Normal subgroup of \({\cal{S}}_4\)

[Grigorios Kostakos]

Prove that the subgroup \[H=\big\{{{\rm{id}},\,({1\,2})\,({3\,4}),\,({1\,3})\,({2\,4}),\,({1\,4})\,({2\,3})}\big\}\] of the symmetric group ${\cal{S}}_{4}$ is normal.

[dmsx]

First of all, $H$ contains the identity permutation, is closed under multiplication and contains every element's inverse, so it is in fact a group.

Obviously, we are not going to prove the result using the definition. Observe that $H$'s elements are exactly all the possible permutations of $S_4$ written as a product of $2$-cycles. From basic group theory, all the conjugates of those elements are of the same form. So, $H$ contains all the $g$-conjugates of its elements, where $g\in S_4$ and this fact proves that $H$ is normal in $S_4$.

This group, usually written $V$ or $K_4$, is the Klein group (or Klein four-group), which is abelian (since it contains $4$ elements), non-cyclic and normal in both $A_4$ and $S_4$. It is usually met when discussing permutation, dihedral and/or Galois groups and possibly elsewhere.

[Grigorios Kostakos]

Although the following solution has the same idea as of dmsx's solution, it's in more detailed way:

We shall need:

$\color{gray}\bullet$ Lemma: Let a $k$-cycle $$\rho=\bigl({\begin{array}{cccccc}\alpha_1&\alpha_2&\cdots&\alpha_{k-1}&\alpha_{k}\end{array}}\bigr)\in{\cal{S}}_{n}\,,\; 1\leqslant k\leqslant n\,,$$ and a permutation $\sigma\in{\cal{S}}_{n}$. Then $$\sigma\,\rho\,\sigma^{-1}=\bigl({\begin{array}{cccccc}\sigma(\alpha_{1})&\sigma(\alpha_{2})&\cdots&\sigma(\alpha_{k-1})&\sigma(\alpha_{k})\end{array}}\bigr)\,.$$
$\color{gray}\bullet$ Let $\sigma,\tau\in{\cal{S}}_{n}$ and $\tau=\rho_1\,\rho_2\ldots\rho_{m}$ is the unique decomposition of $\tau$ into pairwise disjoint cycles $\rho_i\,,\; i=1,2,\ldots m$. Then \begin{align*}
\sigma\,\tau\,\sigma^{-1}&=\sigma\,\rho_1\,\rho_2\ldots\rho_{m}\,\sigma^{-1}\\
&=\underbrace{\sigma\rho_1\,\sigma^{-1}}_{\mu_1}\,\underbrace{\sigma\,\rho_2\,\sigma^{-1}}_{\mu_2}\ldots\underbrace{\sigma\,\rho_{m}\,\sigma^{-1}}_{\mu_m}\\
&=\mu_1\,\mu_2\ldots\mu_{m}\,.
\end{align*}
By the above lemma, for every $i=1,2,\ldots,m$, the cycles $\mu_i$ are of the same order as of $\rho_i$ and, also, the cycles $\mu_i$ are pairwise disjoint because $\tau$ is 1-1 map.

So, we have that for every permutation - except of the identity ${\rm{id}}$ - of subgroup
\[H=\big\{{{\rm{id}},\,({1\,2})\,({3\,4}),\,({1\,3})\,({2\,4}),\,({1\,4})\,({2\,3})}\big\}\] of ${\cal{S}}_{4}$ and for every permutation $\sigma\in{\cal{S}}_{4}$, the permutation $\sigma\,\tau\,\sigma^{-1}$ is a composition of two disjoint $2$-cycles. But all the permutations which are a composition of two disjoint $2$-cycles belong to $H$. Therefore for every $\sigma\in{\cal{S}}_{4}$ we have $$\sigma\,H\,\sigma^{-1}\subseteq H\,; $$ i.e. $H\trianglelefteq{\cal{S}}_{4}$.




Note: Indeed, the $H$ is isomorphic to Klein four-group ${\cal{V}}_{4}$. But, there are three more subgroups in ${\cal{S}}_{4}$, which are isomorphic to ${\cal{V}}_{4}$, but they are not normal subgroups of ${\cal{S}}_{4}$. These subgroups are $\langle({1\,2}),\,({3\,4})\rangle$, $\langle({1\,3}),\,({2\,4})\rangle$ and $\langle({1\,4}),\,({2\,3})\rangle$.

P.S. There are more different solutions of the exercise.

[Grigorios Kostakos]

2nd solution: Observe that the second power of every $4$-cycle of ${\cal{S}}_{4}$ are exactly the non-identity elements of
\[H=\big\{{\rm{Id}},\,({1\,2})\,({3\,4}),\,({1\,3})\,({2\,4}),\,({1\,4})\,({2\,3})\big\}\,.\] The map $f:{\cal{S}}_{4}\longrightarrow{\cal{S}}_{3}$ defined for the cycles of ${\cal{S}}_{4}$ as
$$\left\{\begin{array}{l}
f({\rm{Id}})={\rm{Id}}\,, \\
f\big((a_1\;a_2)\big)=(a_1\;a_2)\,, \\
f\big((a_1\;a_2\,a_3)\big)=(a_1\;a_2\;a_3)\,, \\
f\big((a_1\;a_2\;a_3\;a_4)\big)=(a_1\;a_2)\,,
\end{array}\right.\quad a_i\in\{1,2,3,4\}\,,\; i=1,2,3,4\,,$$
and for every permutation $\sigma\in{\cal{S}}_{4}$ as
$$f(\sigma)=f(\rho_1)\,f(\rho_2)\,\ldots \,f(\rho_{m})\,,$$
where $\sigma=\rho_1\,\rho_2\ldots\rho_{m}$ is the unique decomposition of $\sigma$ into pairwise disjoint cycles $\rho_i\,,\; i=1,2,\ldots m$, is well a defined epimorphism with $\ker{f}=H$. So $H\trianglelefteq{\cal{S}}_{4}$.