I want to prove that the symmetric group $S_n$ has a subgroup isomorphic to $\mathbb{Z}_7 \times \mathbb{Z}_7$ iff $n \ge 14$.
One direction seems clear. |$\mathbb{Z}_7 \times \mathbb{Z}_7| = 49$, and the factorizations of $49$ are $49*1$ and $7*7$. $|S_n| = n!$, and $7$ is prime, and if $n<14$ we have only one factor of $7$ and no factors of $49$, so there can't be an isomorphism.
But I'm having trouble with the other direction. By Cayley's Theorem, I know that $\mathbb{Z}_7 \times \mathbb{Z}_7$ is isomorphic to a subgroup of $S_{49}$. I also know that $S_n$ has two factors of $7$ is it is divisible by $49$. But I don't see how to go from there to saying that $\mathbb{Z}_7 \times \mathbb{Z}_7$ is isomorphic to a subgroup of $S_n$ with $n\ge14$. Does anyone have any hints?
Best Answer
You can take the subgroup generated by $$(1234567)$$ and $$(8\ 9\ 10\ 11\ 12\ 13\ 14)$$ These are both of order $7$ and they commute, so they generate a subgroup isomorphic to $\mathbb Z_7\times\mathbb Z_7$.
Conversely, if $S_n$ has a subgroup isomorphic to $\mathbb Z_7\times \mathbb Z_7$, then $7^2\mid n!$, so $n\geq 14$.