Subgroups of $A_5$ have order at most $12$? How does one prove that any proper subgroup of $A_5$ has order at most $12$?
I have seen that there are $24$ $5$-cycles and $20$ $3$-cycles. What do the other members of $A_5$ look like?
 A: Zach's comment gives a good hint for the canonical solution.

 The cycle structure of a non-trivial element of $A_5$ is either $5$, $3$, or $2$-$2$.  If you've got a $5$-cycle $a$ multiplied by a $3$-cycle $b$ for which $ab$ is a $5$-cycle, then $ab^2$ will be of type $2$-$2$ and thus have order $2$, so there can be no subgroup of order $15$. For a subgroup $H$ of order $20$, note first no $4$-cycles exist in $A_5$, so a Sylow $2$-subgroup $H_2\in \operatorname{Syl}_2(H)$ must be a Klein $V$ group.  Then a Sylow $5$-subgroup $H_5\in\operatorname{Syl}_5(H)$ is not normal in $H$, as this would require that $H_2$ embed into $\operatorname{Aut}(H_5)\cong \mathbb{Z}_4$.  But noting that $n_5$ should divide $[H:H_5] = 4$ and $n_5\equiv 1\pmod 5$ we see by Sylow's theorem that $H_5$ should be normal in $H$, so this is a contradiction.

Here's a fun alternative proof. The only (proper) divisors of $|A_5|=60=2^2\cdot 3\cdot 5$ greater than $12$ are $15=3\cdot 5, 20=2^2\cdot 5,$ and $30=2\cdot 3 \cdot 5$.  Any subgroup of index $2$ is necessarily normal, and groups whose order is the product of three primes are solvable, so since $A_5$ is not solvable so it may not contain a subgroup of order $30$.  If $H$ is a subgroup of order $15$, then $H$ is cyclic, but an easy computation shows that in $A_5$ elements of order $5$ and elements of order $3$ do not commute.  Therefore it remains to be shown that $A_5$ can't have a subgroup of order $20$.
Now, notice that $A_5$ does have a subgroup $K$ of order $12$ - in particular, $K$ is an isomorphic copy of $A_4$.  Suppose that there exists a $J\leqslant A_5$ of order $20$ and notice that $[A_5:J]$ and $[A_5:K]$ are coprime.  Then $A_5=JK$ and $[J:J\cap K]=[A_5:K]$.  So $|J\cap K|=4$.  Thus $J\cap K\in\operatorname{Syl}_2(J)\cap\operatorname{Syl}_2(K)$.  We observe that any $H_3\in\operatorname{Syl}_3(K)$ normalizes $J \cap K$, and in particular, $H_3$ acts fixed point freely on $J \cap K$.  We see that the Sylow $5$-subgroup $H_5$ of $J$ must normal in $J$, so $J\cap K$ acts nontrivially on $H_5$.  It follows that $H_3$ acts faithfully on $H_5$, but this is impossible since $\operatorname{Aut}(H_5)$ has order $4$.  This completes the proof.
