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Let $H,L$ be groups acting on $K,G$. Show that $K\rtimes H\cong G\rtimes L$ and $K\ncong G$ is not true in general.

I found a solution online using $\times$: Counterexample: $G \times K \cong H \times K \implies G \cong H$ . I know the following: If $\{1\}\rtimes H \trianglelefteq K \rtimes H$ it follows that $K \rtimes H=K \times H.$ So am I right that this solutions works for my problem as well if $\mathbb{Z}$ and $\{1\}$ are acting trivial on $\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}\times...$? Any help is greatly appreciated!

Arturo Magidin
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hannah2002
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    Yes, the duplicate also gives an example for your title question, by just taking the trivial semidirect product. – Dietrich Burde Nov 21 '22 at 14:14
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    Shaun, I think you messed up the subject line. The subject line right now is silly; what the OP wanted was an example where this happened, not an example that it didn't. – Arturo Magidin Nov 21 '22 at 15:48
  • At the moment the title and the question both look silly. I can take $H,L,K,G$ to be random completely unrelated groups with trivial actions, and then $K \rtimes H \cong G \rtimes L$ is very unlikely to be true! – Derek Holt Nov 21 '22 at 16:51
  • Oh, I'm sorry, @ArturoMagidin; I'm not sure how to fix it. I'll leave it to you if that's alright. – Shaun Nov 21 '22 at 16:56
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    $\newcommand{\Set}[1]{\left{ #1 \right}}\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$If you wish to see an example where both semidirect products are non-trivial, you may consider the dihedral group of order $8$, say $D = \Span{ (1234), (13)} = \Set{1, (1234), (13)(24), (1432), (13), (24), (12)(34), (14)(23)}$. Then take $K = \Set{1, (13)(24), (12)(34), (14)(23)}$, $G = \Span{(1234)} = \Set{1, (1234), (13)(24), (1432)}$, $H = \Span{(13)} = \Set{1, (13)}$. One has $D = K \rtimes H = G \rtimes H$, but $K \not\cong G$. – Andreas Caranti Nov 22 '22 at 12:02

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