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Suppose I am trying to construct a semidirect product and I have two homomorphisms $\varphi_1:K\rightarrow\text{Aut}(H)$ and $\varphi_2:K\rightarrow\text{Aut}(H)$. Furthermore, suppose that $\ker\varphi_1\cong\ker\varphi_2$. Are the semidirect products constructed by these homomorphisms equivalent?

For an explicit example, suppose I wanted to construct $Z_7\rtimes (Z_2\times Z_2)$, Aut($Z_7)\cong Z_6$. If we have $Z_2\times Z_2=\langle a\rangle\times\langle b\rangle$, the following homomorphisms both work (defined on the generators): $$\varphi_1(a)=1\quad \varphi_1(b)=x$$ $$\varphi_1(a)=x\quad \varphi_1(b)=x$$ where $x$ is the automorphism of order $2$ in Aut$(H)$. In both of these cases, the kernel is isomorphic to $Z_4$. Is it true that $Z_7\rtimes_{\varphi_1} (Z_2\times Z_2)\cong Z_7\rtimes_{\varphi_2} (Z_2\times Z_2)$?

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  • $\begingroup$ The thing with your example is that it satisfies a stronger condition: there is an automorphism of $Z_2\times Z_2$ which conjugates your two morphisms. And that is strong enough to guarantee that the two semi-direct products are isomorphic. $\endgroup$ Nov 18, 2019 at 2:09
  • $\begingroup$ In this specific example, it works, but is my statement true in the general sense? $\endgroup$
    – Vasting
    Nov 18, 2019 at 2:16

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The answer to your question is no: actually you can have $\varphi_1$ and $\varphi_2$ both injective (so with the same kernel) and still have non-isomorphic semi-direct products.

For instance, take $K=\mathbb{Z}/2\mathbb{Z}$ and $H=\mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$, and define $\varphi_1$ such that the non-trivial element in $K$ acts non-trivially on the $\mathbb{Z}/3\mathbb{Z}$ factor (and trivially on the other one), and $\varphi_2$ such that it acts non-trivially on $\mathbb{Z}/4\mathbb{Z}$ (and trivially on the other one).

Then $\varphi_1$ and $\varphi_2$ are injective, but the semi-direct products are respectively $D_6\times \mathbb{Z}/4\mathbb{Z}$ and $D_8\times \mathbb{Z}/3\mathbb{Z}$ (where $D_{2n}$ is the dihedral group with $2n$ elements) which are not isomorphic (for instance we can look at the center).

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  • $\begingroup$ I can see how we can map the nonidentity element in $K$ to nontrivial automorphisms of $Z_3$ and $Z_4$ respectively, as both of their automorphism groups are $Z_2$. But how did you end up with a direct product with the dihedral group? It isn't obvious to me that $(Z_3\times Z_4)\rtimes Z_2\cong D_6\times Z_4$, for example. $\endgroup$
    – Vasting
    Nov 18, 2019 at 18:36
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    $\begingroup$ You can see that in two steps: first $\mathbb{Z}/n\mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z} \simeq D_{2n}$ when $\mathbb{Z}/2\mathbb{Z}$ acts on $\mathbb{Z}/n\mathbb{Z}$ by $x\mapsto -x$. This is immediate from the definition of the dihedral group. Then you can easily see that if $K$ acts trivially on $H_2$ then $(H_1\times H_2)\rtimes K\simeq (H_1\rtimes K) \times H_2$. $\endgroup$ Nov 18, 2019 at 21:20
  • $\begingroup$ Ah, so since $H_2$ essentially commutes with $K$ like a direct product we can switch the order. Thanks! $\endgroup$
    – Vasting
    Nov 18, 2019 at 21:37

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