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Seeking for one more proof to this claim (namely, that a group of order $255=3.5.17$ is Abelian), I'm looking for an argument (if any) which rules out the case $P\cap Z(G)=\{1\}$, where $P$ is the (one) normal subgroup of order $17$ of $G$. If I could get this, I think that $G/P$ cyclic and $P$ central would lead to the desired result (likewise the usual argument about $G/Z(G)$).

What I know. The nontrivial elements of $P$ must lay in two conjugacy classes (of $G$) of size $3$ and in two other of size $5$ (overall $1+3+3+5+5=17$).

How can I complete this argument?

Nicky Hekster
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    The automorphism group of an group of order $17$ has order $16$; any action by an element of order dividing $255$ must be trivial. – Arturo Magidin Aug 25 '22 at 21:54

1 Answers1

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I've realized myself that the argument works by redirecting the attention from the normal $P_{17}$ to any other normal subgroup between $P_3$ and $P_5$ (at least one of them must be normal), say jointly $P_p$. In fact, the equation$^\dagger$: $$|P_p|=|P_p\cap Z(G)|+3k+5l$$ hasn't got any solution $(k\ge0,l\ge0)$ for $|P_p\cap Z(G)|=1$ in either cases $p=3,5$. But then $P_p$ is central. Moreover, $G/P_p$ is in any case cyclic, because both $5\nmid(17-1)$ and $3\nmid(17-1)$. Therefore, $G$ is Abelian by the usual "$G/Z(G)$ argument".

$^\dagger$Every normal subgroup is the union of conjugacy classes of the parent group.