Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 3 / 5

The orbits of a normal subgroup form a block system: $g(N\omega )=N(g\omega )$, so every group element permutes these orbits. By primitivity the orbits are either all singletons or one whole orbit. In the singleton case $N$ fixes every point, giving $N\le G_{[\Omega ]}$.

In the other case $N$ is transitive. For each $g\in G$ choose $n\in N$ with $n\alpha =g\alpha$. Then $n^{-1}g\in G_{\alpha }$, so $G=N{G}_{\alpha }$. Work in the quotient $G/N$. If $g=nh$ with $h\in G_{\alpha }$, the image of $A^g=g^{-1}Ag$ is the image of $A^h$, which equals the image $\overline{A}$ of $A$ because $A◃G_{\alpha }$. All conjugates of $A$ thus have the same image in $G/N$.

By hypothesis those conjugates generate $G$, so the quotient is generated by the single subgroup $\overline{A}$, and equals it. Since $A$ is abelian, $G/N$ is abelian. The kernel of a group homomorphism of a group homomorphism to an abelian group contains the commutator subgroup, yieldingThis is the normal-subgroup argument behind the Iwasawa simplicity lemma. Faithfulness is not needed for this version: retaining the pointwise kernel of a group homomorphism as the second alternative is essential. In a faithful primitive group action it says every nontrivial normal subgroup contains $G^{\prime }$.

For distinct coordinates $i,j$, let $E_{ij}(t)=I+t{e}_{ij}$, where $e_{ij}$ is the matrix unit. For $t\ne 0$ this is a transvection: it fixes the hyperplane with $j$th coordinate zero pointwise, has determinant one and inverse $E_{ij}(-t)$. We show these elementary transvection matrices generate $SL(n,F)$ over any field.

Let $M\in SL(n,F)$. Row addition is left multiplication by an $E_{ij}(t)$. If the current diagonal pivot is zero, a nonzero entry occurs below it in the same remaining column, since the remaining square block is invertible; adding that row makes the pivot nonzero. Row additions then clear the entries below the pivot. Repeating produces an upper-triangular matrix with nonzero diagonal. Clearing the entries above the diagonal, starting from the last column, gives a diagonal matrix $D=\mathrm{diag}(a_1,\dots ,a_n)$ with ${\prod }_i{a}_i=1$. Thus it remains to realize determinant-one diagonal matrices using elementary transvections.

In a two-coordinate block putDirect multiplication givesThe diagonal matrix $D$ is the product, for $i=1,\dots ,n-1$, of these blocks on coordinates $i,n$, with entries $a_i,a_i^{-1}$. Their final $n$th entry is ${\prod }_{i<n}{a}_i^{-1}=a_n$. Therefore $D$, and hence $M$, is a product of elementary transvections. All row operations used are invertible transvections, and no division except by a nonzero pivot or $a_i$ has been made. This provesThe identity element $E_{ij}(0)$ can be omitted from the generating set. This is transvections generate the special linear group; ordinary row scaling alone would not prove it, because such a scaling can leave $SL(n,F)$.

Let $V=F^n$ and let $G=PSL(n,F)$ act on its one-dimensional subspaces. First identify the kernel of a group homomorphism in $SL(n,F)$. A matrix fixing every coordinate line is diagonal, and fixing every line $⟨e_i+e_j⟩$ forces all diagonal entries equal. Thus the projective kernel of a group homomorphism is $\{\lambda I:{\lambda }^n=1\}$. These scalars are central. Conversely a central matrix commutes with all $E_{ij}(1)$: comparing entries in $M{e}_{ij}=e_{ij}M$ shows all off-diagonal entries of $M$ vanish and all diagonal entries agree. Therefore the kernel of a group homomorphism is exactly $Z(SL(n,F))$. This proves scalar kernel of the projective linear action and faithfulness of the induced $G$-action.

The action is doubly transitive. Given two ordered pairs of distinct lines, choose two independent representatives of each pair and extend them to bases. A linear map taking the first basis to the second sends the desired lines correctly. If its determinant is $d\ne 1$, postcompose with a diagonal map in the target basis that scales its first basis vector by $d^{-1}$ and fixes all the others. It preserves the two target lines and corrects the determinant. Thus an element of $SL(n,F)$ sends either ordered pair to the other. Double transitivity implies primitivity: a block containing two distinct points must contain every point by the transitive point-stabilizer action.

For $\alpha =⟨e_1⟩$, considerProducts add the functionals, so $U$ is abelian. If $g$ fixes $\alpha$, write $g{e}_1=a{e}_1$. Then $g(I+e_1\phi )g^{-1}=I+e_1(a\phi g^{-1})$, and the new functional still vanishes on $e_1$. Thus the image $A$ of $U$ in $G$ is normal in the point stabilizer. Its conjugates contain all elementary transvections: changing the center line $⟨e_1⟩$ to $⟨e_i⟩$ gives every map $I+e_i\psi$ with $\psi (e_i)=0$, including $E_{ij}(t)$. Part (b) then shows that those conjugates generate $G$.

It remains to prove perfectness, not assume it. Use $[x,y]=x^{-1}{y}^{-1}xy$. For $n\ge 3$, distinct $i,j,k$ giveEvery generator is therefore a group commutator, so $SL(n,F{)}^{\prime }=SL(n,F)$. For $n=2$ and $|F|>3$, choose $a\in F^{\times }$ with $a^2\ne 1$, possible because a quadratic polynomial has at most two roots. With $D=\mathrm{diag}(a,a^{-1})$,Both coefficients are nonzero, so every upper and lower elementary transvection is again a group commutator. Part (b) proves perfectness in this case as well. Quotients of perfect groups are perfect, hence $G^{\prime }=G$ in exactly the stated range.

Apply part (a) to the faithful primitive $G$-action and its abelian normal point-stabilizer subgroup $A$. Every nontrivial normal subgroup contains $G^{\prime }=G$, so it is the whole group. A transvection is not scalar, showing $G\ne 1$. We have provedThe proof works for infinite fields too and uses no finite-order count. The excluded two-dimensional fields give the familiar soluble exceptions $PSL(2,2)\cong S_3$ and $PSL(2,3)\cong A_4$.