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 $G\times H$, $[K,G]\leq G$. 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 $[K,G]\leq K$. Thus, $G\leq K$.

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