4
$\begingroup$

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

$\endgroup$
10
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ $7\cdot7\cdot17\equiv2\cdot2\cdot2=8\equiv3\pmod5$. $\endgroup$ – Jyrki Lahtonen Aug 5 '18 at 21:27
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
5
$\begingroup$

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 $G=C_5\times C_{49}\times C_{17}=C_{4165}$

$\endgroup$
2
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.