3
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ @ancientmathematician, yes my mistake $\endgroup$
    – codetalker
    Mar 26, 2020 at 16:47

1 Answer 1

6
$\begingroup$

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 $((12),t)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .