Normal subdirect product of perfect groups equals direct product

Statement
Suppose $$G$$ and $$H$$ are fact about::perfect groups. Then, if $$K$$ is a fact about::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, $$K = G \times H$$.