$[G:H]=4$ where $H\neq 1$ means that $G$ is not simple.

Question

The question is as follows:

Suppose $G$ is a finite group with a nontrivial subgroup of index $4$. Prove that $G$ is not simple.

I am on the hunt for a non-trivial normal subgroup. This proof followed a proof of Cayley's theorem, so it is likely that is involved. So if $|G|=n$ and $\phi_g: G \to G$ is defined by $\phi_g (x) =g\cdot x.$ Then $\phi_g \in S_G$ so that $\psi: G\to S_G$ defined by $\psi (g) = \phi_g$ is an embedding (pretty much an outline of Cayleys thm). This helps us in some way, but I am unsure how. My mental summary of the sitch says that I have a subgroup of $S_n$ that is the "same shape" as $G$, call it Im$\psi$. But Im$\psi$ also has a non-trivial subgroup of index $4$, call it $H$, and this subgroup is a group of permutations of the set of $n$ elements. Overall, we have that $[\text{Im}\psi :H]=4$ and $|\text{Im}\psi|$ divides $n!$ All of this (and perhaps something else I am missing), tells me that Im$\psi$ has a non-trivial normal subgroup.

In general, I feel that I have to exhibit a map from Im$\psi$ to some other group that I am unsure of with nontrivial kernel.

In hindsight, perhaps Cayleys theorem was misleading: if $[G:H]=4$ and $A=\left\{xH: x\in G \right\}$ can I just talk about the map $$\rho: G\to A$$ defined by $$\rho (g) = gH$$ and hope to goodness gracious that it is a homormorphism? Any help is appreciated.

Answer 1

Hints.

• As said in the comments, use the action of $G$ on $G/H$ to define a group morphism $\psi:G\to S_4$.

Justify that $\psi$ is non trivial.

• If $G$ is simple, deduce that $G$ is isomorphic to a subgroup $G'$ of $S_4$ , and that the order of this subgroup is a multiple of $4$.

• Justify that $\vert G'\vert =4,8,12$.

• Using a property of $p$-groups ($p$ prime), rule out $4$ and $8$ (you may also want to identify subgroups of order $4$ and $8$ of $S_4$ if you do not know anything about $p$-groups)

• If $G'$ has order $12$, what is $G'$ ?

• Conclude that $G$ is not simple.

Answer 2

Proof: Suppose $[G:H]=4$ and $|G|>4$ and let $$A=\left\{xH: x\in G \right\}$$ Notice that each $g\in G$ induces the action $g\cdot (xH) =(gx)H$ on $A$. In particular, $\sigma_g : A\to A$ defined by $\sigma_g (xH) = g\cdot (xH)$ is a permutation. Thus $\phi: G\to S_A \cong S_4$ defined by $\phi (g) = \sigma_g$ is a group homomorphism. Edit: Further, if $\phi$ is not injective then we would have that $1 <\text{ker}\phi \triangleleft G$ so that $G$ is not simple. If $\phi$ is injective, in the spirit of Cayley, $$G\cong \text{Im} \phi \leqslant S_A \cong S_4. $$ Let Im$\phi =G'$. Then $|G'|$ divides $4! =24$. But since $G$ is finite, so is $G'$. Further, we know there is some $H'$ such that $[G':H']=4$ Hence, $|H'|=|G'|/4$. This means that $$ |H'| \in \left\{2,3\right\} $$ If $|H'|=2$ then $|G'|=8=2^3$ which we know not to be simple as it is a $p$-group with $n>1$. If $|H'| =3$, then $|G'|=2^2\cdot 3$. Now $n_3 =1\lor 4$. If $n_3=1$ then we found our nontrivial normal subgroup and we are donzo. If $n_3=4$ then there are $8$ nonidentity elements, leaving just enough room for our unique, therefore normal, Sylow $2$-subgroup of order 4. Hence, in either case, $G'$ isn't simple so that neither is $G$. $\nsim$