# When the direct and semi-direct products are isomorphic.

I am not asking what direct/semi-direct products are.

Suppose $$H$$ and $$K$$ are any two groups, and let $$\varphi:K\to\text{Aut}(H)$$ be a homomorphism, and consider the semi-direct product $$H\rtimes K$$ with respect to $$\varphi$$. Let $$K$$ also denote the isomorphic copy of $$K$$ in $$H\rtimes K$$.

(By the isomorphic copy I mean the most natural copy, i.e $$K = \{(1,k)|k\in K\}$$)

The following theorem is true and not hard to prove: the identity map from $$H\rtimes K$$ to $$H\times K$$ is a homomorphism (and hence an isomorphism) if and only if $$K\unlhd H\rtimes K$$.

My question is:

In the above fact, a very specific kind of isomorphism is being used, i.e the natural identity isomorphism. If we are just given that $$H\rtimes K\cong H\times K$$ (where the isomorphism need not be the identity map), is it still true that $$K\unlhd H\rtimes K$$?

I tried proving it but couldn't make any progress. Is this fact true and if yes can you give me a hint?

• @ancientmathematician, yes my mistake Mar 26 '20 at 16:47

Let $$H=S_3$$ and $$K=\langle t \rangle\cong S_2$$.
Let $$\phi:t\mapsto i_{(12)}$$, the inner automorphism induced by $$(12)$$.
Form the semi-direct product $$G=H\rtimes_{\phi} K$$.
Then $$G\cong S_3\times S_2$$, but the "natural" $$K$$ in $$G$$ is not normal; the normal subgroup of order $$2$$ is generated by $$((12),t)$$.