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Show that if there were one Sylow $3$-subgroup, then $A_4$ would be Abelian.

I know the order of $A_4$ is $12$ and therefore $A_4$ can have $1$ or $4$ Sylow $3$-subgroup. However, I don't know how to prove the above statement.

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    $\begingroup$ Maybe this is wrong but I was thinking along the following line. It can be proved that $A_4$ is generated by the 3-cycles. Then if $n_3 = 1$ it means every $3-$cycle is a power of $(1,2,3)$ for example, as every $3-$cycle has order 3 and there is only one 3-sylow. This means the group is generated by $(1,2,3)$, so it is abelian. $\endgroup$ – Leo Lerena Apr 29 '19 at 21:13
  • $\begingroup$ Also, it's worth noting there are 2 groups of order 12 such that have one $3$-Sylow but are not abelian. So you can't use a counting argument to find a contradiction supposing $n_3 = 1$. $\endgroup$ – Leo Lerena Apr 29 '19 at 21:21
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    $\begingroup$ @LeoLerena Thanks for the insight! $\endgroup$ – RandomThinker Apr 29 '19 at 21:34
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Turning the comment into an answer so it doesn't remain unanswered. Knowing that $A_4$ is generated by the $3$-cycles you can arrive at a contradiction. If $n_3 = 1$ it means every $3$ cycle is a power of $(1,2,3)$ for example, as every $3$-cycle has order exactly $3$. Combining these two, you got that $A_4$ is generated by $(1,2,3)$. This means it is abelian.

Actually you can show much easily that there can not be one Sylow 3-subgroup if you just realized that the $3$-cycles $(1,2,3)$ and $(1,2,4)$ are not a power of one another. This means that $n_3 = 4$. It is a very interesting and instructive exercise the classification of groups of order $12$ if you haven't thought of it. Keith Conrad has an amazing article summarizing it, GROUPS OF ORDER 12.

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