Math 371 Lecture #36 §8.1 (Ed.2), 9.1 (Ed.3): Direct Products

We take another step in the classification of groups by way of products of normal sub- groups.

Example. The abelian group U15 = {1, 2, 4, 7, 8, 11, 13, 14} has two (normal) subgroups M = {1, 11} and N = {1, 2, 4, 8}. The “product” MN consists of all products mn where m ∈ M and n ∈ N. We have M ⊆ MN because m = m1 for all m ∈ M, and N ⊆ MN because n = 1n for all n ∈ N. Do 7, 13, 14 ∈ MN? Yes, because 11 · 2 = 7, 11 · 8 = 13, and 11 · 4 = 14.

Thus U15 = MN, the “product” of two normal subgroups of U15.

Notice that every element of U15 can be written uniquely in the form mn for m ∈ M and n ∈ N.

Are the normal subgroups M and N of U15 isomorphic to quotient groups of U15? Since M ∩ N = {1} and both M and N are normal, we have by the Second Isomorphism Theorem that ∼ ∼ ∼ ∼ M = M/(M ∩ N) = NM/N = U15/N, N = N/(N ∩ M) = MN/M = U15/M. ∼ This says that U15/N × U15/M = M × N. ∼ Could it happen that U15 = M × N, the Cartesian product of two normal subgroups? The table for M × N is

· (1, 1) (1, 2) (1, 4) (1, 8) (11, 1) (11, 2) (11, 4) (11, 8) (1, 1) (1, 1) (1, 2) (1, 4) (1, 8) (11, 1) (11, 2) (11, 4) (11, 8) (1, 2) (1, 2) (1, 4) (1, 8) (1, 1) (11, 2) (11, 4) (11, 8) (11, 1) (1, 4) (1, 4) (1, 8) (1, 1) (1, 2) (11, 4) (11, 8) (11, 1) (11, 2) (1, 8) (1, 8) (1, 1) (1, 2) (1, 4) (11, 8) (11, 1) (11, 2) (11, 4) (11, 1) (11, 1) (11, 2) (11, 4) (11, 8) (1, 1) (1, 2) (1, 4) (1, 8) (11, 2) (11, 2) (11, 4) (11, 8) (11, 1) (1, 2) (1, 4) (1, 8) (1, 1) (11, 4) (11, 4) (11, 8) (11, 1) (11, 2) (1, 4) (1, 8) (1, 1) (1, 2) (11, 8) (11, 8) (11, 1) (11, 2) (11, 4) (1, 8) (1, 1) (1, 2) (1, 4)

Could the function f : M × N → MN given by f(m, n) = mn be an isomorphism? We will answer this question in a moment in a more general setting which we now describe.

The Cartesian product of groups G1,G2,...,Gn, denoted by G1 × G2 × · · · × Gn, consists of n-tuples (a1, a2, . . . , an) for ai ∈ Gi, equipped with the component operation

(a1, a2, . . . , an)(b1, b2, . . . , bn) = (a1b1, a2b2, . . . , anbn). Technically, each Gi is NOT a subgroup of the direct product G1 × G2 × · · · × Gn (but each Gi is isomorphic to the subgroup of G1 × G2 × · · · × Gn).

Observe that if all of G1,G2,...,Gn are finite groups, then the order of G1 ×G2 ×· · ·×Gn is the the product of the orders of G1,G2,...,Gn.

Should G1,G2,...,Gn happen to be normal subgroups of a group G, we can consider the normal subgroup G1G2 ··· Gn that consists of products of the form a1a2 ··· an in G for ai ∈ Gi, i = 1, 2, . . . , n. Lemma 9.2. Let M and N be normal subgroups of a group G. If M ∩ N = {e}, then ab = ba for all a ∈ M and all n ∈ N. Proof. Let a ∈ M and b ∈ N. The goal is to show that ab = ba. Since M is normal, we have that b−1ab ∈ M. Since a−1 ∈ M we have that a−1b−1ab ∈ M. Similarly, by the normality of N we have that a−1b−1a ∈ N, so that a−1b−1ab ∈ N. −1 −1 Since M ∩ N = {e}, we obtain a b ab = e, and hence ab = ba. 

