# Show that if there were one Sylow 3-subgroup, then A4 would be Abelian.

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.

• 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. – Leo Lerena Apr 29 '19 at 21:13
• 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 $n_3 = 1$. – Leo Lerena Apr 29 '19 at 21:21
• @LeoLerena Thanks for the insight! – RandomThinker Apr 29 '19 at 21:34

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.