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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.

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    $\begingroup$ 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. $\endgroup$ Nov 11, 2022 at 3:53
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    $\begingroup$ I think the above argument actually works for all $SL_n(\Bbb{F}_q)$. All $n>1$ and $q$. $\endgroup$ Nov 11, 2022 at 3:59
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    $\begingroup$ 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 $Z\big(SL_n(\mathbb F_q)\big)=\{cI_n: c^n=1\}$. $\endgroup$
    – user1551
    Nov 11, 2022 at 5:05

2 Answers 2

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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.

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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: $$1+1+6+4+4+4+4$$.

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