View source for Equivalence of internal and external direct product

{{definition equivalence|direct product}}

==The definitions we have to prove as equivalent==

===Internal direct product===

Suppose <math>G</math> is a group, and <math>N_1, N_2</math> are normal subgroups. We say that <math>G</math> is an internal direct product of <math>N_1</math> and <math>N_2</math> if <math>N_1 \cap N_2</math> is trivial, and <math>N_1N_2 = G</math>.

===External direct product===

==What we need to prove==

===Every internal direct product is naturally isomorphic to the external direct product===

We need to prove that given a group <math>G</math> that is the internal direct product of normal subgroups <math>N_1, N_2</math>, we can identify <math>G</math> naturally with the external direct product <math>N_1 \times N_2</math>.

''Proof'': Consider the map:

<math>\alpha: N_1 \times N_2 \to G, \qquad \alpha(a,b) = ab</math>

We first must verify that <math>\alpha</math> is a group homomorphism. For this, we first observe that if <math>a \in N_1, b \in N_2</math>, then:

<math>aba^{-1}b^{-1} = a(ba^{-1}b^{-1}) \in N_1; \qquad aba^{-1}b^{-1} = (aba^{-1})b^{-1} \in N_2</math>

Combining these, we get:

<math>aba^{-1}b^{-1} \in N_1 \cap N_2 = \{ e \}</math>

Hence, we get:

<math> ab = ba</math>.

Thus, any element of <math>N_1</math> commutes with any element of <math>N_2</math>. Thus, if we pick two pairs <math>(a,b), (c,d) \in N_1 \times N_2</math>, then:

<math>\alpha(a,b)\alpha(c,d) = abcd = a(bc)d = a(cb)d = (ac)(bd) = \alpha(ac,bd) = \alpha((a,b)(c,d))</math>

In the intermediate step, we use that <math>b</math> and <math>c</math> commute. Thus, <math>\alpha</math> is a group homomorphism. It's clear that <math>\alpha</math> preserves the identity element and inverses, too.

Next we note that:

# <math>\alpha</math> is injective because <math>\alpha(a,b) = e \implies ab = e \implies b = a^{-1} \implies a,b \in N_1 \cap N_2 = \{ e \}</math>.
# <math>\alpha</math> is surjective because by assumption, <math>N_1N_2 = G</math>.

Thus, <math>\alpha</math> defines an isomorphism from <math>N_1 \times N_2</math> to <math>G</math>.

===Every external direct product is naturally realized as an internal direct product===

We need to show that if <math>A</math> and <math>B</math> are abstract groups, then we can find subgroups in <math>A \times B</math> isomorphic to <math>A</math> and <math>B</math> respectively, whose internal direct product is <math>A \times B</math>. Indeed:

# The subgroup isomorphic to <math>A</math> is <math>A \times \{ e \}</math> (i.e., elements of the form <math>(a,e)</math>).
# The subgroup isomorphic to <math>B</math> is <math>\{ e \} \times B</math> (i.e., elements of the form <math>(e,b)</math>).

Let's now check the conditions for an internal direct product:

# Both the subgroups are normal: This is a direct check using the coordinate-wise multiplication.
# The subgroups intersect only at <math>(e,e)</math>: This is set-theoretically obvious.
# The product of the subgroups is <math>A \times B</math>: Any element <math>(a,b)</math> can be written as the product of <math>(a,e)</math> and <math>(e,b)</math>.