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$\begingroup$

I want to show that a group G of order 345 is Abelian.

I used Sylow's theorem to find Syl(5)=Syl(23)=1. but i was unable to conclude Syl(3)=1 because i found Syl(3)=1 or 115. I'm not sure how to proceed from here..

Please help, thank you!

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Note that $(5,23)=1$ and $syl(5)=syl(23)=1$ shows that $G$ has a subgroup $H$ with order $5\times 23=115$, of course it is cyclic. And because it has index 3, which is the smallest divisor of the order of $G$, so it is normal subgroup, then we get a homomorphism $\phi$ from $C_3\to Aut(C_{115})$, while $|Aut(C_{115})|=115(1-\frac{1}{5})(1-\frac{1}{23})=88$, which is not divisable by 3, so the $\phi$ is trivial, so $G$ is a cyclic group.

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    Centre, sorry: don't know why I typed centralizer,2012-12-20