$SL(2, \mathbb F_3)$ does not have a subgroup of order $12$

Question

Using the characteristic polynomial I can prove that $SL(2, \mathbb F_3)$ does not has an element of order $12$, but how can I prove that $SL(2, \mathbb F_3)$ does not has a subgroup of order $12$?

Answer 1

Since $G := \text{SL}(2,\mathbb{F}_3)$ has order 24, any subgroup of order $12$ would be normal (because of index 2), and hence $G$ would have an abelian quotient isomorphic to $C_2$. In particular we obtain an epimorphism $G^{ab} \to C_2$, where $G^{ab} := G/G'$.

To get a contradiction, you could proof that $G^{ab}$ does not contain an element of order 2 (indeed, $G^{ab} \cong C_3$).

Answer 2

This is Theorem $2$ in George Mackiw's article :http://www.maa.org/sites/default/files/George_Mackiw20823.pdf.

Answer 3

The sizes of conjugacy classes are $1,1,4,4,4,4,6$ (see http://groupprops.subwiki.org/wiki/Special_linear_group:SL(2,3)).

Now the two elements with conjugacy class of size $1$ are $I, 2I$. These have order $1,2$ respectively.

Any other element lies in a conjugacy class of size $4$ or $6$, so has centralizer group of order $6$ or $4$. In particular an element lies in its own centralizer so the order of any such element must divide $4$ or $6$ so cannot be $12$.

Answer 4

Outline for a proof:

With the characteristic polynomial, you can see that $A^3 = I$ in $G = \operatorname{SL}(2,3)$ if and only if $tr(A) = -1$. Count that there are $8$ elements of order $3$ in $G$.

A subgroup $H$ of order $12$ would have to contain every element of order $3$. Conclude $H \cong A_4$.

On the other hand $H$ must contain $-I$, so $H$ has a subgroup of order $6$. This is a contradiction since $A_4$ has no subgroup of order $6$.