a problem on fixed element of symmetric group of order 4 Let $S_4$ denote the group of permutations of $\{1 , 2 , 3 , 4\}$ and let $H$ be a subgroup of $S_4$ of order $6$. Show that there exists an element $i \in \{1, 2, 3, 4\}$ which is fixed by each element of $H$.

can anyone help me please how can I tackle this problem. I have no idea.thanks for your help
 A: Hint:
1) $\,A_4\,$ has no subgroup of order $\,6\,$ (this is a pretty easy exercise even if you write down specifically the elements of $\,A_4\,$ and check directly)
2) Thus, if  $\,H\le S_4\,\,\wedge\,\,|H|=6\Longrightarrow\,\;A_4\cap H=3$ (why?) . Write down the elements in this intersection (further hint: there are no many possibilities...)
3) Together with above, reason out what the other elements of $\,H\,$ must be so that we still have the two given conditions on $\,H\, $ (i.e., subgroup of $\,S_4\,$ and of order six)
A: It is a rather strong constraint that $H$ cannot contain any cycle of length $4$, since $4$ does not divide $6$. Also it must contain an element of order $3$ (by Cauchy's theorem), which must therefore be a $3$-cycle, and that we may take to be $(1~2~3)$. Now throwing in and transposition $(x~4)$ would produce a $4$-cycle as product, and throwing in any $3$-cycle involving the element $4$ would lead to generating the full alternating group $A_4$, which is too large. If you dislike doing the calculation to check that, you may use: in any group of order six, any pair of distinct elements of order $3$ are inverses of one another (check for the $2$ possible isomorphism types, though of course you know that you won't run into a cyclic group of order $6$ inside $S_4$). Conclusion, you've got the subgroup of all permutations that fix $4$.
A: This is straightforward if you know orbit-stabiliser theorem.
