I am trying to show that for $q$ odd prime number, the groups $SL_2(q)$ and $PGL_2(q)$ are not isomorphic. On a similar post:Is $PGL_2(q)$ isomorphic to $SL_2(q)$ . Stephan and Andres stated that studying the center is a way to show that they are not isomorphic. The center of $SL_2(q)$ is clear to me. But how to show that $PGL_2(q)$ is centerless?
Any other approach is welcomed.
Thanks
Your intuition about using the orders of elements is a great approach, and your argument regarding the eigenvalues is on the right track. Let's tidy it up a bit:
Given [ A = \begin{pmatrix} 1 & 0 \\ 0 & a \end{pmatrix} ] in ( PGL_2(q) ), where ( a ) is a generator of ( F_q^\times ), the order of ( A ) is ( q-1 ).
For an element ( B ) in ( SL_2(q) ) with order ( q-1 ), we have ( B^{q-1} = I ). Given that the determinant of ( B ) is 1, the eigenvalues of ( B ) are either ( \pm 1 ).
However, the polynomial ( x^{q-1} - 1 ) has ( q-1 ) distinct roots in ( F_q ) since ( F_q ) is a field. Thus, no ( 2 \times 2 ) matrix in ( SL_2(q) ) can satisfy the equation ( B^{q-1} = I ) without having distinct eigenvalues. This is a contradiction, as our hypothetical matrix would need to have both eigenvalues as 1 or both as -1.
Regarding the centers:
The center of ( SL_2(q) ) consists of scalar matrices with entries in ( F_q^\times ). [ PGL_2(q) ] is the quotient of ( SL_2(q) ) by its center, so it "collapses" this center to the identity element, making it centerless.
This approach, combined with your element order observation, provides a solid argument that ( SL_2(q) ) and ( PGL_2(q) ) are not isomorphic. Well done!