Showing this subgroup of $S_4$ equals $S_4$: need help understanding the proof I need help understanding a proof of the following exercise:
Let $H$ be a subgroup of $S_4$ containing $(12)$ and $(234)$. Show that $H = S_4$.
The proof goes as follows: Note that $H$ contains $(234)$ and $(1234)$ hence both $3$ and $4$ divide $|H|$ and $|H|$ divides $24$.
Now the step that puzzles me:
If $|H| =12$ then the even permutations in $H$ would be a subgroup of $A_4$ of order $6$.
So, in other words, the above says that $|H \cap A_4|=6$. 
But why this should be so I do not understand.
The cycle $(234)$ is even and its square is also even. From this we have that $|H \cap A_4|\ge 3$.

How does it follow that it has to be $6$?

 A: Suppose $H\subseteq S_n$ is any subgroup of $S_n$, and consider the sign homomorphism $\mathrm{sgn}:H\to \{-1,1\}:\sigma\mapsto\mathrm{sgn}(\sigma)$. The kernel of this map is either the entire group, in which case $H$ consists solely of even permutations, or is a subgroup of index $2$, in which case exactly half the elements of $H$ are even permutations.
In the case you present, if $|H|=12$, since $(1234)\in H$ is odd, it follows that exactly half the elements of $H$ are odd and half even, so that $|H\cap A_4| = 6$.
A: As you shown $|H|$ is divisible by $3$ and $4$, hence by $12$, and it divides $|S_4|$. If $|H|=12$, then $H$ will contain only even permutations ($A_4$ is the only subgroup of order $12$ in $S_4$); but $H$ contains odd permutation $(12)$. Hence $|H|$ can not be $12$; it should be $24$; so $H=S_4$.

Comment:
I think, showing $|H\cap A_4|=6$ is not helpful (unless we are obtaining some contradiction finally) since $A_4$ has no subgroup of order $6$. 
consider $(12)$ and conjugate it by $(234)$:
$$ (234).(12).(234)^{-1}=(13).$$
Again conjugate new element $(13)$ by $(234)$:
$$(234).(13).(234)=(14).$$
This convinces that subgroup containing $(12)$ and $(234)$ also contains $(13)$ and $(14)$. 
Next you try to show that every permutation in $S_4$ can be written as product of $(12)$, $(13)$ and $(14)$. For this, note that $(1i)(1j)(1i)=(ij)$. In other words, any transposition $(ij)$ can be obtained from the transpositions of the form $(12), (13),\cdots (1n)$. 
Since every permutation in $S_4$ is product of transpositions, it is also product of specific transpositions $(12)$, $(13)$ and $(14)$; these elements can be obtained from $(12)$ and $(234)$. Hence your result follows.
