Non-Abelian simple group of order $120$ Let's assume that there exists simple non-Abelian group $G$ of order $120$. How can I show that $G$ is isomorphic to some subgroup of $A_6$?
 A: Hint: How many Sylow 5-subgroups will $G$ have? Do you know of a way $G$ acts on them?
A: A group of order 120 can not be simple. Let's assume that there exists simple non-abelian group $G$ of order 120. Then we know the number Sylow 5-subgroups of $G$ is 6. Hence, the index of $N_{G}(P)$ in $G$ is 6 ($P$ is a Sylow 5-subgroup of $G$). Now there exists a monomorphism  $\phi$ of $G$ to $S_{6}$. We claim that $\operatorname{Im\phi}\leq A_{6}$. Otherwise $\operatorname{Im\phi}$ has an odd permutation and so $G$ has a normal subgroup of index 2, a contradiction.  Hence,  $G\cong \operatorname{Im(\phi)}\leq A_{6}$. 
A: Suppose for the sake of contradiction that your group $G$ of order $120$ is simple. Note that $120=2^3\times3\times5$. By the third Sylow theorem, the number of Sylow $5$-subgroups of $G$ must divide $120$ and be congruent to $1 \pmod{5}$, so there can be either $1$ or $6$ Sylow $5$-subgroups. But if there is only $1$, then it'll be normal in $G$ contradicting $G$ being simple. So there must be $6$ Sylow $5$-subgroups. Let $X$ denote this set of Sylow $5$-subgroups. By the second Sylow theorem, $G$ acts on this set $X$ by conjugation, permuting the subgroups. This action gives us a map $\varphi\colon G \to S_6$ with each $g\in G$ being sent to the permutation that describes its action on $X$. But again, if $G$ is simple and doesn't have a normal subgroup then $\mathrm{Ker}(\varphi)$ must be trivial, and $\varphi$ must be injective, telling us $G \cong \mathrm{Im}(\varphi) < S_6$. 
But we can strengthen this and say $G \cong \mathrm{Im}(\varphi) < A_6$. Recall that for $H < S_n$ either $H < A_n$ or exactly half of the elements of $H$ are contained in $A_n$. But in the latter case $\mathrm{Im}(\varphi) \cap A_n$ would be an index $2$ subgroup, and index $2$ subgroups are normal subgroups, contradicting $G \cong \mathrm{Im}(\varphi)$ being simple. So we have $G \cong \mathrm{Im}(\varphi) < A_6$. 
Now since $|A_6| = 6\times5\times4\times3 = 360$, $\mathrm{Im}(\varphi)$ will be an index three subgroup of $A_6$. Using the same trick as in the first paragraph, $A_6$ will act on the set of left cosets $A_6 / \mathrm{Im}(\varphi)$ and we'll get a nonzero homomorphism $A_6 \to S_3$. But $A_6$ is bigger than $S_3$, so this homomorphism has a nontrivial kernel, which will be a normal subgroup of $A_6$. We know $A_6$ is simple though, so this is our contradiction. $\Rightarrow\!\Leftarrow
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