Number of subgroups of $S_4$ isomorphic to $K_4$

Question

I was trying to find the number of subgroups in $S_4$ which are isomorphic to the Klein's four group $K_4$.

I know for doing this, I will have to find the subgroups of the type {$e, a, b, ab$} in $S_4$ where $e$ is identity of $S_4$, $a$ and $b$ are in $S_4$ and of order $2$.

Now the elements in $S_4$ of order $2$ will either be $2$-cycles precisely being $\left((12), (13), (23), (24), (14), (34)\right)$ or of type $(ab)(cd)$ which are precisely $(12)(34), (13)(24), (14)(23)$.

Now I know that disjoint cycles commute and I also know that In a group $G$, $a,b$ $\in$ $G$ if $ab$ = $ba$ and $<a>$ $\cap$ $<b>$ = {$e$} then order of $ab$ is lcm(O($a$), O($b$)).

Using this fact, I have found $3$ subgroups of $S_4$ isomorphic to $K_4$ which are

{$e, (12), (34), (12)(34)$}

{$e, (13), (24), (13)(24)$}

{$e, (23), (14), (23)(14)$}.

Now how to find the remaining subgroups isomorphic to $K_4$?

If I look at each of those $9$ elements of order $2$ then the whole Commuting process is becoming really lengthy.

Answer 1

The elements of order $2$ of $S_4$ are all those with cyclic structure $(1,1,2)$ or $(2,2)$, namely:

\begin{alignat}{1} &\sigma_1=(12) &\space\space\space\space\space \sigma_2=(13) &\space\space\space\space\space \sigma_3=(14) \\ &\sigma_4=(23) &\space\space\space\space\space\sigma_5=(24) &\space\space\space\space\space \sigma_6=(34) \\ &\sigma_7=(12)(34)=\sigma_1\sigma_6 &\space\space\space\space\space\sigma_8=(13)(24)=\sigma_2\sigma_5 &\space\space\space\space\space \sigma_9=(14)(23)=\sigma_3\sigma_4 \\ \tag 1 \end{alignat}

A subgroup $K\le S_4$ is isomorphic to Klein's $4$-group if and only if it is made of:

1. the unit $\iota=()$;
2. any pair of commuting elements $\sigma_i, \sigma_j$ from $(1)$;
3. the element $\sigma_i\sigma_j=\sigma_j\sigma_i$ (in fact: $\sigma_i\sigma_j\in K$ by closure, and $\sigma_i\sigma_j\ne \iota,\sigma_i,\sigma_j$),

namely if and only if $K$ is of the form:

$$K_{ij}:=\{\iota,\sigma_i,\sigma_j,\sigma_i\sigma_j\mid \sigma_i\sigma_j=\sigma_j\sigma_i, \space1\le i<j\le 9\} \tag 2$$

So, the point is to single out of $(1)$ all the pairs of commuting elements (with $i<j$). The inventory brings to:

• $$(\sigma_1,\sigma_6), \space (\sigma_2,\sigma_5), \space (\sigma_3,\sigma_4) \tag 3$$

because their elements have disjoint support; then:

• \begin{alignat}{1} &(\sigma_1,\sigma_7), \space (\sigma_6,\sigma_7), \space (\sigma_2,\sigma_8), \space (\sigma_5,\sigma_8), \space (\sigma_3,\sigma_9), \space (\sigma_4,\sigma_9) \\ \tag 4 \end{alignat}

because of $(1)$ and the previous point; finally:

• $$(\sigma_7,\sigma_8), \space (\sigma_7,\sigma_9), \space (\sigma_8,\sigma_9) \tag 5$$

because $\space\sigma_7\sigma_8=\sigma_9, \space\space\sigma_7\sigma_9=\sigma_8, \space\space\sigma_8\sigma_9=\sigma_7$.

Now, if we denote $\sigma_k:=\sigma_i\sigma_j \in K_{ij}$, then we have: $\sigma_k\sigma_i=\sigma_j$ and $\sigma_k\sigma_j=\sigma_i$. So:

\begin{alignat}{1} &\text{if} \space k<i<j, \space\text{then}\space K_{ij}=K_{ki}=K_{kj} \\ &\text{if} \space i<k<j, \space\text{then}\space K_{ij}=K_{ik}=K_{kj} \\ &\text{if} \space i<j<k, \space\text{then}\space K_{ij}=K_{ik}=K_{jk} \\ \tag 6 \end{alignat}

Therefore, by $(3)$ to $(6)$, the number of (distinct) subgroups of $S_4$ isomorphic to Klein's $4$-group is:

\begin{alignat}{1} n_K &= \frac{1}{3}\cdot 12 = 4 \\ \tag 7 \end{alignat}

Explicitly, according to the labelling $(1)$:

\begin{alignat}{1} K_{16} &= \{\iota,\sigma_1,\sigma_6,\sigma_7\}\space (=K_{17}=K_{67}) \\ K_{25} &= \{\iota,\sigma_2,\sigma_5,\sigma_8\}\space (=K_{28}=K_{58}) \\ K_{34} &= \{\iota,\sigma_3,\sigma_4,\sigma_9\}\space (=K_{39}=K_{49}) \\ K_{78} &= \{\iota,\sigma_7,\sigma_8,\sigma_9\}\space (=K_{79}=K_{89}) \\ \tag 8 \end{alignat}