Hi I need help with proving a group of order 343 is not simple. Since a group of order 343 is a 7-group, I believe the proof will involve the use of G's nontrivial center and class equation. I am just not sure how to start this proof. Thanks.
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2Well, it has a nontrivial centre, which is a normal subgroup. – Angina Seng Apr 30 '19 at 20:10
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Finite $p$-groups are nilpotent. Also, as was pointed out above, the center of a group is always normal. – anomaly Apr 30 '19 at 20:15
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https://en.wikipedia.org/wiki/Feit%E2%80%93Thompson_theorem – Tsemo Aristide Apr 30 '19 at 20:16
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Note that the center being normal is not enough, since it can be equal to $G$. Of course if $Z(G)=G$ then it is easy to prove that $G$ is not simple, but still. – Mark Apr 30 '19 at 20:17
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For every group $G$ we have $Z(G)\trianglelefteq G$. Here as we know $Z(G)\ne\{e\}$. So if $Z(G)\ne G$ then $G$ has a non trivial normal subgroup and hence is not simple. If $Z(G)=G$ then $G$ is abelian. In that case each subgroup of $G$ is normal in it. By Cauchy's theorem $G$ has a subgroup of order $7$, and this is a non trivial normal subgroup. Again, $G$ is not simple.
Mark
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Well $343= 7^3$ and it is a well known fact that the only simple $p$-groups are the cyclic $\mathbb{Z}/p\mathbb{Z}$. To prove this just look to the class equation $$ |G|= |Z(G)| + \sum_{i} |G : C_G(g_i)| $$ that becomes in our case $$ p^n = |Z(G)| + \sum_i p^{k_i} $$ with $k_i \geq 1$ unless $n=1$. Therefore $p$ divides $|Z(G)|$ and $G$ has nontrivial center.
Alan Muniz
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