Some Methods for Proving That $A_4$ Contains No Subgroup of Order $6$

Definition 1. Let $\sigma\in S_n$, and write $\sigma$ as a product of disjoint cycles. We say that the form of $\sigma$ is
$$
1^{\lambda_1}2^{\lambda_2}\cdots n^{\lambda_n}
$$
if $\sigma$ has exactly $\lambda_r$ cycles of length $r$ for each $1\le r\le n$.

Example 1. The form of the permutation
$$
\sigma=(1\ 2\ 3)(4\ 5)
$$
in $S_7$ is
$$
1^22^13^14^05^06^07^0=1^22^13^1.
$$

Definition 2. Let $\alpha,\beta\in S_n$. We say that $\alpha$ and $\beta$ are conjugate in $S_n$ if there exists $\sigma\in S_n$ such that
$$
\sigma^{-1}\alpha\sigma=\beta.
$$

Proposition 1. Let $\sigma,\sigma’\in S_n$. Then $\sigma$ and $\sigma’$ are conjugate in $S_n$ if and only if they have the same form.

Proof : First suppose that $\sigma$ and $\sigma’$ are conjugate in $S_n$. Then there exists $\tau\in S_n$ such that
$$
\tau^{-1}\sigma\tau=\sigma’.
$$
Write $\sigma$ as a product of disjoint cycles:
$$
\sigma=(a\ b\ \cdots\ c)\cdots(\alpha\ \beta\ \cdots\ \gamma).
$$
Then
$$
\sigma’=\tau^{-1}\sigma\tau=((a)\tau\ (b)\tau\ \cdots\ (c)\tau)\cdots((\alpha)\tau\ (\beta)\tau\ \cdots\ (\gamma)\tau).
$$
Hence $\sigma$ and $\sigma’$ have the same form.

Conversely, suppose that $\sigma$ and $\sigma’$ have the same form, say
$$
\sigma=(a\ b\ \cdots\ c)\cdots(\alpha\ \beta\ \cdots\ \gamma),
$$
$$
\sigma’=(a’\ b’\ \cdots\ c’)\cdots(\alpha’\ \beta’\ \cdots\ \gamma’).
$$
Define
$$
\tau=
\begin{pmatrix}
a&b&\cdots&c&\cdots&\alpha&\beta&\cdots&\gamma
\\\
a’&b’&\cdots&c’&\cdots&\alpha’&\beta’&\cdots&\gamma’
\end{pmatrix}.
$$
Then one checks directly that
$$
\tau^{-1}\sigma\tau=\sigma’.
$$
Therefore $\sigma$ and $\sigma’$ are conjugate in $S_n$.

Corollary 1. A subgroup $H$ of $S_n$ is normal if and only if, whenever $\alpha\in H$, every permutation having the same form as $\alpha$ also belongs to $H$.

More precisely, the normal subgroups of $S_n$ are exactly the subgroups that are unions of conjugacy classes, that is, unions of sets of permutations having the same form.

Proposition 2. If $G$ is a group of even order, then $G$ contains an element of order $2$.

Proof : Suppose the contrary. Then for every $g\in G\setminus\{e\}$, we have
$$
g\neq g^{-1}.
$$
Therefore the nonidentity elements of $G$ can be partitioned into pairs of the form $\{g,g^{-1}\}$. Hence
$$
G=\{e\}\cup \bigcup_{g\in G\setminus\{e\}}\{g,g^{-1}\}.
$$
This shows that $|G|$ is odd, contradicting the assumption that $|G|$ is even. Therefore $G$ contains an element of order $2$.

Method 1. $A_4$ contains no subgroup of order $6$.

Proof : Suppose the contrary, and let $H$ be a subgroup of $A_4$ of order $6$. Then $H$ has index $2$ in $A_4$, so by Lagrange’s theorem it is normal in $A_4$.

Since $|H|=6$ is even, Proposition 2 implies that $H$ contains an element of order $2$.

