Show that $\{1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)\}$ is a subgroup of $S_4$

Question

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?

Answer 1

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.

Answer 2

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. $$ \begin{array}{c|ccc} \circ & \sigma_1 & \sigma_2 & \sigma_3 & \sigma_4 \\ \hline \sigma_1 & \sigma_1 & \sigma_2 & \sigma_3 & \sigma_4 \\ \sigma_2 & \sigma_2 & \sigma_1 & \sigma_4 & \sigma_3 \\ \sigma_3 & \sigma_3 & \sigma_4 & \sigma_1 & \sigma_2\\ \sigma_4 & \sigma_4 & \sigma_3 & \sigma_2 & \sigma_1 \\ \end{array} $$

Answer 3

It's a subgroup of $S_4$ if it's a group (since it's obviously a subset of $S_4$, with the same group operation (composition of permutations).

Now, even though it's probably not what you are supposed to do, here is how you could see this group: take a rectangle whose vertices are labeled 1, 2, 3, 4, and consider the group of transformations that make this rectangle invariant.