# An intelligent way to determine the center of the special linear group of order 2 over the field of order 3.

The Problem: Show that the center of $$SL_2(\mathbb{F}_3)$$ is the group of order $$2$$ consisting of $$\pm\mathit{I}$$, where $$\mathit{I}$$ is the identity matrix.

My Question: Obviously one can list all the elements in $$SL_2(\mathbb{F}_3)$$ and compute the center using brute force; but is there an easy way? I tried to make use of The Class Equation: $$|G|=|Z(G)|+\sum_{i=1}^r|G: C_G(g_i)|$$ ($$g_i$$'s are the representatives of the distinct conjugacy classes of $$G$$ not contained in $$Z(G)$$), to no avail.

Any help would be greatly appreciated.

• One way would be to look at the action of $SL_2(\Bbb{F}_3)$ on the set of four 1-dimensional subspaces of $\Bbb{F}_3^2$. The fixed points of this action are the eigenspaces of individual matrices. There are matrices with just one 1-dimensional eigenspace $V$. For $g$ to commute with such a matrix, it needs to have $V$ as an eigenspace as well. So the center needs to have every 1-dimensional subspace as an eigenspace. That is possible only for scalar matrices. Nov 11, 2022 at 3:53
• I think the above argument actually works for all $SL_n(\Bbb{F}_q)$. All $n>1$ and $q$. Nov 11, 2022 at 3:59
• Since $SL_n(\mathbb F_q)$ spans $M_n(\mathbb F_q)$, we have $Z\big(SL_n(\mathbb F_q)\big)\subseteq Z\big(M_n(\mathbb F_q)\big)=\{cI_n: c\in\mathbb F_q\}$. Hence $Z\big(SL_n(\mathbb F_q)\big)=\{cI_n: c^n=1\}$. Nov 11, 2022 at 5:05

Let $$E_{ij}$$ be the matrix with $$1$$ on position $$(i,j)$$ and zeros elsewhere. A matrix $$A\in\mathrm{SL}_n(K)$$ (where $$n\ge 1$$ and $$K$$ is any field) commutes with all $$I_n+E_{ij}$$ where $$i\ne j$$. Hence, $$A$$ commutes with $$E_{ij}$$. This implies that $$A$$ is a scalar matrix.
If you happen to know (and I don't expect you to; I didn't) that the inner automorphism group of $$\mathcal {SL}(2,3)$$, also known as the projective special linear group, $$\mathcal {PSL}(2,3)$$, is isomorphic to $$A_4$$, and hence has order $$12$$... then the center must have order $$2$$.
And of course $$\pm I$$ are in the center.
Another way is to look at the class equation, which contains two $$1$$'s: $$1+1+6+4+4+4+4$$.