# Normal subdirect product of perfect groups equals direct product

## 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, $K = G \times H$.