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!

  • $\begingroup$ I posted a solution to your question. If you have any questions, let me know. I'll be happy to help. $\endgroup$
    – user729424
    Jan 12, 2020 at 3:15
  • $\begingroup$ @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. (= $\endgroup$ Jan 12, 2020 at 4:54

1 Answer 1


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 $|\ker\phi|=1$.

It follows that

$$H\cong H/\ker\phi\cong\text{Im }\phi=\mathbb{Z}_2,$$

because of the first isomorphism theorem.


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