MHB Show that the group is not simple

[mathmari]

Hey!

I want to show that:

If $G$ contains a subgroup with index at most $4$ and $G$ has not a prime order, then $G$ is not a simple group. In my notes I found the following proposition:

$$H\leq G, \ [G]=m \text{ and } |G|\not\mid m! \text{ then } G \text{ is not simple. }$$

We have that $H\leq G$ and $[G]=m, \ 1\leq m\leq 4$.

Suppose that $|G|\mid m!$.

• If $m=1$, then $G=H$, or not? Can that be? (Wondering)
• If $m=2$, the possible values for $|G|$ are $1$ and $2$. It cannot be that $|G|=2$, since $G$ has not a prime order.
• If $m=3$, then $|G|\mid 3!=6$, then the possible values for $|G|$ are $1,2,3,6$.
  The cases $2,3$ are rejected, since $G$ has not a prime order.
  Can the order of $G$ be $6$ ? (Wondering)
• If $m=4$, then $|G|\mid 4!=24$, then the possible values for $|G|$ are $1,2,3,4,6,8,12,24$.
  The cases $2,3$ are rejected, $G$ has not a prime order.
  Are the other cases possible? (Wondering)

Or do we not use this proposition? (Wondering)

 

[Deveno]

The action of $G$ by left-mulitplication on the left coset space of $G/N$ is given by:

$g\cdot xN = (gx)N$.

This gives a homomorphism $\phi: G \to S_{[G:N]}$.

I will do the case $[G:N] = 4$, which should tell you how to proceed for $n = 3$ (we don't typically allow the case $n = 1$, since the trivial group is usually not counted as simple, much like 1 is not considered prime, and if $[G:N] = 2$, then $N \lhd G$).

If we assume (in order to find a contradiction) $G$ is of non-prime order, but simple, then $\text{ker }\phi = \{e_G\}$

(We know that $\text{ker }\phi \neq G$ because if $xN \neq N$, then $x$ sends $N \to xN$, so $x \not\in \text{ker }\phi$).

Therefore, $G$ is isomorphic to a subgroup of $S_4$. So, in order to show a contradiction, we need to show no subgroup of $S_4$ of non-prime order is simple.

The possible subgroups we need to investigate have orders: $4,6,8,12,24$.

There is only one subgroup of order 24, $S_4$ itself, which is not simple since $A_4 \lhd S_4$. So $|G| \neq 24$.

The only subgroup of order 12 of $S_4$ is $A_4$, which is not simple since its Sylow 2-subgroup $V$ is normal.

There are 3 subgroups of $S_4$ of order 8, each is isomorphic to $D_4$, the dihedral group of order 8, and $D_4$ is not simple (the subgroup of an isomorph of $D_4$ in $S_4$ generated by a $4$-cycle, is of index 2, thus normal).

This rules out $|G| = 8,12$.

$S_4$ has no elements of order 6, and so no subgroup isomorphic to $\Bbb Z_6$. It does have subgroups of order 6 isomorphic to $S_3$ (the subgroups which fix a particular letter), but $S_3$ is not simple, since it has a subgroup of index 2.

So $|G| \neq 6$.

That only leaves $|G| = 4$, but any group of order 4 is abelian, and being of non-prime order, any non-trivial subgroup of prime order (in this case of order 2) is normal.

So $G$ cannot be simple, QED.

 

[mathmari]

The action of $G$ by left-mulitplication on the left coset space of $G/N$ is given by:

$g\cdot xN = (gx)N$.

This gives a homomorphism $\phi: G \to S_{[G:N]}$.

Why do we look at the action of $G$ on the left coset space of $G/N$ ? (Wondering)

 

[Deveno]

To form a group homomorphism from $G \to \text{Sym}(G/N)$, when there are $4$ cosets, then:

$\text{Sym}(G/N) \cong S_4$.

 

[mathmari]

To form a group homomorphism from $G \to \text{Sym}(G/N)$, when there are $4$ cosets, then:

$\text{Sym}(G/N) \cong S_4$.

So, do we look at the left cosets, because we have an information for $[G:N]=|G/N|$, where $G/N$ is defined as $\{gN\mid g\in G\}$, which is the set of left cosets? (Wondering)

 

[Deveno]

So, do we look at the left cosets, because we have an information for $[G:N]=|G/N|$, where $G/N$ is defined as $\{gN\mid g\in G\}$, which is the set of left cosets? (Wondering)

Yes, because that gives us a possible left action.

We could look at the right cosets, if we wanted to, but we'd have to use a right action:

$Nx\cdot g = N(xg)$ (the index is the same, either way).