Normal subdirect product of perfect groups equals direct product

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Statement

Suppose G and H are Perfect group (?)s. Then, if K is a Subdirect product (?) of G and H such that K is normal inside G \times H, then K = G \times H.

Proof

Given: Perfect groups G,H, with projections p_1:G \times H \to G, p_2:G \times H \to H. A normal subgroup K of G \times H such that p_1(K) = G, p_2(K) = H.

To prove: K = G \times H.

Proof: View G and H as subgroups in G \times H by the embeddings as G \times \{ e \} and \{ e \} \times H respectively.

Since G is normal in G \times H, [K,G] \le G. In particular, p_1([K,G]) = [K,G].

Further, p_1([K,G]) = [p_1(K),p_1(G)] = [G,G] = G. Thus [K,G] = G.

But K was assumed to be normal in G \times H, so [K,G] \le K. Thus, G \le K.

A similar argument shows that H \le K, so K contains both G and H. Hence, K = G \times H.