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For a field $\mathbb F$ the group $SL(2, \mathbb F)$ is the group of all invertible linear maps on a $2$-dimensional $\mathbb F$-vector space with determinant $1$. If we view them as matrices, its center is $Z := \{ aI : a \in \mathbb F^{\ast} \}$. The projective special linear group is $PSL(2, \mathbb F) := SL(2, \mathbb F) / Z$.

Is there an easy way to show that $PSL(2,\mathbb F)$ has trivial center, i.e. there is no nontrivial element commuting with every other element? I guess it to be true, but I do not see any way to proof it.

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If $g \in {\rm SL}(2,F)$ is not scalar, then there exist linearly independent vectors $v,w\in F^2$ with $gv=w$. Let $h \in {\rm SL}(2,F)$ with $hv=v$, $hw=v+w$. Then $ghv = w$, but $hgv = v+w$, so the images of $g$ and $h$ in ${\rm PSL}(2,F)$ do not commute.

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  • $\begingroup$ Thanks for your answer. I have also posted an answer just to offer a solution I came up with recently. But yours is preferable! Maybe you want to take a look and say if my solution is okay as it stands. $\endgroup$ – StefanH Dec 23 '15 at 20:10
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Derek's proof is great, and short. But just for completeness I want to add what I got in the last ~2h.

Let $A, B \in SL(2, \mathbb F)$, then they commute in $PSL(2, \mathbb F)$ iff $[A,B] = A^{-1} B^{-1} A B \in Z$. Let $$ A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \quad B = \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}. $$ As the determinants are $1$, we have $$ A^{-1} = \begin{pmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{pmatrix} \quad B^{-1} = \begin{pmatrix} b_{22} & -b_{12} \\ -b_{21} & b_{11} \end{pmatrix}. $$ We compute \begin{align*} A^{-1} B^{-1} A B & = \begin{pmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{pmatrix} \begin{pmatrix} b_{22} & -b_{12} \\ -b_{21} & b_{11} \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} \\ & = \begin{pmatrix} a_{22} b_{22} + a_{12}b_{21} & -a_{22}b_{12} - a_{12}b_{11} \\ -a_{21}b_{22} - a_{11}b_{21} & a_{21}b_{12} + a_{11} b_{11} \end{pmatrix} \begin{pmatrix} a_{11} b_{11} + a_{12}b_{21} & b_{12}a_{11} + a_{12}b_{22} \\ a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22} \end{pmatrix} \\ & = \begin{pmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{pmatrix} \end{align*} with \begin{align*} c_{11} & = (a_{22}b_{22} + a_{12}b_{21})(a_{11}b_{11} + a_{12}b_{21}) - (a_{22}b_{12} + a_{12}b_{11})(a_{21}b_{11} + a_{22}b_{21}) \\ c_{12} & = (a_{22}b_{22} + a_{12}b_{21})(b_{12}a_{11} + a_{12}b_{22}) - (a_{22}b_{12} + a_{12}b_{11})(a_{21}b_{12} + a_{22}b_{22}) \\ c_{21} & = -(a_{21}b_{22} + a_{11}b_{21})(a_{11}b_{22} + a_{12}b_{21}) + (a_{21}b_{12} + a_{11}b_{11})(a_{21}b_{11} + a_{22}b_{21}) \\ c_{22} & = -(a_{21}b_{22} + a_{11}b_{21})(b_{12}a_{11} + a_{12}b_{22}) + (a_{21}b_{12} + a_{11}b_{11})(a_{21}b_{12} + a_{22}b_{22}). \end{align*} Now $AZ$ (the coset of $A$ with respect to $Z$) is in the center of $PSL(2, \mathbb F)$ iff for every $B$ we have $[A,B] \in Z = \{ \alpha I : \alpha \in \mathbb F^{\ast} \}$. So we choose the special matrices $$ B = \begin{pmatrix} b_{11} & 0 \\ 0 & b_{11}^{-1} \end{pmatrix} $$ then \begin{align*} c_{11} & = a_{22}b_{22} a_{11} b_{11} - a_{12} b_{11} a_{21} b_{11} \\ c_{12} & = a_{22}b_{22} a_{12} b_{22} - a_{12} b_{11} a_{22} b_{22} \\ c_{21} & = -a_{21}b_{22}a_{11}b_{22} + a_{11}b_{11}a_{21}b_{11} \\ c_{22} & = -a_{21}b_{22}a_{12}b_{22} + a_{11}b_{11}a_{22} b_{22}. \end{align*} Now requiring $$ \begin{pmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{pmatrix} = \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} $$ this gives after simplifying \begin{align*} \alpha & = a_{11} a_{22} - a_{12} a_{21} b_{11}^2 \\ 0 & = a_{22} a_{12} b_{11}^{-2} - a_{12}a_{22} = a_{22}a_{12}(b_{11}^{-2} - 1) \\ 0 & = a_{11}a_{21}b_{11}^{-2} - a_{11}a_{21} b_{11}^2 = a_{11}a_{21}(b_{11}^{-2} - b_{11}^2) \\ \alpha & = a_{11}a_{22} - a_{21}a_{12} b_{11}^{-2}. \end{align*} The first and last equation give $b_{11}^2 = b_{11}^{-2} \Leftrightarrow b_{11}^4 = 1$. But as the equations have to hold for every $B$, just choose some matrix $B$ with $b_{11}^4 \ne 1$ and this gives a contradiction if the field is such that we can choose such an element. Otherwise the ground field is the finite field of order $2$, $3$ or $5$, see here. But $PSL(2,2) \cong S_3, PSL(2,3) \cong A_4, PSL(2,5) \cong A_5$ and all these groups have trivial center.

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