Theorem 9.1. Let N1,N2,...,Nk be normal subgroups of a group G such that every element in G can be written uniquely in the form a1a2 ··· ak for ai ∈ Ni, i = 1, 2, . . . , k. Then G is isomorphic to the direct product N1 × N2 × · · · × Nk.

Proof. We define the map f : N1 × N2 × · · · × Nk → G by

f(a1, a2, . . . , ak) = a1a2 ··· ak.

The map f is surjective because by hypothesis, each element of G can be written in the form a1a2 ··· ak for some ai ∈ Ni, i = 1, 2, . . . , k.

The map f is injective because when f(a1, a2, . . . , ak) = f(b1, b2, . . . , bk), we obtain that a1a2 ··· ak = b1b2 ··· bk, so that by the uniqueness of writing each element of G as a product, we have a1 = b1, a2 = b2,..., ak = bk.

Before we show that f is a homomorphism, we show for i 6= j that Ni ∩ Nj = {e}.

To do this we take a ∈ Ni ∩ Nj, which a can be written as product of elements of N1,N2,...,Nk in two different ways:

a = e ··· e a e ··· e e e ··· e ↑ ↑ ↑ ↑ N1 Ni Nj Nk

a = e ··· e e e ··· e a e ··· e ↑ ↑ ↑ ↑ N1 Ni Nj Nk

Uniqueness of writing a as a product of the elements of Ni implies that a = e, so that Ni ∩ Nj = {e}.

By Lemma 9.2, we then have that aibj = bjai for all ai ∈ Ni and bj ∈ Nj when i 6= j. Repeated using this commutativity we show that f is a homomorphism:

f((a1, a2, . . . , ak)(b1, b2, . . . , bk)) = f(a1b1, a2b2, . . . , akbk)

= a1b1a2b2 ··· akbk

= a1a2b1b2 ··· akbk

= a1a2 ··· akb1b2 ··· bk

= f(a1, a2, . . . , ak)f(b1, b2, . . . , bk).

Therefore f : N1 × N2 × · · · × Nk → G is an isomorphism. 

Definitions. Suppose that normal subgroups N1,N2,...,Nk of a group G have the property that every element of G can be written uniquely in the form a1a2 ··· ak for ai ∈ Ni, i = 1, 2, . . . , k.

Then G = N1N2 ··· Nk and we say that G is an internal direct product of N1,N2,...,Nk.

By Theorem 9.1, the internal direct product N1N2 ··· Nk is isomorphic to N1 ×N2 ×· · ·× Nk, and we say that G is the external direct product of N1,N2,...,Nk.

Each normal subgroup Ni in the internal or external direct product is called a direct factor of G. Theorem 9.3. If M and N are normal subgroups of a group G such that G = MN and M ∩ N = {e}, then G ∼= M × N where M ∼= G/N and N ∼= G/M. Proof. Suppose every element of G is of the form mn for m ∈ M and n ∈ N.

We want to show that this form mn is unique, so suppose mn = m1n1 for some m1 ∈ M and n1 ∈ N. −1 −1 −1 −1 −1 −1 Then m1 mn = m1 m1n1 = n1, and so n1n = m1 mnn = m1 m. −1 −1 Since M and N are subgroups, we have m1m ∈ M and n1n ∈ N, −1 −1 −1 −1 Since m1 m = n1n and M ∩ N = {e}, then m1 m = e and n1n = e, and hence m1 = m and n1 = n. Thus every element of G can be written uniquely as mn for m ∈ M and n ∈ N. Since M and N are normal subgroups of G, Theorem 9.1 implies that G ∼= M × N. Because M∩N = {e}, the Second Isomorphism Theorem implies that M ∼= M/(M∩N) ∼= ∼ ∼ MN/N = G/N and N = N/(M ∩ N) = MN/M = G/M.  In practice we typically identity the exterior direct product and the interior direct product because they are isomorphic.

Example (Continued). The group U15 = {1, 2, 4, 7, 8, 11, 13, 14} has two normal subgroups M = {0, 11} and N = {1, 2, 4, 8}. Because G = MN and M ∩ N = {e}, we have by Theorem 9.3 that G ∼= M × N. ∼ ∼ ∼ Since M = Z2 and N = Z4, we have that U15 = Z2 × Z4.

The group Z2 × Z4 is one of five isomorphism classes of groups of order 8.