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A function f : AmapsB 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 HnormalG and K < G.

(1) HK < G.

(2) HnormalHK; HintersectionK normal K.

(3) HK/Hisomorphic with K/H intersection K.

Let HnormalG, KnormalG, K < H. Then (G/K)/(H/K)isomorphic toG/H. The set residue classes mod n. The set of integers. If f:GmapsH is a homomorphism. The kernel of f is ker(f) = {x element of G | f(x) = 1H}. A function f : AmapsB is said to be surjective (or onto) if for every y element of B there exists x element of 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 : GmapsH is a homomorphism, then G/ker(f)isomorphic to im(f) Let G1 and G2 be groups. An isomorphism from G1 to G2 is an injection phi:G1mapsG2 such that phi(a · b) = phi(a) · phi(b)  (a,b element ofG1).

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 HnormalG (or GnormalH).

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

Let H be a subgroup of G such that

G > H > K.

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

(hK)(h'K)-1 = hh'-1K element of 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 element of G | gK element of H}.

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

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

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

Proposition 2: Let HnormalG, KnormalG, K < H. Then H/KnormalG/K.

Theorem 3 (Second Isomorphism Theorem): Let HnormalG, KnormalG, K < H. Then

(G/K)/(H/K)isomorphic toG/H.

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


Since i and j are surjective, the homomorphism

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

is surjective. Moreover,

ker(j) = H/K.

Note that if g element of G, then

g element of ker(ij) is equivalent to (ji)(g) = 1M
is equivalent to j(i(g)) = 1M
is equivalent toi(g) element of ker(j) = H/K     [by (3)]
is equivalent to g element of 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 HnormalG, KnormalG. 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)isomorphic toZ2,

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 element of H, k element of K}. Let us first determine when HK is a group.

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

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

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

(hk)(h'k')-1 = h(kk'-1h'-1)
= hh''k'' element of 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 HnormalG. If h element of H, k element of K, then k-1hk element of H, say k-1hk = h'. Thus, hk = kh' element of KH. Thus, HK subset of KH. A similar argument shows that KH subset of HK, so that HK = KH and the result follows from Proposition 4.

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

normal subgroups

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

(1) HK < G.

(2) HnormalHK; HintersectionK normal K.

(3) HK/Hisomorphic with K/H intersection K.

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


defined by

i(k) = k      (k element of 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


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

(ji)(k) = hkH.

Indeed, since HnormalG, we see that k-1hk element of 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 intersection K = 2Z intersection 3Z = 6Z. The third isomorphism theorem then asserts that

Z/2Zisomorphic to3Z/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 HnormalG, then Theorem 6 already provides us with the answer. For HK/H has |HK|/|H| elements, and K/(H intersection K) has |K|/|H intersection K| elements. Therefore, by Theorem 6, we have

|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 element of H, k element of 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

|H\HK| = .


|H intersection K\K| = .

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

|H\HK| = |H intersection K\K|

Let us define a function

psi: H\HKmapsHintersectionK\K


psi(Hhk) = (HintersectionK)k,     h element of H, k element of K.

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

(HintersectionK)k = (HintersectionK)hk' = (HintersectionK)k',
implies psi(Hhk) = psi(Hh'k'),
implies psi is well defined.

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