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A group $G$ is perfect when $G=G'$.

$\textbf{Question.}$ Prove that there are no perfect groups of order $180$.

$\textbf{My attempt.}$ I assumed there is such a $G$, so it is not solvable. I was trying to analyze the number of $5$-Sylow subgroups of $G$. If $n_5=1$, then there is a normal subgroup of order $5$ and the quotient has order $36$, so $G$ would be solvable. If $n_5=36$, then $P=N_G(P)$, so using Burnside's normal complement theorem, $G$ would be $5$-nilpotent and it would be possible to show also that $G$ is solvable. Now if $n_5=6$, I don't know how to proceed.

Shaun
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Zed
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The image of the conjugation action on the six Sylow $5$-subgroups is a perfect subgroup of $A_6$ with order divisible by $30$. It can't be order $180$ because that would be a subgroup of index $2$ in $A_6$. It can't be order $30$ or $90$, because they have twice odd order and hence have subgroups of index $2$.

So the image has order $60$ and the kernel $K$ has order $3$. Since $G$ is perfect, we must have $K \le Z(G)$.

But now we can use 10.1.6 of Robinson's book. A Sylow $3$-subgroup $P$ of $G$ is abelian, and since $G$ is perfect the transfer homorphism $\tau:G \to P$ is the trival map, but then $C_P(N_G(P)) = 1$, contradicting $K \le C_P(N_G(P))$.

Derek Holt
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  • I couldn't understand why the image has order divisible by $30$ and why $K \leq Z(G)$. Could you explain to me please? – Zed Sep 29 '21 at 11:35
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    These are very routine arguments! It is transitive and hence divisible by $6$, and the Sylow $5$-subgroups act non-trivially, so also divisible by $5$. If $K$ wasn't central, then its centralizer would have index $2$ in $G$. – Derek Holt Sep 29 '21 at 12:10
  • I was able to understand the first part, but I still can't see why $C_G(K)$ would have index $2$ in $G$. – Zed Sep 30 '21 at 19:23
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    $K = {1,x.x^2}$ is cyclic of order $3$. What are the possible values of $g^{-1}xg$ for $g \in G$? It also follows from the more general result that, for any subgroup $H$ of any group $G$, $N_G(H)/C_G(H)$ is isomorphic to a subgroup of ${\rm Aut} H$. That is in Chapter 1 (1.6.13) of Robinson's book but you are attempting problems in Chapter 10. – Derek Holt Sep 30 '21 at 20:11
  • I got it now, thank you very much! Just one thing, this $P$ has to be taken so that $K \leq P$. – Zed Oct 01 '21 at 14:58
  • Again, I feel that you should know already that $K \le P$ for all Sylow $3$-subgroups of $G$. – Derek Holt Oct 01 '21 at 17:12
  • True, sorry! Because $K \trianglelefteq G$. – Zed Oct 01 '21 at 17:27