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Prove that $A_{5}$ has a subgroup of order $12$.

What I have so far: The order of an alternating subgroup $A_{n}$ = $n!/2$. So the order of $A_{5}$ is $60$. We know that the order of a subgroup must be divide the order of the group, and this is the case since $12$ divides $60$.

Also, I have another question. Are all alternating groups cyclic?

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Consider the Alternating group $A_{4}$, it has order 12, and it's in $A_{5}$. Aletnaring group $A_{n}$ is not cyclic for $n >3$

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  • $\begingroup$ Why is $A_{4}$ a subgroup of $A_{5}$? Is it always the case that a smaller alternating group will be a subgroup of a larger alternating group? $\endgroup$ – EmaLee Apr 13 '15 at 0:16
  • $\begingroup$ Never mind, someone else answered this question. $\endgroup$ – EmaLee Apr 13 '15 at 0:17
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Are alternating groups cyclic? Well, all cyclic groups are abelian (you should prove this). Is the alternating group abelian?

As a hint to your subgroup of order $12$ question, note that $A_n \subset A_m$ for all $n < m$.

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