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I am trying to classify the non-abelian group of order $20$.

We know that $K$ is a subgroup of order $5$ is normal. $H$ is a subgroup of order $4$.

Case $1:$ let $H \cong \mathbb{Z_2} \times \mathbb{Z_2}$

Then we need to find a homomorphism $\varphi:H \to Aut(\mathbb{Z_5})$

We need to find an element of order $2$ in Aut($\mathbb{Z_5}$).

The element of order $2$ in given by $\sigma_4:(1_5) \to 4_5$.

Now possible nontrivial homomorphisms are

$\varphi_1(1_2,0) \mapsto \sigma_4$ and $\varphi_1(0,1_2) \mapsto id$

Another possible homomorphism is

$\varphi_2(1_2,0_2) \mapsto \sigma_4$ and $\varphi_2(0_2,1_2) \mapsto \sigma_4$

Then $K \rtimes_{\varphi_1} H$ and $K \rtimes_{\varphi_2} H$ are two possible groups. Is this ok? Or are the two groups isomorphic?

Case $2:$ let $H \cong \mathbb{Z_4}$

Then we need to find a homomorphism using the following

$\varphi'_1:H \to Aut(K)$

Where $\varphi'_1(1_4) \mapsto \sigma_2$.

Then $\sigma_2:1_5 \to 2_5$ is the only element of order $4$.

Then we consider the map $\varphi'_1(1_4) \to \sigma_2$.

Then $\mathbb{Z_5} \rtimes_{\varphi'_1} \mathbb{Z_4}$

Shaun
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Antimony
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  • They are isomorphic. For the group to be non-abelian, two of $(1,0),$ $(0,1)$ and $(1,1)$ will go to $\sigma_4.$ And any two of those generate $\mathbb Z_2\times\mathbb Z_2.$ – Thomas Andrews May 03 '22 at 13:06
  • How do you understand that the two are isomorphic. One of the homorphism sends $(1,0),(1,1)$ to $\sigma_4$ and the other $(0,1),(1,0)$ to $\sigma_4$. – Antimony May 03 '22 at 13:10
  • Also is there anything more we can saay about the group using semi direct product? – Antimony May 03 '22 at 13:13
  • Because there is an automorphism on the $4$ element group that equates the two. If $\phi_i:H_i\to \operatorname{Aut}(G)$ for $i=1,2,$ and an isomorphism $f:H_1\to H_2$ with $\phi_2\circ f=\phi_1,$ then the two semi direct products are also isomorphic. – Thomas Andrews May 03 '22 at 13:15
  • Yes, you can show this group is isomorphic to $\mathbb Z_2\times D_{10},$ where $D_{10}$ is the dihedral group on a pentagon. – Thomas Andrews May 03 '22 at 13:21
  • Can you go through the second case once?These are the only possible cases right?How do you understand that the last group is isomorphic to $Z_2 \times D_{10}$? – Antimony May 03 '22 at 13:25
  • Let us continue this discussion in chat. – Antimony May 03 '22 at 13:26

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