A function f : AB is said to be injective (or one to one) if whenever f(x) = f(y), we have x = y. A function that is both one-to-one and onto, that is if f(x) = f(y) then x = y, and for every y of the domain there is an x of the range so that f(x) = y. Assume that HG and K < G.

(1) HK < G.

(2) HHK; HK K.

(3) HK/H K/H K.

Let HG, KG, K < H. Then (G/K)/(H/K)G/H. The set residue classes mod n. The set of integers. If f:GH is a homomorphism. The kernel of f is ker(f) = {x G | f(x) = 1H}. A function f : AB is said to be surjective (or onto) if for every y B there exists x A such that f(x) = y. If H is a subgroup of G then aH = {a · h | a in G, h in H} is a left coset Ha = {h · a | a in G, h in H} is a right coset If f : GH is a homomorphism, then G/ker(f) im(f) Let G1 and G2 be groups. An isomorphism from G1 to G2 is an injection :G1G2 such that (a · b) = (a) · (b)  (a,b G1).

### The Isomorphism Theorems of Group Theory

Previously we proved the first isomorphism theorem of group theory. Let us now complement our previous result with two additional isomorphism theorems which actually follow from the first. Throughout the following sections on group theory the following notation will be in effect: G is a group; if H is a subgroup, then we will write H < G (or G > H); if H is a normal subgroup of G, then we will write HG (or GH).

Let KG. The elements of the factor group G/K are the cosets gK (g G). Let us begin by describing the subgroups of G/K.

Let H be a subgroup of G such that

(1)
G > H > K.

Since KG, we see that KH, so that the factor group H/K makes sense. The elements of H/K are cosets hK (h H), so that H/K G/K. If hK,h'K H/K, then

(hK)(h'K)-1 = hh'-1K H/K,

so that H/K is a subgroup of G/K. Thus, we have described a procedure for constructing subgroups of G/K.

Proposition 1: Let H < G/K. Then there exists a subgroup H of G, satisfying (1), such that H = H/K.

Proof: Set

H = {g G | gK H}.

If h,h' H, then hK, h'K H, so that

hh'-1K = (hK)(h'K)-1 H,

since H is a subgroup of G/K. Therefore, hh'-1 H and H is a subgroup of G. Moreover, if k K, then kK = K H, so that K < H, and (1) is satisfied. It is clear from definition of H that H/K = H.

Proposition 2: Let HG, KG, K < H. Then H/KG/K.

Theorem 3 (Second Isomorphism Theorem): Let HG, KG, K < H. Then

(2)
(G/K)/(H/K)G/H.

Proof: By Proposition 2, H/KG/K, so that the left side of (2) makes sense. Let i and j denote the natural homomorphisms defined by

GG/K
G/K(G/K)/(H/K).

Since i and j are surjective, the homomorphism

ji:G(G/K)/(H/K) = M

is surjective. Moreover,

(3)
ker(j) = H/K.

Note that if g G, then

g ker(ij) (ji)(g) = 1M
j(i(g)) = 1M
i(g) ker(j) = H/K     [by (3)]
g i-1(H/K) = H.

Therefore, ker(ij) = H, so that (2) follows from the first isomorphism theorem.

Example 1: Let G = Z, H = 2Z, K = 6Z. Then G is abelian, so that HG, KG. Thus the hypotheses of Theorem 3 are satisfied. Now G/H = Z/2Z = Z2, G/K = Z/6Z = Z6, Therefore, by Theorem 3,

Z6/(2Z/6Z)Z2,

which can be easily verified directly.

Throughout the remainder of this section, let H and K be subgroups of G. Recall that HK = {hk | h H, k K}. Let us first determine when HK is a group.

Proposition 4: HK is a group if and only if HK = KH.

Proof: Assume that HK is a group and let h H, k K. Then h-1k-1 HK, and since HK is a group kh = (h-1k-1)-1 HK, Therefore KH HK. By a similar argument, HK KH, so that HK = KH.

