# Simple subgroup of $S_n$ containing at least one odd permutation is isomorphic to $\mathbb{Z}_2$

Let $$H$$ be a simple subgroup of a finite symmetric group that contains at least one odd permutation. Prove that $$H \cong \mathbb{Z}_2$$.

Here are my thoughts so far:

Let $$G = S_n$$ for some $$n$$. Since $$H$$ contains at least one odd permutation, $$H$$ cannot be contained in $$A_n$$, the set of all even permutations of $$G$$. Further, it's an easy exercise to show that if $$H$$ is a subgroup of $$S_n$$, then either all elements of $$H$$ are even or exactly half are even and half are odd. Thus, it must be that $$H$$ contains an equal number of odd and even permutations.

But I'm not sure how to use the fact that $$H$$ is a simple subgroup of $$G$$, here. Why must it follow that if $$H$$ has at least one odd permutation, and contains no proper nontrivial normal subgroups, that $$H$$ must only contain the identity element (which is even) along with exactly one transposition?

Any help would be appreciated. Thanks!

• I posted a solution to your question. If you have any questions, let me know. I'll be happy to help. Jan 12, 2020 at 3:15
• @AndrewOstergaard I appreciate the help. The solution is of great help -- I didn't think to use the First Isomorphism Theorem, but your solution makes a lot of sense after that idea is put into place. Thanks for the assistance, stranger. (= Jan 12, 2020 at 4:54

Let $$\phi:S_n\to \mathbb{Z}_2$$ be the homomorphism that sends even permutations to $$0$$, and odd permutations to $$1$$.
If we restrict this homomorphism to $$H$$ we get a homorphism $$\phi:H\to \mathbb{Z}_2$$ that must be surjective, since $$H$$ contains an odd permutation.
Since the kernel of $$\phi$$ is normal in $$H$$, and $$H$$ is simple, we must have either $$\ker\phi$$ is trivial or $$\ker\phi=H$$. But we can't have the latter, because $$H$$ contains an odd permutation. Hence $$|\ker\phi|=1$$.
$$H\cong H/\ker\phi\cong\text{Im }\phi=\mathbb{Z}_2,$$