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Take the convention that $D_{2n}$ is the dihedral group of order $2n$, and $\text{Dic}_{n}$ is the dicyclic group of order $4n$.

I want to show that if $p$ is a prime congruent to $3 \bmod 4$, $p>3$, then there are only two non-abelian groups of order $4p$ (which therefore must be $D_{4p}$ and $\text{Dic}_{p}$, since these are non-isomorphic and non-abelian. In fact, there will be a total of $4$ groups, since the only abelian ones can be $C_2 \times C_2 \times C_p$, $C_4 \times C_p$).

This reference confirms that for $p=7$, $p=11$, $p=19$ and $p=23$, the smallest $5$ primes with this property, that this result holds. Note it doesn't hold for all primes. For $p=5$, there are certainly more than $2$ non-abelian groups of order $20$, for example.

This page on groupprops confirms the result. May I have a hint as to how to start to prove this?

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  • $\begingroup$ Are you familiar with character theory? $\endgroup$
    – Fotis
    Jun 4, 2023 at 10:55
  • $\begingroup$ @Fotis No. If that's the best way to work this out, I could try look into it. Is there anywhere you recommend I start? $\endgroup$
    – Robin
    Jun 4, 2023 at 10:56
  • $\begingroup$ Sorry I meant representation Theory, characters are not directly needed. I'm not sure if the Theorems provided by Representation Theory are the best way to work this out to be honest, but if you're interested in studying it, the book I've been using is Representation Theory: A first course, by Fulton (I'm not saying this is the best book out there of course) $\endgroup$
    – Fotis
    Jun 4, 2023 at 11:03
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    $\begingroup$ @Fotis Thanks. I'll take a look. I'm doing a course on Representation Theory at my university next year as well, so hopefully that will help. $\endgroup$
    – Robin
    Jun 4, 2023 at 11:09
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    $\begingroup$ I would expect the easiest way to do this would be directly using Sylow's theorem. Any such group is a semidirect product $C_p \rtimes C_4$ or $C_p \rtimes (C_2 \times C_2)$, and $C_p$ has no automorphism of order $4$ when $p \equiv 3 \bmod 4$, sso there is just one nonabelian semidirect product of each of those two types. $\endgroup$
    – Derek Holt
    Jun 4, 2023 at 12:50

1 Answer 1

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Here is an expansion of the comment by Derek Holt.

Let $G$ be a group of order $4p$, where $p$ is a prime $>3$ and $\equiv 3\pmod 4$. By Sylow's theorems, we have Sylow subgroups $S_p$ (of order $p$) and $S_2$ (of order $4$) of $G$. The number of such $S_p$ is a divisor of $4p$ and $\equiv 1\pmod p$, hence is $1$, namely $S_p$ is a unique and normal subgroup of order $p$. $S_2$ acts on $S_p$ by $$\varphi\colon S_2\rightarrow \mathrm{Aut}(S_p),\ g\mapsto (x\mapsto gxg^{-1}).$$ Clearly $S_p\cap S_2=\{1\}$ and $S_p$ and $S_2$ generate $G$, so $G$ is a semi-direct product $G\simeq S_p\rtimes S_2$. Note that $S_p=\langle x_1\rangle$ is a cyclic group with $\mathrm{Aut}(S_p)\simeq C_{p-1}$ a cyclic group of order $p-1$ whose generator $\sigma$ maps $x_1$ to $x_1^k$, where $k$ is a primitive root modulo $p$.

Next, we investigate the structure of this semi-direct product. If the map $\varphi$ is trivial, $G$ is abelian (since $S_2$ is abelian), so we omit them. If $S_2=\langle g_1\rangle$ is cyclic, the only nontrivial homomorphism $\varphi$ is $g_1\mapsto \sigma^{(p-1)/2}$ because $p\equiv 3\pmod 4$, and this gives the dicyclic group. If $S_2=\langle g_1,g_2\rangle$ is the $2$-elementary group, then the only nontrivial homomorphism $\varphi$ is such that $g_1\mapsto \mathrm{id}, g_2\mapsto \sigma^{(p-1)/2}$ up to change of generators. This gives the group $D_{4p}\simeq C_2\times D_{2p}$.

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