# Show that the group is not simple

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.

Let $N\leq G$ with $[G:N]\leq 4$.

We have that $|G|=x\cdot y, \ 1<x,y<|G|$.

Could yout give me a hint how we could conclude that $G$ is not simple?

Do we maybe use Sylow subgroups?



EDIT:

In my notes I found the following proposition:

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

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

Suppose that $|G|\mid m!$. Then $G$ is simple. Or isn't the above proposition an off statement?

• If $m=1$, then $G=H$ and since $|G|\mid 1\Rightarrow |G|=1$. Therefore, $G=H=1$. In that case the group is not simple.
• If $m=2$, the possible values for $|G|$ are $1$ and $2$. The case $|G|=1$ is rejected. 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 $1, 2,3$ are rejected, since $G$ has not a prime order.

If $|G|=6$ then $G$ is isomorphic to $\mathbb{Z}_6$ or to $S_3$. Both of them are not simple since $\langle 2\rangle$ is normal in $\mathbb{Z}_6$ and $A_3$ is normal in $S_3$. Therefore, $G$ is not simple, a contradiction. Is this correct?

• 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 $1,2,3,6$ are rejected.

If $|G|=4$ we have that $G$ is abelian. We have that every subgroup of an abelian group is normal. Therefore, $G$ is not simple, a contradiction.

What can we say about the cases $|G|=8$, $|G|=12$ and $|G|=24$ ?

• If p is the minimum prime that divides to order of G, and you know that any subgroup H have index p, H must be normal – Martín Vacas Vignolo Apr 2 '16 at 3:05
• Do we suppose that there are such $H$ ? Why would it imply that $H$ is normal? @vvnitram – Mary Star Apr 2 '16 at 3:10
• Just a nitpick, you would want $N$ to be a proper subgroup. – Arun Kumar Apr 2 '16 at 3:51
• @MaryStar If we let $N=G$, then $[G:N]=1$ and this condition is satisfied by all groups, then the proposition cannot be true as there are simple groups of non-prime order. – Arun Kumar Apr 3 '16 at 13:06
• If you look through it you will see that matt samuel's answer basically uses the proposition you have stated. What you have tried will also boil down to showing that all simple groups of size less than 24 are of prime order. – Arun Kumar Apr 4 '16 at 12:10

• There is a mistake in your argument (from EDIT), you assumed $|G| | m!$ then $G$ is simple. Here, to get your argument fixed, you should first suppose by contradiction (original question) that $G$ is simple, then use the counterpositive of your Proposition. That is, knowing that $H \leq G$ and $[G:H] = m$ and assuming $G$ simple, you have $|G| | m!$. And work on each case as you did so far to get a contradiction.

• Yes, $\mathbb Z_6$, is abelian so every subgroup is normal in $\mathbb Z_6$ and $S_3$, are not simple because of what you have pointed out.

• Case $|G| = 8$ we have that $G$ is not simple because it is a $2$-group and has centre $Z(G)$ non-trivial, which is normal in $G$.

• Case $|G| = 12 = 2^2 \cdot 3$. Looking at the number Sylow $3$-subgroups of $G$ we have that $n_3 = 1$ or $4$ (). If it is 1, then it is normal ($2^{nd}$ - Sylow Theorem) so $G$ is not simple. We assume then $n_3 = 4$. Then as each of those subgroups has order $3$, which is prime, thus they must be disjoint, therefore there $8 \cdot 2 = 8$ elements of order $3$. Which leaves us with $4$ elements left, and they must be all the elements of the Sylow $2$-subgroup (again $2^{nd}$ - Sylow Theorem). We conclude that latter must be unique and therefore normal.

• Case $|G| = 24 = 2^3\cdot 3$ similarly we have that $$\begin{cases}n_3 \equiv 1 \mod 3 , n_3 | 8 \\ n_8 \equiv 1\mod 8 , n_8 | 3\end{cases} \implies \begin{cases}n_3 = 1 \,\,\ \text{or} \,\,\ n _3 =4 \\ n_8 = 1\,\,\, \text{or} \,\,\, n_8 = 3\end{cases}$$

Now if either $n_3 = 1$ or $n_8 =1$ we are done. Now if $n_8 = 3$ then the action of G by conjugation on its subgroups of order $8$ determines $\varphi : G \to S_3$. By the $2^{nd}$-Sylow Theorem the image of $G$ acts in $S_3$ transitively on the set $3$ subgroups of order $8$. Then the image is not trivial and $\ker \varphi \neq G$. As $|S_3|= 6 < |G| = 24$ we have that $\ker \phi \neq \{e\}$. We may conclude that $G$ is not simple.

• Ok... Thanks a lot!! :-) – Mary Star Apr 11 '16 at 18:02
Suppose $G$ is simple and let $1<m\leq 4$ be the index of a subgroup $H$. Then the action of $G$ on the set of left cosets of $H$ yields a nontrivial homomorphism $G\to S_m$, and since $G$ is simple the homomorphism is injective, so $|G|\le 24$. The only simple groups with order this small are of prime order.
• Can it not be that $m=1$ ? – Mary Star Apr 2 '16 at 9:19
• I found inmy notes the following proposition: $$H\leq G, \ [G:H]=m \text{ and } |G|\not\mid m! \text{ then } G \text{ is not simple. }$$ Could we use this one and reject all the possible cases for $m$ ? – Mary Star Apr 2 '16 at 9:47
• @Mary if $m=1$ the action is trivial so the proof doesn't work. – Matt Samuel Apr 2 '16 at 13:39