Assume that HK = KH, and let hk,h'k' HK. Then, since HK = KH, there exist h'' H, k'' K such that (kk'-1) · h'-1 = h''k''. But then

(hk)(h'k')-1 = h(kk'-1h'-1)
= hh''k'' HK.

Therefore, HK is a subgroup of G.

Corollary 5: If one of H, K is a normal subgroup of G, then HK is a subgroup of G.

Proof: Assume that HG. If h H, k K, then k-1hk H, say k-1hk = h'. Thus, hk = kh' KH. Thus, HK KH. A similar argument shows that KH HK, so that HK = KH and the result follows from Proposition 4.

Let us now suppose that HG and K < G. By Corollary 5, HK < G. Moreover, since 1 K, we see that H < HK. In fact, HHK, since HG. Similarly, K < HK. Further we see that H K is a subgroup of K and of H. Actually, if k K, h H K, then khk-1 K, since h K and K is a group; moreover, since k G and HG, we have khk-1 H K. Thus we have shown that H KK. We may summarize the situation in the following diagram, where single lines denote subgroups and double lines denote normal subgroups:

Theorem 6 (Third Isomorphism Theorem): Assume that HG and K < G.

(1) HK < G.

(2) HHK; HK K.

(3) HK/H K/H K.

Proof: We have already proved (1) and (2). In order to prove (3), let us consider the homomorphisms

i:KHK,
j:HKHK/H,

defined by

i(k) = k      (k K),
j = the natural homomorphism.

Note that i is an injection, so that ker(i) = {1}, while ker(j) = H. Let us now consider the homomorphism

ji:KHK/H.

Then it is clear that k ker(ji) if and only if k ker(j) if and only if k H. Therefore, ker(ji) = H k. Therefore, by the first isomorphism theorem, it suffices to prove the ji is surjective. Let hk HK. We claim that

(ji)(k) = hkH.

Indeed, since HG, we see that k-1hk H. Therefore,

hkH = k(k-1hk)H = kH.

But (ji)(k) = kH. Thus our claim is proved and hkH is contained in the image of ji. Therefore, ji is surjective and our result is proved.

Example 2: Let G = Z with respect to addition and let H = 2Z, K = 3Z. Then HK = 2Z + 3Z = Z, H K = 2Z 3Z = 6Z. The third isomorphism theorem then asserts that

Z/2Z3Z/6Z,

a result which can be trivially checked.

Let H and K be subgroups of a finite group G. As for our last result in this section, let us compute the number of elements in the set HK. If HG, then Theorem 6 already provides us with the answer. For HK/H has |HK|/|H| elements, and K/(H K) has |K|/|H K| elements. Therefore, by Theorem 6, we have

(4)
|HK| =

If H is not a normal subgroup of G, then HK is not necessarily a group. However, we will prove that (4) still holds in this case.

Even is HK is not a group, it still makes sense to consider the set of right cosets H\HK. A typical coset belonging to H\HK is Hhk, where h H, k K. Each coset contains |H| elements. Moreover, we have already seen that two cosets intersect if and only if they are equal. Therefore, the elements of HK are distributed into nonoverlapping cosets, each having |H| elements so that |HK| is a multiple of |H| and the number of cosets on H\HK is given by

(5)
|H\HK| = .

Similarly,

(6)
|H K\K| = .

Thus, to prove (4), it suffices to prove that

(7)
|H\HK| = |H K\K|

Let us define a function

: H\HKHK\K

by

(Hhk) = (HK)k,     h H, k K.

is well defined, but since if Hhk = Hh'k', then Hk = Hk' k = hk' for some h H. But since h = kk'-1,h K and thus h HK.

(HK)k = (HK)hk' = (HK)k',
(Hhk) = (Hh'k'),
is well defined.

If (HK)k = (HK)k', then k = k*k' (k* HK). Therefore, if (Hhk) = (Hh'k'), then Hhk = Hk = Hk*k' = Hk' = Hh'k'. And therefore is injective. It is clear that is surjective. Therefore, is a bijection and (7) is proved.