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 $H$.
Please give some required hints to approach the problem.
The problem suggests that $H$ is isomorphic to the permutation group of some three elements. Find an element of order $3$ in $H$, an element of order $2$. Show that they must be, for example, $(123)$ and $(12)$, otherwise the order of $H$ is more than $6$.
elements of $H$ can be of order $1,2,3,6$ only. now if element is identity it is obvious. if order of an element is $2$ or $3$, then they are $2$ cycle or $3$ cycle resp. so will definitely leave $2$ or $1$ element fixed resp. Now if $a \in S_4$ and |$a$|=$6$ is not possible in $S_4$ as it will have to be either a $6$ cycle, absurd or product of disjoint $2$ and $3$ cycle, again absurd in group of permutation of $4$ elements.
