$(C_2 )^3$ is not a subgroup of $S_4$

Question

Prove $(C_2)^3$ is not a subgroup of $S_4$. (Using group actions.)

I could think of a permutation argument that $(C_2)^3$ is not a subgroup of $S_4$. But I would like to argue it by considering group actions.

I could prove in general, by the left action of $G$ on to $\Omega$ the set of all subsets of $G$ of size $p^k$ there is a subgroup of size $p^k$, where $k$ is the largest power of $p$ that divides $G$. Hence, there exists a subgroup of order $8$ (which in $S_4$, one obvious case is $D_8$).

How do I then prove that $(C_2)^3$ is not a subgroup of $S_4$ using group actions? I was thinking of using the fact that we have $D_8$...

Thank you!

EDIT 2: I have attempted according to comments but is still stuck.

Let $\Omega$ be the set of all subgroups of $G$ of order $8$. Then consider the left action by conjugation of $G$ on $\Omega$.

So we have $H \mapsto gHg^{-1}$ when acted by an element $g \in G$. Now since the orbits partition the set, we have $$ |\Omega| = \sum_{i > \in I } |Orb(S_i)| = \sum_{i \in I} \frac{|G|}{|Stab(S_i)|}$$ Where $I$ is index set for a family of orbits which paritions $\Omega$.

I don't know what I can deduce further from here...I do not know how to use the fact that $\Omega$ is a set of subgroup of order $8$ nor I have a $D_8$.

Answer 1

You are almost there! Let $S$ be a subgroup of $S_4$ of order $8$. The orbit of $S$ under the conjugation of $S_4$ has size $24/|Stab(S)|$. Now note that the stabilizer of $S$ for this action is $N(S)$, the normalizer of $S$ in $S_4$ (i.e., the biggest subgroup of $S_4$ in which $S$ is normal). In particular $N(S)$ has size $8$ or $24$ since it contains $S$, and $|Orb(S)|$ is prime to $2$.

Now, let $T$ be any other subgroup of $S_4$ of size $8$ act on the orbit of $S$, again by conjugation. The number of fixed points for this action has to be equal to $|Orb(S)|$ modulo 2 (this is another application of the orbit-stabilizer formula). As stated above, $|Orb(S)|=1\bmod 2$, so there exists a fixed point for this action, say $U$. Being a fixed point for the action of $T$ means that $T\subseteq N(U)$. You probably have seen already that for $p$-groups this implies that $T\subseteq U$. Since $T$ and $U$ have the same cardinality, it follows that $T=U$. It follows that $T$ is in the orbit of $S$ under the conjugation of $S_4$, and since $T$ was arbitrary this means that this action is transitive.

It would probably be a good exercise for you to generalize this argument, for an arbitrary finite group $G$:

• Prove that if $U$ is a $p$-Sylow subgroup of $G$ and $T\subseteq N_G(U)$ is a $p$-subgroup, then $T\subseteq U$.
• Prove that if a $p$-group $T$ acts on a set $X$, the number of fixed point of the action is equal to $X$ modulo $p$.
• Prove that the cardinal of $Orb(S)$, for $S$ being a $p$-Sylow subgroup of $G$, is prime to $p$.
• Finally, prove that all the $p$-Sylow subgroups of $G$ are conjugate (and are in particular isomorphic).

Answer 2

Here is another answer using the standard action of $S_4$ on the set $\{1,2,3,4\}$. We use the following lemma (proof below)

Lemma: If $(C_2)^3$ acts on a set $X$ with four elements, then there is at least one nontrivial element of $(C_2)^3$ which acts trivially.

If $(C_2)^3$ were a subgroup of $S_4$, then $(C_2)^3$ would act on $\{1,2,3,4\}$ by restriction of the action of $S_4$. By the lemma, at least one nontrivial element of $(C_2)^3$ would act trivially. But there is no element of $S_4 - \{ 1 \}$ which acts trivially on $\{1,2,3,4\}$, and this is a contradiction.

Proof of the lemma: Note that $X$ has to decompose into a disjoint union of orbits and each orbit is isomorphic as a $(C_2)^3$-set to $(C_2)^3/H$ for some subgroup $H$ of $(C_2)^3$. In particular, the cardinality of each orbit has to divide $8$. So we have four cases.

1) $X$ is the disjoint union of four orbits of cardinality $1$. In this case, the action of $(C_2)^3$ is trivial.

2) $X$ is the disjoint union of two orbits of cardinality $1$ and one orbit of cardinality $2$, that is $X \cong (C_2)^3/(C_2)^3 \coprod (C_2)^3/(C_2)^3 \coprod (C_2)^3/H$ for some subgroup $H$ of order $4$. In this case, any nontrivial element of $H$ acts trivially.

3) $X$ is the disjoint union of two orbits of cardinality $2$, that is $X \cong (C_2)^3/H \coprod (C_2)^3/K$ where $H$ and $K$ are both subgroups of order $4$. The union of $H$ and $K$ has seven elements, so the subgroup generated by $H$ and $K$ is $(C_2)^3$. The remaining element in $(C_2)^3$ has to be of the form $hk$ with $h \in H$ and $k \in K$. Now consider $h' \neq h,1$ in $H$ and $k' \neq k,1$ in $K$. This product has to be equal to $hk$, to an element of $H$ or to an element of $K$. In any case, this implies that there is a nontrivial element in $H \cap K$. This element acts trivially on $X$.

4) $X$ is an orbit of cardinality $4$, that is, $X \cong (C_2)^3/H$, where $H$ is a subgroup of order $2$. Then the nontrivial element of $H$ acts trivially on $X$.