Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 2 / 4 / c

Let $S=S{L}_n(F)$ and let $Z$ be its scalar center. Its action on the lines of $F^n$ has kernel exactly $Z$: a linear map fixing every line is diagonal in a basis, and fixing the lines spanned by $e_i+e_j$ makes all its diagonal entries equal. The action is two-transitive. Indeed an ordered pair of distinct lines can be represented by two independent vectors and extended to a basis. A linear map between two such bases can be adjusted to determinant one by scaling one target basis vector; this preserves both target lines. Therefore $S/Z$ has a faithful primitive group action.

Fix $L=Fv$. The mapsform an abelian subgroup: all products of the rank-one terms vanish. It is normal in the line stabilizer, since conjugation replaces $v$ by a scalar multiple and transports $f$. Its conjugates contain every transvection, and hence generate $S$ by 4(b). The same statements hold for its image in $S/Z$.

It remains to prove perfectness rather than assume it. Use the group commutator convention $[x,y]=xy{x}^{-1}{y}^{-1}$. If $n\ge 3$, distinct $i,j,k$ giveEvery elementary generator is consequently in $S^{\prime }$. If $n=2$ and $|F|>3$, choose $a\in F^{\ast }$ with $a^2\ne 1$. There are at most two roots of $a^2=1$, so such a choice exists. For $h=\mathrm{diag}(a,a^{-1})$,Varying $b$ gives every upper transvection; conjugating by $w(1)$ gives every lower one. Again $S^{\prime }=S$, and its projective quotient is a perfect group.

Apply 4(a) to this faithful primitive projective action. Every nontrivial normal subgroup contains the commutator subgroup, which is the whole quotient. ThusThe restriction is real: $PS{L}_2(2)\cong S_3$ and $PS{L}_2(3)\cong A_4$ are not simple.