# Homomorphisms with the same kernel has the same semidirect product?

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 $$Z_7\rtimes_{\varphi_1} (Z_2\times Z_2)\cong Z_7\rtimes_{\varphi_2} (Z_2\times Z_2)$$?

• 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. Nov 18, 2019 at 2:09
• In this specific example, it works, but is my statement true in the general sense? Nov 18, 2019 at 2:16

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).
• 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. Nov 18, 2019 at 18:36
• 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 $(H_1\times H_2)\rtimes K\simeq (H_1\rtimes K) \times H_2$. Nov 18, 2019 at 21:20
• Ah, so since $H_2$ essentially commutes with $K$ like a direct product we can switch the order. Thanks! Nov 18, 2019 at 21:37