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Equivalence of internal and external direct product

This article gives a proof/explanation of the equivalence of multiple definitions for the term direct product
View a complete list of pages giving proofs of equivalence of definitions


The definitions we have to prove as equivalent

Internal direct product

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

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 G that is the internal direct product of normal subgroups N_1, N_2, we can identify G naturally with the external direct product N_1 \times N_2.

Proof: Consider the map:

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

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

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

Combining these, we get:

aba^{-1}b^{-1} \in N_1 \cap N_2 = \{ e \}

Hence, we get:

 ab = ba.

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

\alpha(a,b)\alpha(c,d) = abcd = a(bc)d = a(cb)d = (ac)(bd) = \alpha(ac,bd) = \alpha((a,b)(c,d))

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

Next we note that:

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

Thus, \alpha defines an isomorphism from N_1 \times N_2 to G.

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

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

  1. The subgroup isomorphic to A is A \times \{ e \} (i.e., elements of the form (a,e)).
  2. The subgroup isomorphic to B is \{ e \} \times B (i.e., elements of the form (e,b)).

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

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