Let $G$ be a Group of order 117. Then is $G$ abelian?
Note that $117 = 13*3^2$ and that $13 \equiv 1 \bmod 3$, so A group of order $p^2q$ will be abelian doesn't help.
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8The statement is false. There are non-abelian groups of order $117$. – Tobias Kildetoft Feb 06 '19 at 12:57
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4This link contains a classification, which includes non-abelian examples. – lulu Feb 06 '19 at 13:01
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3$\langle, a,b\mid a^{13}=b^3=1, ab=b^3a,\rangle$ is non-abelian of order $39$. – Hagen von Eitzen Feb 06 '19 at 13:05
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2I'm going to edit the question so it contains a statement which can be proved to be not true. – Sam Feb 06 '19 at 18:09
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No. $117$ is not in OEIS/A051532. See also https://math.stackexchange.com/questions/1556811/does-there-exist-an-n-such-that-all-groups-of-order-n-are-abelian. – lhf Feb 06 '19 at 21:05
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The statement is false.
Notice that $C_{13}<G$ and by Sylow's theorem there is a subgroup $P$ of size $9$ as well. Consider $\textrm{Aut}(C_{13})\cong C_{12}$, we can easily find a non-trivial homomorphism $\phi$ from $P$ to $\textrm{Aut}(C_{13})$. Thus we can form a non-trivial semidirect product of $C_{13}$ with $P$ using $\phi$.
Explicitly let $P=C_3^2$ so elements are tuples $(g,h)$ with integer entries taken $\mod 3$ and let $\textrm{Aut}(C_{13})=\langle x\rangle$. We can then define $\Phi:C_3^2\rightarrow \textrm{Aut}(C_{13})$ by $(g,h)\mapsto x^{4g}$.
Sam
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