Prove that there is a fixed point in any subgroup $H$ of $S_4$ of order $6$. Prove that in every subgroup $H$ of $S_4$ of order 6  there is a fixed point in {$1,2,3,4$}, i.e, there exists $1\le i\le 4$ such that $h(i)=i$ $\forall h\in H$. 
$Start$: Suppose there is a subgroup $H$ of order 6 such that for every $i\in$ {$1,2,3,4$} there exists $\sigma\in H$ such that $\sigma (i)\ne i$. If there is a $4-cycle$ we are done by contradiction.Otherwise, 
case 1: There are a $3-cycle$ of the form $(abc)$ and a transposition of the form $(dk)$ where $k\in$ {$a,b,c$} and $a,b,c,d$ are all different. WLOG $k=a$, and $(abc)\cdot (da)= (abcd)\in H$, a contradiction. 
case 2: There are four different $2-cycles$. I am pretty stuck here $:($ . 
How is the first case to your assessment? How shall I continue?  
 A: Suppose there is no fixed point, then the subgroup is transitive on four points, and hence has order divisible by $4$.
You simply need to prove transitivity. The cyclic group generated by $(1 2)(34)$ has order two, moves every point and is not transitive. But it is easy to show that this is the only case. Take a $3$-cycle and an element which moves the other point. The subgroup is transitive on the first three points, and by conjugation by the $3$-cycle, the fourth point can be moved to any of the first three.
Conjugation in permutation groups is well worth understanding, because it can be made to do a huge amount of work.
A: The order of $H$ is $6$, hence $H\cong \mathbb{Z}_{6}$ or $H \cong S_3$. Since there are no elements of order $6$ in $S_4$, $H \cong S_3$. Let $f:S_3\rightarrow H$ isomorphism. Let $\sigma=f\left(\left(123\right)\right)=\left(jts\right),\rho=f\left(\left(12\right)\right)$. Notice that $\sigma$ is a $3$-cycle, so let $i$ its fixed point, and the order of $\rho$ is $2$.
$$(12)(123)=(23)$$
hence $f\left((12)(123)\right)=f\left((23)\right)$ is of order $2$.


*

*if $\rho=(ij)$, then $f\left((23)\right)=(ij)(jts)=(ijts)$ 

*if $\rho=(it)$, then $f\left((23)\right)=(it)(jts)=(itsj)$

*if $\rho=(is)$, then $f\left((23)\right)=(is)(jts)=(isjt)$

*if $\rho=(ij)(ts)$, then $f\left((23)\right)=(ij)(ts)(jts)=(ijs)$

*if $\rho=(it)(js)$, then $f\left((23)\right)=(it)(js)(jts)=(itj)$

*if $\rho=(is)(tj)$, then $f\left((23)\right)=(is)(tj)(jts)=(ist)$


And in al cases $f\left((23)\right)$ is not of order $2$. Hence, $i$ is fixed point of $f\left((12)\right)$, and because $H=\left<f\left((12)\right),f\left(\left(123\right)\right)\right>, i$ is fixed point of every $h\in H$.
A: You can prove the claim using the elements like you tried to. First, as $|H| = 6$, we have a $3$-cycle in $H$, say $(a \ b \ c) \in H$, where $\{a,b,c,d\} = \{1,2,3,4\}$. If now $\exists \ h \in H$ s.t. $h(d) \neq d$, then 
1) $h$ is a transposition (we may choose $(a \ d)$): But then $(a \ d)(a \ b \ c) = (a \ b \ c \ d) \in H$ which is not possible as $4 \not \mid 6$.
2) $h$ is a $3$-cycle: Calculate products of elements in $H$ to get a contradiction (too many).
3) $h = (d \ e)(f \ g)$: Same here.
