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?

Let $S=\{1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)\}$, and relabel the elements $\sigma_1=1$, $\sigma_2=(1, 2)(3,4)$, $\sigma_3=(1, 3)(2,4)$, $\sigma_4=(1,4)(2, 3)$, where we note $\sigma_1=1=(1)(2)(3)(4)$ is the identity permutation.
Consider the product $(1,3)(2,4){\circ}(1,3)(2,4)$ in $S_4$. Remember when computing products one reads from right to left. So first look at the permutation $(1,3)$ on the RHS of $\circ$, this maps $1$ to $3$ (we can just ignore the permutation $(2,4)$ for the moment since $1$ and $3$ do not belong to it). Now consider the composition $(1,3){\circ}(1,3)$. Well the permutation $(1,3)$ on the RHS of $\circ$ maps $1$ to $3$, then the permutation $(1,3)$ on the LHS of $\circ$ maps $3$ back to $1$. Doing likewise with $3$ we see the mapping $3\mapsto1\mapsto3$, and so $(1,3){\circ}(1,3)=(1)(3)$, and $1$ and $3$ are fixed.
To work out $(1,3)(2,4){\circ}(1,3)(2,4)$ simply trace where each element on the RHS of $\circ$ is mapped to on the LHS of $\circ$:
$1\mapsto3\mapsto1$
so $1$ is fixed.
$3\mapsto1\mapsto3$
so $3$ is fixed.
$2\mapsto4\mapsto2$
so $2$ is fixed.
$4\mapsto2\mapsto4$
so $4$ is fixed.
Hence $(1,3)(2,4){\circ}(1,3)(2,4)=(1)(2)(3)(4)$, which is the identity permutation, and so the element $(1,3)(2,4)$ is self-inverse. In fact doing this with the other two nonidentity permutations reveals they are also self-inverse, and we can write $\sigma_i\circ\sigma_i^{-1}=1$, for $i=2,3,4$. Note to get the cycle decomposition of $\sigma^{-1}$ just write the numbers in each cycle back-to-front, e.g., $\sigma_2=(1, 2)(3,4)$, and $\sigma_2^{-1}=(2, 1)(4,3)$, this makes it clearer where things end up. You have already worked out $\sigma_2\circ\sigma_3=\sigma_3\circ\sigma_2=\sigma_4$: Check:
$\sigma_2\circ\sigma_4=\sigma_4\circ\sigma_2=\sigma_3$
and
$\sigma_3\circ\sigma_4=\sigma_4\circ\sigma_3=\sigma_2$.
This gives you a subgroup of $S_4$ isomorphic to the Klein $4$-group.
$$
\begin{array}{c|ccc}
\circ    & \sigma_1 & \sigma_2 & \sigma_3 & \sigma_4 \\ \hline
\sigma_1 & \sigma_1 & \sigma_2 & \sigma_3 & \sigma_4  \\
\sigma_2 & \sigma_2 & \sigma_1 & \sigma_4 & \sigma_3 \\
\sigma_3 & \sigma_3 & \sigma_4 & \sigma_1 & \sigma_2\\
\sigma_4 & \sigma_4 & \sigma_3 & \sigma_2 & \sigma_1  \\
\end{array}
$$
