# Thread: quick question on SL2(F3)

1. ## quick question on SL2(F3)

i know $SL_2(F_3)$ is solvable.

it's sylow 2-subgroup is isomorphic to $Q_8$, which pretty much wraps it up (this group contains the center, to finish off the composition series).

is there an easier way to demonstrate this group is solvable, without using the above isomorphism? can we show, for example, the sylow 2-subgroup is normal, without ever calculating any matrix products, and can we prove the center is contained in this group?

2. ## Re: quick question on SL2(F3)

Originally Posted by Deveno
i know $SL_2(F_3)$ is solvable.

it's sylow 2-subgroup is isomorphic to $Q_8$, which pretty much wraps it up (this group contains the center, to finish off the composition series).

is there an easier way to demonstrate this group is solvable, without using the above isomorphism? can we show, for example, the sylow 2-subgroup is normal, without ever calculating any matrix products, and can we prove the center is contained in this group?
i think it's much more rewarding to prove that every group of order $2^m \times 3$ is solvable. this can be done quite easily by induction over $m$ and using this elementary fact that if $H$ is a subgroup of $G$ and $[G:H]=n < \infty,$ then there exists a normal subgroup $N$ of $G$ such that $N \subseteq H$ and $[G:N] \mid n!.$ (you can find the proof of this fact here although i'm quite sure you already know it!)

3. ## Re: quick question on SL2(F3)

I guess it all depends upon what you're assumed to know.

You can trivially check that the Sylow $2$-subgroup $Q$ is normal and $|\text{SL}_2(\mathbb{F}_3)/Q|=3$ which evidently implies that $\text{SL}_2(\mathbb{F}_3)/Q$ is solvable. That said, $Q$ is of order eight, and regardless of $Q$'s isomorphism type we know it's solvable--namely because it is a fourth-week-of-group theory result that the only groups of order $8$ are the three abelian and $D_4,Q$ all of which are solvabe. So, $Q$ is solvable. Since solvability respects extensions we may conclude that $\text{SL}_2(\mathbb{F}_3)$ is solvable.

Of course, the above really shows that any group of order $24$ with normal Sylow $2$-subgroup is solvable, which is consistent with the common theorem that NonCommAlg stated.

4. ## Re: quick question on SL2(F3)

Originally Posted by Drexel28
I guess it all depends upon what you're assumed to know.

You can trivially check that the Sylow $2$-subgroup $Q$ is normal and $|\text{SL}_2(\mathbb{F}_3)/Q|=3$ which evidently implies that $\text{SL}_2(\mathbb{F}_3)/Q$ is solvable. That said, $Q$ is of order eight, and regardless of $Q$'s isomorphism type we know it's solvable--namely because it is a fourth-week-of-group theory result that the only groups of order $8$ are the three abelian and $D_4,Q$ all of which are solvabe. So, $Q$ is solvable. Since solvability respects extensions we may conclude that $\text{SL}_2(\mathbb{F}_3)$ is solvable.

Of course, the above really shows that any group of order $24$ with normal Sylow $2$-subgroup is solvable, which is consistent with the common theorem that NonCommAlg stated.
you don't need to know which groups have order $8$ to show that groups of order $8$ are solvable. actually every $p$-group is solvable because we know that every group of order $p^n$ has a normal subgroup of order $p^{n-1}$ and so we can use induction.

5. ## Re: quick question on SL2(F3)

Originally Posted by NonCommAlg
you don't need to know which groups have order $8$ to show that groups of order $8$ are solvable. actually every $p$-group is solvable because we know that every group of order $p^n$ has a normal subgroup of order $p^{n-1}$ and so we can use induction.
Right, I forgot I was talking to Deveno and was trying to give an easier result haha.

6. ## Re: quick question on SL2(F3)

Originally Posted by NonCommAlg
i think it's much more rewarding to prove that every group of order $2^m \times 3$ is solvable. this can be done quite easily by induction over $m$ and using this elementary fact that if $H$ is a subgroup of $G$ and $[G:H]=n < \infty,$ then there exists a normal subgroup $N$ of $G$ such that $N \subseteq H$ and $[G:N] \mid n!.$ (you can find the proof of this fact here although i'm quite sure you already know it!)
yes, this is what i wanted. thanks!

and without even looking, let's see...left-action on the coset space, right?