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Using the Sylow theorem, I know that there exists normal subgroup of order $5$ and subgroup of order $4$. Letting them $Q$ and $P$, and taking semidirect product $Q\rtimes P$, I get homomorphism

$$\varphi: P \to {\rm Aut}(Q) $$

I have two cases when $P\cong \mathbb{Z}_2 \times\mathbb{Z}_2 $ and $P\cong \mathbb{Z}_4$.

Specifically when $P$ is isomorphic to $\mathbb{Z}_4$, I have homomorphism $\mathbb{Z}_4 \to \mathbb{Z}_4$. I can see that there are three cases when the order kernel of this homomorphism is $1,2,4$.

When order of kernel of this homomorphism is $2,$ the textbook says that $\varphi $ is an "inverse map." I can't seem to understand what it means by having inverse map. Can anybody explain so that I can explicitly see what group this is?

Obviously I can picture the homorphism $\varphi$ from $\Bbb Z_4$ to "$\Bbb Z_4$" (when the order of kernel is $2$) such that $0,2$ goes to $0$ and $1,3$ goes to $2$, but then how can I construct the semidirect product based on this?

able20
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  • NB: The symbol $\ast$ is used for the free product of groups. Use $\times$ for $\times$, the direct product. – Shaun Dec 15 '19 at 17:30
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    In this case $Q\cong C_5$ and $Aut(Q)\cong C_4$. Note that $Aut(Q)$ has a unique element of order $2$, and this unique automorphism of $C_5$ of order $2$ is "inversion", that is, the map that sends every element to its inverse. So, when the kernel has order $2$, the generator of $P$ is sent to an element of order $2$, which must be this inversion map. I believe this is what they meant. – verret Dec 15 '19 at 19:46
  • Thank you! I can now understand – able20 Dec 16 '19 at 20:20
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    Does this answer your question? Find four groups of order 20 not isomorphic to each other. – verret Dec 16 '19 at 22:21
  • It seems to me that the proposed duplicate, while related to the classification problem, does little to address the meaning of "inversion map" that is spotlighted here. See the exchange of comments above. – hardmath Dec 17 '19 at 02:09

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