About semidirect product and faithful actions

Question

I need some help with the following question:

We have $H$ acting by automorphisms on $N$, and let $\rho:H\to Aut(N)$ the associated representation by automorphisms.

Suppose that $G=H[N]_{\rho}$ is a semidirect product, and $K=\ker(\rho)$.

Prove that $K\unlhd G$ and that $G/K$ is also a semidirect product.

Thanks a lot in advance for any help!

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Edit: I deleted the part that was unclear (in fact bad formulated). The answers of DonAntonio and user finally solved the question.

Answer 1

Directly:

We use the notation $\;N\rtimes H\;$ , and the operational notation in the semidirect product (=s.p.):

$$(n_1,h_1)(n_2,h_2):=(n_1n_2^{h_1},h_1h_2)\;\;,\;\;n_2^{h_1}:=\rho(h_1)(n_2)$$

So suppose $\;k\in K\le H\iff \rho(k)=\text{Id}_N\;$ . In the s.p. we can denote this element as $\;(1,k)\;$ , and thus we get that

$$(n,h)^{-1}(1,k)(n,h)=(n^{-h^{-1}},h^{-1})(1,k)(n,h):=$$

$$(n^{-h^{-1}},h^{-1})(n^k,kh):=\left(n^{-h^{-1}}(n^k)^{h^{-1}}\,,\,h^{-1}kh\right)$$

Now, obviously $\;h^{-1}kh\in K\;$ since clearly $\;K\lhd H\;$, and in the first coordinate we have (remember that $\;H\;$ acts by automorphisms on $\;N\;$!):

$$n^{-h^{-1}}n^{kh^{-1}}=\left(n^{h^{-1}}\right)^{-1}n^{h^{-1}}=1$$

because $\;k\in K=\ker\rho\iff n^k=n\;\;\;\forall\,n\in N\;$ !

The above is, of course, after you already prove you have a semidirect product here, but that's easy and, in fact, some parts above can help you to see this.

Answer 2

I don't understand very well the last part of your question, but for me things look like this:

Take $\bar{\rho}:H/K\to Aut(N)$ defined by $\bar{\rho}(hK)=\rho(h)$. (This is well defined since $K=\ker\rho$.) Then define $\varphi:N\rtimes_{\rho}H\to N\rtimes_{\bar{\rho}}H/K$ by $\varphi(n,h)=(n,hK)$. Now it's easy to prove that $\varphi$ is a surjective group homomorphism and $\ker\varphi=\{1\}\times K$. In particular, $G/K\cong N\rtimes_{\bar{\rho}}H/K$.