sm
. Define
0
0 0
Φ : Sm 7→ An as follows: for any σ ∈ Sm , Φ(σ) = σσ (For example, Φ(1, 2) = (1, 2)(1 , 2 )).
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For any τ1 , τ2 ∈ Sm , Φ(τ1 τ2 ) = τ1 τ2 τ10 τ20 . Since {1, 2, ..., m} and {10 , 20 , ..., m0 } are disjoint, τ2 and
τ10 are disjoint and so Φ(τ1 τ2 ) = τ1 τ2 τ10 τ20 = τ1 τ10 τ2 τ20 = Φ(τ1 )Φ(τ2 ). Hence Φ is a homomorphism.
Let e, e0 denote the identity of Sm and An , respectively. It remains to show that Ker(Φ) = {e}.
Suppose that for some σ ∈ Sm , Φ(σ) = e0 . As Φ(σ) = σσ 0 , and as σ and σ 0 are on disjoint sets of
elements, we must have σ = e, which proves that Φ is a monomorphism.
13. If H < G and H ≤ G such that [G : H] < ∞, then G has a normal subgroup N with
[G : N ] < ∞.
Proof: Since [G : NG (H)][(NG (H) : H] = [G : H] < ∞, [G : NG (H)] = m < ∞. Thus there are m
conjugates of H: H1 , · · · , Hm . Use the fact that
if A < G and B < G, then [A : A ∩ B] ≤ [G : B]
to get
m
m
[G : ∩m
i=1 Hi ] ≤ Πi=1 [G : Hi ] = [G : H] < ∞.
−1
G.
Note that N = ∩m
i=1 Hi = ∩x∈G xHx
Remark: One can also obtain N by letting G act on the set of all left cosets of H in G. Then the
kernel of the action if N .
14. If |G| = pn, with p > n, p prime, and H is a subgroup of order p, then H G.
Proof: Let Ω be the set of left cosets of H in G. Let G acts on Ω by left multiplication, and let
N = Ker(G/Ω) be the kernel of the action. Then N G.
Recall that N = {n ∈ G : ∀aH ∈ Ω, (ga)H = aH}. Therefore, ∀n ∈ N , nH = n(eH) = eH = H,
and so N ≤ H. If N = H, then H = N G, and the proof is completed.
Therefore, we assume that N 6= H, and so |N | < |H|. By Lagrange Theorem, |N | is a factor
of |H| = p. Since p is a prime, we must have N = {e}. Therefore, G is isomorphic to a subgroup
of Sn0 , where n0 = |Ω| = [G : H] = (np)/p = n. By Lagrange Theorem again, np = |G| is a factor
of |Sn | = n!. But p > n is a prime, and so (pn)|n! is impossible. This contradiction shows that we
must have H = N .
Supplemental Problems
16. Show that if |G| = p2 with p prime, then G ∼
= Zp2 or G ∼
= Zp × Zp .
∼ Zp2 . Thus we assume that ∀a ∈ G − {e},
Proof: If G has a element a with |a| = p2 , then G =
|a| = p.
Claim 1 G is abelian. Since |G| = p2 , C(G) 6= {e}. If C(G) = G, then G is abelian and the claim
holds. Suppose that |C(G)| = p. Then |G/C(G)| = p, and so G/C(G) is cyclic. By Exercise 9 in
this section, G = C(G), a contradiction. Hence G must be abelian.
Pick x ∈ G − {e}, and let X = hxi. Then X ∼
= Zp , as |x| = p. Since |G/X| = p, we must have
∼
G/X = Zp , and so we may assume that for G/X = hyi. By Exercise II-2.3, there exists y ∈ y such
that |y| = p. Let Y = hyi.
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Claim 2 G = XY . ∀g ∈ G. Then gX = y i X, for some integer i, as G/X = hyi. It follows by Claim
1, that for some integer j, g = y i xj = xj y i ∈ XY . This proves G = XY .
Claim 3 X G and Y G. This follows from the fact that G is abelian.
Claim 4 X ∩ Y = {e}. Suppose that g = xj = y i ∈ X ∩ Y , for some integers 0 ≤ i, j ≤ p − 1. Then
y i = X, and so p|i. Since 0 ≤ i ≤ p − 1, we must have i = 0, and so g = e.
By Claims 1-4, G ∼
=X ×Y ∼
= Zp × Zp .
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