Assume $H\le S_n$ contains an element of order $n$ and a transposition. Given that $n$ is prime, does $H=S_n$?

Question

Let $H\leq S_n$ be a subgroup such that $H$ contains an element of order $n$. Assume that $n$ is a prime and prove that $H=S_n$.

Thoughts:

Since $n$ is prime, the element of order $n$ in $H$ must be a cycle of order $n$. So we have a cycle of order $n$ and we have a trnasposition, say $(x_1 \space x_2)$. What I want is to prove that we can get any simple transposition by composition of enough times of the cycle on our transposition. Not sure how to do it yet.

Any help would be appreciated. Thanks in advance.

In the text, you forgot to mention that $H$ contains a transposition. — Dietrich Burde, Dec 24 '21 at 11:59

Carl Schildkraut · Accepted Answer · 2021-12-24T22:44:46.703

3

Hint: Say the transposition transposes $1$ and $k+1$, and let $\sigma$ be the cycle $(123\cdots n)$. Then $\sigma^k$ is a cycle (this part uses $n$ prime), so you can relabel the elements so that the transposition is $(12)$. Can you finish from here?

edited Dec 24 '21 at 22:44

answered Dec 24 '21 at 09:21

Carl Schildkraut

32,931

Isn't $\sigma^n$ the identity? since the order is $n$? – FreeZe Dec 24 '21 at 09:48
@FreeZe Oops -- I meant $\sigma^k$. The error is fixed now. – Carl Schildkraut Dec 24 '21 at 22:44

Assume $H\le S_n$ contains an element of order $n$ and a transposition. Given that $n$ is prime, does $H=S_n$?

1 Answers1