Now
$$
A_4=\{(1),(123),(132),(124),(142),(134),(143),(234),(243),(12)(34),(13)(24),(14)(23)\}.
$$
The elements of order $2$ in $A_4$ are
$$
(12)(34),\quad (13)(24),\quad (14)(23).
$$
Suppose that
$$
(12)(34)\in H.
$$
Since $H$ is normal in $A_4$, Corollary 1 implies that every element of the same form also lies in $H$. Hence
$$
(13)(24),(14)(23)\in H.
$$
Of course,
$$
(1)\in H.
$$
Let
$$
V_4=\{(1),(12)(34),(13)(24),(14)(23)\}.
$$
Then $V_4$ is a subgroup of $H$. But
$$
|V_4|=4,
$$
so by Lagrange’s theorem we must have
$$
4\mid |H|=6,
$$
which is impossible. This contradiction shows that $A_4$ contains no subgroup of order $6$.

Proposition 3. Let $H$ be a normal subgroup of a group $G$ with index $n$. Then for every $g\in G$, we have
$$
g^n\in H.
$$

Proof : Consider the quotient group $G/H$. Since
$$
|G/H|=[G:H]=n,
$$
every element of $G/H$ has order dividing $n$. In particular, for any $g\in G$, the coset $gH\in G/H$ satisfies
$$
(gH)^n=H.
$$
Therefore
$$
g^nH=H,
$$
which implies
$$
g^n\in H.
$$

Method 2. $A_4$ contains no subgroup of order $6$.

Proof : Suppose the contrary, and let $H$ be a subgroup of $A_4$ of order $6$. Then $H$ has index $2$ in $A_4$, and hence is normal in $A_4$.

By Proposition 3, for every $a\in A_4$, we must have
$$
a^2\in H.
$$
Now
$$
A_4=\{(1),(123),(132),(124),(142),(134),(143),(234),(243),(12)(34),(13)(24),(14)(23)\},
$$
and a direct computation gives
$$
\{a^2\mid a\in A_4\}=\{(1),(123),(132),(124),(142),(134),(143),(234),(243)\}.
$$
Thus
$$
\{a^2\mid a\in A_4\}\subseteq H.
$$
But the set on the left has $9$ elements, contradicting the fact that $|H|=6$. Therefore $A_4$ contains no subgroup of order $6$.

Method 3. $A_4$ contains no subgroup of order $6$.

Proof : Suppose the contrary, and let $H$ be a subgroup of $A_4$ of order $6$. Then $H$ has index $2$ in $A_4$, and hence is normal in $A_4$.

Write
$$
A_4=H_0\cup H_1\cup H_2\cup H_3\cup H_4\cup H_5,
$$
where
$$
H_0=\{(1)\},
$$
$$
H_1=\{(123),(132)\},
$$
$$
H_2=\{(124),(142)\},
$$
$$
H_3=\{(134),(143)\},
$$
$$
H_4=\{(234),(243)\},
$$
$$
H_5=\{(12)(34),(13)(24),(14)(23)\}.
$$
Since $H$ is normal in $A_4$, Corollary 1 implies that $H$ must be a union of subsets consisting of elements of the same form. Since $|H|=6$ and $(1)\in H$, the only possibility is
$$
H=H_0\cup H_i\cup H_5
$$
for some $1\le i\le 4$.

However, the elements in $H_1,H_2,H_3,H_4$ are all $3$-cycles, and hence they are conjugate in $S_4$. Since these are all even permutations, they are also conjugate in $A_4$ in the sense relevant here for normality considerations inside $A_4$. Therefore, if one of the sets $H_i$ is contained in $H$, then in fact all of them must be contained in $H$. It follows that
$$
H_1\cup H_2\cup H_3\cup H_4\subseteq H.
$$
Together with $H_0\subseteq H$, this yields
$$
A_4=H,
$$
contradicting $|H|=6$. Therefore $A_4$ contains no subgroup of order $6$.

The cover image of this article was taken in Irkutsk Oblast, Russia.