Equivalence Relations
Theorem 2.119 (Second Isomorphism Theorem) If H and K are subgroups
of a group G with H G, then HK is a subgroup, H∩ K K , and K /(H∩ K ) ∼= HK/H.
Proof. We begin by showing first that HK/H makes sense and then describing its elements. Since H G, Proposition 2.117 shows that HK is a subgroup. Normality of H in HK follows from a more general fact: if H ≤ S ≤ G and if H is normal in G, then H is normal in S (if ghg−1 ∈ H for every g ∈ G, then, in particular, ghg−1 ∈ H for every g ∈ S).
We now show that each coset x H ∈ HK/H has the form k H for some k∈ K . Of course, x H = hk H, where h ∈ H and k ∈ K . But hk = k(k−1hk)= kh0for some h0∈ H, so that hk H = kh0H = k H.
It follows that the function f : K → HK/H, given by f : k 7→ k H, is surjective. Moreover, f is a homomorphism, for it is the restriction of the natural map π: G → G/H. Since ker π = H, it follows that ker f = H ∩ K , and so H ∩ K is a normal subgroup of K . The first isomorphism theorem now gives K /(H∩ K ) ∼= HK/H. •
The second isomorphism theorem gives the product formula in the special case when one of the subgroups is normal: if K /(H ∩ K ) ∼= H K /H, then |K /(H ∩ K )| = |H K /H|, and so |H K ||H ∩ K | = |H||K |.
Theorem 2.120 (Third Isomorphism Theorem). If H and K are normal subgroups of a group G with K ≤ H, then H/K G/K and
(G/K )/(H/K ) ∼= G/H.
Proof. Define f : G/K → G/H by f : aK 7→ a H. Note that f is a (well- defined) function, for if a0 ∈ G and a0K = aK , then a−1a0 ∈ K ≤ H, and so a H = a0H . It is easy to see that f is a surjective homomorphism.
Now ker f = H/K , for aK = H if and only if a ∈ H, and so H/K is a normal subgroup of G/K . Since f is surjective, the first isomorphism theorem gives (G/K )/(H/K ) ∼= G/H. •
The third isomorphism theorem is easy to remember: the K ’s in the fraction (G/K )/(H/K ) can be canceled. One can better appreciate the first isomorphism theorem after having proved the third one. The quotient group (G/K )/(H/K ) consists of cosets (of H/K ) whose representatives are themselves cosets (of G/K ). A direct proof of the third isomorphism theorem could be nasty.
The next result, which describes the subgroups of a quotient group G/K , can be regarded as a fourth isomorphism theorem. Recall that a function f: X → Y sets up a correspondence, using direct and inverse images, between subsets of X and subsets of Y . We now adapt this viewpoint to the special case when
f : G → H is a homomorphism.
If G is a group and K G, let Sub(G; K ) denote the family of all those subgroups S of G containing K , and let Sub(G/K ) denote the family of all the subgroups of G/K .
Proposition 2.121 (Correspondence Theorem). If G is a group and K G, then S7→ S/K is a bijection Sub(G; K ) → Sub(G/K ). Denoting S/K by S∗, we have
(i) T ≤ S ≤ G in Sub(G; K ) if and only if T∗≤ S∗in Sub(G/K ), in which case[S : T ] = [S∗: T∗];
(ii) T S in Sub(G; K ) if and only if T∗ S∗in Sub(G/K ), in which case S/ T ∼= S∗/T∗.
Proof. Let 8: Sub(G; K ) → Sub(G/K ) denote the function 8 : S 7→ S/K (it is routine to check that if S is subgroup of G containing K , then S/K is a subgroup of G/K ).
To see that 8 is injective, we begin by showing that if K ≤ S ≤ G, then π−1π(S) = S, where π : G → G/K is the natural map. As always, S ⊆ π−1π(S), by Proposition 2.14(iii). For the reverse inclusion, let a ∈ π−1π(S), so that π(a)= π(s) for some s ∈ S. It follows that as−1 ∈ ker π = K , so that a= sk for some k ∈ K . But K ≤ S, and so a = sk ∈ S, as desired.
Assume now that π(S) = π(S0), where S and S0 are subgroups of G con- taining K (note that π(S)= S/K ). Then π−1π(S)= π−1π(S0), as we have just proved in the preceding paragraph, and so S= S0; hence, 8 is injective.
To see that 8 is surjective, let U be a subgroup of G/K . By Example 2.90(iv) π−1(U ) is a subgroup of G containing K = π−1({1}), and π(π−1(U ))= U , by Proposition 2.14(ii).
Proposition 2.14(i) shows that T ≤ S ≤ G implies T /K = π(T ) ≤ π(S) = S/K . Conversely, assume that T /K ≤ S/K . If t ∈ T , then t K ∈ T /K ≤ S/K and so t K = s K for some s ∈ S. Hence, t = sk for some k ∈ K ≤ S, and so t ∈ S.
In the important special case when G is finite, we prove[S : T ] = [S∗: T∗] as follows. [S∗: T∗] = |S∗|/|T∗| = |S/K |/|T /K | = (|S|/|K |) / (|T |/|K |) = |S|/|T | = [S : T ].
To prove that[S : T ] = [S∗ : T∗] in the general case, it suffices to show that there is a bijection from the family of all cosets of the form sT , where s ∈ S, and the family of all cosets of the form s∗T∗, where s∗∈ S∗, and the reader may check that sT 7→ π(s)T∗is such a bijection.
