When there exists an element of $S_4$ in $H$ such that it is fixed by each element of $H$. ($H$ is a subgroup of $S_4$)

Question

The following question was part of my Group Theory assignment and I am not able to get an idea on which result I should use to solve this question.

So, I am posting it here.

Let $S_4$ be the group of 4 element permutations and $H$ be any subgroup of $S_4$. Then which of following options are true. Give proper reasons.

1. If order of $H$ is 4, then there exists $i\in X$ such that $i$ is fixed by each element of $H$.

2. If order of $H$ is 6, then there exists $i\in X$ such that $i$ is fixed by each element of $H$.

3. If order of $H$ is 8, then there exists $i\in X$ such that $i$ is fixed by each element of $H$.

4. If order of $H$ is 12, then there exists $i\in X$ such that $i$ is fixed by each element of $H$.

I can't think of which result to use despite thinking a lot and would really like someone helping me. I have studied abstract algebra from Thomas Hungerford and did a lot of problems but couldn't did it.

Answer 1

Hints:

If $H$ fixes $i\in X$, that means that $H$ must be a subgroup of the stabilizer of $i$ which is just $S_{X\setminus\{i\}}\cong S_3$.
This can be directly applied to get answer for problems 1., 3. and 4.

For problem 2., you can use Cauchy's theorem to obtain a 3-cycle $\sigma\in H$ and also a permutation $\tau\in H$ of order 2. Without loss of generality, we can assume $\sigma=(1\,2\,3)$.
Then show that either if $\tau$ is a transposition involving $4$ or if it is a product of two disjoint transpositions, then the subgroup generated by $\sigma$ and $\tau$ has more than 6 elements.