Classify all groups of order $4165=5(7^2)17$.

I've determined the following possibilities for each of the sylow subgroups

$r_5 = 1$

$r_7 = 1$ or $5(17)$

$r_{17} = 1$ or $5(7)$

I'm trying to show either the sylow $7$ subgroup or the sylow $17$ subgroup is normal so that I can create a subgroup of index $5$. Then I would use semi-direct product theorem. But maybe this is not necessary and maybe there is a simpler solution.

Source: Spring 1992

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    $\begingroup$ None of $7, 7^2$ or $7^2\cdot17$ are $\equiv1\pmod5$, so it looks like you have a normal Sylow $5$ $\endgroup$ – Jyrki Lahtonen Aug 5 '18 at 21:24
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    $\begingroup$ $7\cdot7\cdot17\equiv2\cdot2\cdot2=8\equiv3\pmod5$. $\endgroup$ – Jyrki Lahtonen Aug 5 '18 at 21:27
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    $\begingroup$ And the automorphism group of $C_5$ is $\simeq C_4$, so that Sylow $5$-subgroup must be central. $\endgroup$ – Jyrki Lahtonen Aug 5 '18 at 21:29
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    $\begingroup$ Yes, a subgroup being central means it's contained in the centre of the group. If we have a normal subgroup $N \subset G$, then conjugation gives a homomorphism $G \to \operatorname{Aut}(N)$. If $G$ is finite and its order is coprime to the order of $\operatorname{Aut}(N)$ - as is the case here, where $N$ is the $5$-Sylow subgroup whose automorphism group has order $4$ while the order of $G$ is odd - this homomorphism must be trivial, and that says nothing other than that the elements of $N$ commute with all elements of $G$, i.e. $N$ is contained in the centre of $G$. $\endgroup$ – Daniel Fischer Aug 5 '18 at 21:50
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    $\begingroup$ And continuing this line of thought, since the $5$-Sylow subgroup is central, it is contained in the normalisers of the $7$-Sylow and the $17$-Sylow subgroups, hence … $\endgroup$ – Daniel Fischer Aug 5 '18 at 21:56

Using Sylow theory, we have a normal Sylow $5$-subgroup $N$. As Daniel noted, this gives an automorphism $G\to \operatorname{Aut}(N)$ where $G$ acts by conjugation. The map must be trivial, which means $gng^{-1}=n$ for all $g\in G$ and $n\in C_5$, i.e., $N\in Z(G)$.

Now $G/Z(G)$ must have order dividing $7^2\cdot 17$, and all of these groups are abelian (you can show this using Sylow theory). For any group, if $G/Z(G)$ is abelian, then $G$ is nilpotent.

Since $G$ is nilpotent, it is a product of its Sylow groups. Thus there are two possibilities: $G=C_5\times C_7\times C_7\times C_{17}=C_{595}\times C_7$, or $G=C_5\times C_{49}\times C_{17}=C_{4165}$

  • $\begingroup$ Interesting, I did not know about these results regarding nilpotent groups. Is there anywhere I can read more? It might be useful for more of these types of problems. In class I believe we briefly mentioned the definition of nilpotent groups, but that was about it. $\endgroup$ – iYOA Aug 5 '18 at 23:07
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    $\begingroup$ The five equivalent formulations of finite nilpotent groups on wikipedia is useful. Combined with the fact that we have classified all groups of order $p^2$ (and $p^3$), this can go a long way. en.wikipedia.org/wiki/Nilpotent_group $\endgroup$ – Elliot G Aug 5 '18 at 23:57

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