The third isomorphism theorem shows that if T S, then T /K S/K and (S/K )/(T /K ) ∼= S/T ; that is, S∗/T∗∼= S/T . It remains to show that T S if T∗ S∗; that is, if t∈ T and s ∈ S, then sts−1 ∈ T . Now
π(st s−1)= π(s)π(t)π(s)−1∈ π(s)T∗π(s)−1 = T∗, so that st s−1∈ π−1(T∗)= T . •
When dealing with quotient groups, one usually says, without mentioning the correspondence theorem explicitly, that every subgroup of G/K has the form S/K for a unique subgroup S ≤ G containing K . For example, if G is a finite p-group, that is, if|G| = pn for some prime p, then Theorem 2.146 says that if G 6= {1}, then the center is nontrivial; that is, Z(G) 6= {1}. Hence, if G is not abelian, then Z (G/Z (G)) 6= {1}. An important role in the investigation of finite p-groups is played by Z2(G) which is, by definition, the inverse image of Z (G/Z (G)). Note that Z2(G) G, by the correspondence theorem,
Proposition 2.122. If G is a finite abelian group, then G has a subgroup of order d for every divisor d of|G|. In particular, if p is a prime divisor of |G|, then G contains an element of order p. (Compare Proposition 2.97.)
Proof. We begin by proving, by induction on n = |G|, that for every prime divisor p of|G|, there is an element of order p in G. The base step n = 1 is true, for there are no prime divisors of 1. For the inductive step, choose a ∈ G of order k > 1. If p | k, say k = p`, then Exercise 2.34 on page 143 says that a` has order p. If p k, consider the cyclic subgroup H = hai. Now H G, because G is abelian, and so the quotient group G/H exists. Note that |G/H| = n/k is divisible by p, and so the inductive hypothesis gives an element b H ∈ G/H of order p. If b has order m, then (bH)m = bmH = H in G/H, and so Lemma 2.53 gives p| m. We have returned to the first case.
We prove the general result by induction on d ≥ 1. The base step d = 1 is obviously true, and so we may assume that d > 1; that is, we may assume that d has a prime divisor, say, p. By induction, G contains a subgroup H of order p. Since G is abelian, H G, and so the quotient group G/H is defined. Moreover,|G/H| = |G|/ p, so that (d/ p) | |G/H|. The inductive hypothesis gives a subgroup S∗ ≤ G/H with |S∗| = d/ p. By the correspondence theorem, there is a subgroup S (where H ≤ S ≤ G) with S∗ = S/H. Therefore, |S| =
p|S∗| = p · (d/ p) = d. •
A theorem of Cauchy, Theorem 2.145, says that if p is a prime divisor of |G|, where G is any finite, not necessarily abelian, group, then G has an element of order p.
Here is another construction of a new group from two given groups.
Definition. If H and K are groups, then their direct product, denoted by H×K ,
is the set of all ordered pairs (h, k) with h ∈ H and k ∈ K equipped with the operation
(h, k)(h0,k0)= (hh0,kk0).
It is routine to check that H × K is a group [the identity is (1, 1) and (h, k)−1 = (h−1,k−1)]. Note that H × K is abelian if and only if both H and K are abelian.
Example 2.123.
The four-group V is isomorphic to
2×
2. The reader may check that the function f : V → 2× 2, defined by f: (1) 7→ ([0], [0]), f: (1 2)(3 4) 7→ ([1], [0]), f: (1 3)(2 4) 7→ ([0], [1]), f: (1 4)(2 3) 7→ ([1], [1]), is an isomorphism.
We now apply the first isomorphism theorem to direct products.
Proposition 2.124. Let G and G0be groups, and let K G and K0 G0be normal subgroups. Then K × K0is a normal subgroup of G× G0, and there is an isomorphism
Proof. Let π: G → G/K and π0: G0 → G0/K0 be the natural maps. The reader may check that f: G × G0→ (G/K ) × (G0/K0), given by
f: (g, g0)7→ (π(g), π0(g0))= (gK , g0K0)
is a surjective homomorphism with ker f = K × K0. The first isomorphism theorem now gives the desired isomorphism. •
Here is a characterization of direct products.
Proposition 2.125. If G is a group containing normal subgroups H and K with
H ∩ K = {1} and H K = G, then G ∼= H × K .
Proof. We show first that if g ∈ G, then the factorization g = hk, where h∈ H and k ∈ K , is unique. If hk = h0k0, then h0−1h= k0k−1 ∈ H ∩ K = {1}. Therefore, h0 = h and k0 = k. We may now define a function ϕ : G → H × K by ϕ(g) = (h, k), where g = hk, h ∈ H, and k ∈ K . To see whether ϕ is a homomorphism, let g0 = h0k0, so that gg0 = hkh0k0 = hh0kk0. Hence, ϕ(gg0)= ϕ(hkh0k0), which is not in the proper form for evaluation. If we knew that if h∈ H and k ∈ K , then hk = kh, then we could continue:
ϕ(hkh0k0)= ϕ(hh0kk0) = (hh0,kk0) = (h, k)(h0,k0) = ϕ(g)ϕ(g0).
Let h ∈ H and k ∈ K . Since K is a normal subgroup, (hkh−1)k−1 ∈ K ; since H is a normal subgroup, h(kh−1k−1) ∈ H. But H ∩ K = {1}, so that hkh−1k−1 = 1 and hk = kh. Finally, we show that the homomorphism ϕ is an isomorphism. If (h, k)∈ H × K , then the element g ∈ G defined by g = hk satisfies ϕ(g)= (h, k); hence ϕ is surjective. If ϕ(g) = (1, 1), then g = 1, so that ker ϕ= 1 and ϕ is injective. Therefore, ϕ is an isomorphism. •
All the hypotheses in Proposition 2.125 are needed. For example, let G = S3, H = h(1 2 3)i, and K = h(1 2)i. Now S3= H K , {1} = H ∩ K , and H S3, but K is not a normal subgroup. It is not true that S3 ∼= H × K , for S3 is not abelian, while the direct product H× K of abelian groups is abelian.