Show that $SL(3,2)$ is simple

Question

The group $SL(3,2)$ has $6$ conjugation classes with cardinalities $1$, $21$, $56$, $42$, $24$ e $24$. Deduce that $SL(3,2)$ is simple.

I guess I should be able to tell the group is simple from the cardinalities of its conjugation classes, but how?

Answer 1

A normal subgroup is closed under conjugation by any element of the group. So if $N$ is a normal subgroup and it contains $x$, an element of a conjugacy class, it also contains every conjugate of $x$, so the entire conjugacy class. Consequently, any normal subgroup is a union of conjugacy classes. The order of a normal subgroup is the sum of the sizes of the conjugacy classes it contains.

The order of a subgroup divides the order of the group. So a normal subgroup's order is the sum of some of the sizes of the conjugacy classes and also divides the order of the large group.

Notice that the conjugacy class of size $1$ is the identity of the group and is included in every subgroup. Also, since it doesn't matter which of the two $24$ element classes we keep, we only track having $0$, $1$, or $2$ of them.

Candidate orders of $N$ (trailing minus signs will be explained after the table): \begin{array}{c|c|c|c|c|c} 1 & 21 & 56 & 42 & 24 & \text{sum} \\ \hline * & & & & & 1 \\ * & * & & & & 22 \\ * & & * & & & 57 \\ * & & & * & & 43 \\ * & & & & 1 & 25 \\ * & & & & 2 & 49 \\ * & * & * & & & 78 \\ * & * & & * & & 64 \\ * & * & & & 1 & 46 \\ * & * & & & 2 & 70 \\ * & & * & * & & 99 - \\ * & & * & & 1 & 101 \\ * & & * & & 2 & 125 - \\ * & & & * & 1 & 67 \\ * & & & * & 2 & 91 \\ * & * & * & * & & 128 - \\ * & * & * & & 1 & 102 - \\ * & * & * & & 2 & 126 - \\ * & * & & * & 1 & 88 \\ * & * & & * & 2 & 112 \\ * & & * & * & 1 & 123 - \\ * & & * & * & 2 & 147 - \\ * & * & * & * & 1 & 142 - \\ * & * & * & * & 2 & 168 \end{array}

And we should notice that none of those sums divide $168$, except for the trivial subgroup and the whole group.

This can be done a little faster. We have no interest in sums in the interval $(168/2, 168) = (84,168)$. So we need not consider any combination of the $56$ and the $42$ or $56 + 24 + [\text{anything bigger than $4$}]$. Combinations rejected by these two observations are marked with trailing minus signs; they eliminate a little less than half the table. With a little more trickiness, we can eliminate a few more, but there is a tradeoff in time spent construcitng and checking for such conditions versus just adding and testing for divisibility.

Answer 2

Let $N \triangleleft SL(3,2)$ be a normal subgroup. Then $|N|$ must divide $|SL(3,2)|=168$. Since $N$ is normal, it is a union of conjugacy classes, so its order is a sum with some of $1, 21, 56, 42, 24, 24$ as summands. Since it contains the neutral element, $1$ must be a summand.

If $|N|$ is odd, it must divide $168/8=21$. The smallest possible value $|N|$ could take is $1$. The second smallest is $1+21=22$. Thus, $|N|=1$.

Now, let $|N|$ be even. Since $56,42,24,24$ are even and $1$ is odd, $21$ must be one of the summands.

Either $N=SL(3,2)$ or $|N| \le 168/2=84$. In the latter case, the even summands must add up to at most $84-21-1=62$. The only possibilities for that are as follows:

$|N|=1+21+24+24=70 \nmid 168$

$|N|=1+21+42=64 \nmid 168$

$|N|=1+21+56=78 \nmid 168$

Since none of these are valid, it follows that $N=SL(3,2)$ or $|N|=1$. Therefore, $SL(3,2)$ is simple.