Is $PGL_2(q)$ isomorphic to $SL_2(q)$ Let $F_q$ denote the field of order $q$. 
Define:


*

*$GL_2(q)$ to be the group of invertible $2$ by $2$ matrices over $F_q$.

*$SL_2(q)$ to be its subgroup consisiting of invertible $2$ by $2$ matrices with determinant $1$. 

*$PGL_2(q)=GL_2(q)/D_1$ where $D_1=\{\pmatrix{a& 0\\0& a}:a\in   
   F_q^{\times}\}$.

*$D_2=\{\pmatrix{a& 0\\0& a}:a=\pm1\}$.
It can be shown that $|SL_2(q)|=|PGL_2(q)|=n$ for $n=q(q^2-1)$.
Now I wish to see whether $PGL_2(q)$ is isomorphic to $SL_2(q)$. For $q=2$ it is easy to see that this is true. How do I prove (or disprove) it for $q>2$. I want to use the first isomorphism theorem but I cannot think of any surjective homomorphism between $GL_2(q)$ and $SL_2(q)$.
 A: There are natural maps $\mathrm{SL}_2(q) \hookrightarrow \mathrm{GL}_2(q) \twoheadrightarrow \mathrm{PGL}_2(q)$; composing these gives a map from $\mathrm{SL}_2(q)$ to $\mathrm{PGL}_2(q)$. For $q$ not a power of $2$ this map definitely has a kernel, the subgroup of order $2$ generated by the diagonal matrix $-1$. For $q$ a power of $2$ the kernel is trivial, since in that case $a^2=1$ implies $a=1$ for $a \in F$ a field of char. $2$, so your calculation of the orders shows that over a field of char. $2$ (for $q$ any power of $2$) the groups are isomorphic.
In general, the center of $\mathrm{SL}_2(q)$ consists of scalar matrices $a$ for which $a^2=1$; meanwhile, the group $\mathrm{PGL}_2(q)$ has trivial center: if $g$ commutes with every matrix module scalars, $g h g^{-1}=a$, then $g$ in particular normalizes the torus and hence is either diagonal or anti-diagonal, but a quick calculation shows that $g$ cannot be anti-diagonal and then another that it must itself be scalar.
Added: Andreas just beat me to this, with a much pithier answer.
A: $\operatorname{PGL}_{2}(q)$ is centreless, while $\operatorname{SL}_{2}(q)$ has a centre $D_{2}$ of order $2$ for $q$ odd.
As a reference, see the Wikipedia article about the projective special linear group, or Steve's post ;-)
A: I can answer the question like this:
Looking at the Fratini subgroups using GAP shows that the Fratini subgroup of $SL_2(3)$ is of order $2$, where the one of $PGL_2(3)$ is trivial.
Assuming the algorithms are correct (which I guess, since they have been testet much), the groups are not isomorphic. You might want to prove this by hand.
So you get a counterexample for your claim.
