# 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 $A_6$?

Hint: How many Sylow 5-subgroups will $G$ have? Do you know of a way $G$ acts on them?

• I know that G acts on them by conjugation, but that's all :( – Martin Jan 17 '13 at 11:08
• Good! So if $X$ is the set of Sylow 5-subgroups, then conjugation action gives you a group homomorphism from $G$ to $S_n$, where $n=|X|$. Can you work with that? – Jyrki Lahtonen Jan 17 '13 at 11:11
• Yes, I think I can. Thanks a lot :) – Martin Jan 17 '13 at 11:13
• Great! If you can complete the argument, I would encourage you to type your solution as an answer to this question. Then the rest of us can comment on it and upvote it. You get more feedback that way. You can also accept that answer. It is not clear that a hint like this should be the accepted answer, for the readers who join the party later may benefit more from a full solution. – Jyrki Lahtonen Jan 17 '13 at 11:16
• No problem :) When I get complete solution, it will be posted – Martin Jan 17 '13 at 11:21

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, $G\cong \operatorname{Im(\phi)}\leq A_{6}$.

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. $$\Rightarrow\!\Leftarrow$$