I've been battling at this for a bit, $S_4$ isn't cyclic so that makes things a bit more difficult. I know by Lagrange a subgroup G with order 6 exists. I actually found some examples. Then found by research that each order 6 group is precisely the $S_3$. I'm failing to see why this is important, hence failing to prove the actual headline of this question. Any help is much appreciated.
2026-03-25 18:57:31.1774465051
Given $G,L \leq S_4$, if $|G| = |L| = 6$, do we have $L \cong G$
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There are two groups of order 6, up to isomorphism: $S_3$ and $\mathbb{Z}_6$.
Is it possible for a subgroup of $S_4$ to be isomorphic to $\mathbb{Z}_6$?
Hint: Think about the orders of elements of $S_4$.