Show that $\{1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)\}$ is a subgroup of $S_4$ Show that {$1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)$} is a subgroup of $S_4$ 
My attempt:
Let $H = \{1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)\}$, where $1$ is the identity permutation. Then, 
$1\circ(1, 2)(3,4) = (1, 2)(3,4) $
$1\circ(1, 3)(2,4) =(1, 3)(2,4) $
$1\circ(1,4)(2, 3)=(1,4)(2, 3) $
$(1, 2)(3,4)\circ1=(1, 2)(3,4) $ 
$(1, 3)(2,4) \circ1=(1, 3)(2,4) $
$(1,4)(2, 3)\circ1=(1,4)(2, 3) $
$(1,2)(3,4)\circ(1,3)(2,4)=(1,4)(2,3)$
$(1,3)(2,4)\circ(1,2)(3,4)=(1,4)(2,3)$
and so on.... (I will multiply all the elements with each other)
$H$ is finite and for all $a,b\in H$, $ab\in H$. Therefore, $H$ is a subgroup of $S_4$
Side note: What is 
$(1,3)(2,4)\circ(1,3)(2,4)=?$
Do I need to include all the elements composed with itself?

You can simplify your proof by setting $x,y,z$ the non-unit elements of your set. Show that $x^2=y^2=z^2=1$, $xy=z$ and $yz=x$. From this you conclude that $xz=xxy=y$. Therefore an arbitrary product of $x,y,z$ (and their inverses) will be in $\{1,x,y,z\}$, which is therefore a subgroup.
