Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 3 / 2

Extend each element of $G$ to fix $\omega$. A one-point extension of a permutation group is a transitive permutation group $G^{+}\le S_{{\Omega }^{+}}$ such thatwith the induced action on $\Omega$ equal to the prescribed action. The stabilizer subgroup equality is the substantive condition: merely adjoining an element that moves $\omega$ does not suffice. If $|\Omega |=n$, the orbit-stabilizer theorem gives $|G^{+}|=(n+1)|G|$.

Here is the double-coset criterion for a one-point extension. Let $H=G_{\alpha }$, and suppose $x$ swaps $\omega$ and $\alpha$. Then $⟨G,x⟩$ is a one-point extension if and only ifFor necessity, in an extension $H$ fixes both $\omega$ and $\alpha$, so $x$ normalizes it and $x^2$ lies in it. Since $G$ is transitive on $\Omega$, the extension has exactly two double cosets relative to $G$: $G$ and $GxG$. For $g\notin H$, $xgx$ moves $\omega$ into $\Omega$ and is in the latter double coset.

For sufficiency, the displayed conditions make $G\cup GxG$ closed under multiplication. Products with middle element in $H$ reduce using $xhx=(xh{x}^{-1})x^2\in G$; those with middle element outside $H$ remain in $GxG$. A finite nonempty multiplication-closed set of permutations containing the identity is a group. It contains $G$ and $x$, hence equals $⟨G,x⟩$. Every element in $GxG$ moves $\omega$, while $G$ fixes it, giving the required stabilizer subgroup. The group is transitive because $G$ is transitive on $\Omega$ and $x$ moves the additional point.

For $1\le t\le |\Omega |$, a group is sharply t-transitive when any two ordered $t$-tuples of distinct points are related by exactly one group element. Equivalently its action on the set of such tuples is regular. In the finite caseand the stabilizer subgroup of an ordered $t$-tuple is trivial. The condition includes both existence and uniqueness, not just transitivity.

Put $n=|\Omega |\ge 2$, and let $H=G_{\alpha }$. Sharp two-transitivity gives $|G|=n(n-1)$ and $|H|=n-1$. A nonidentity element fixes at most one point. Counting the nonidentity elements in the $n$ stabilizer subgroups shows that there are $n-1$ fixed-point-free elements. Let $N$ be this set together with the identity. We first prove it is a normal subgroup, rather than presuming that fixed-point-free elements are closed under multiplication.

For $n=2$, $G=C_2$ and $N=G$ directly. Otherwise use complex characters of a finite group. Let $\pi$ be the permutation character and $\psi =\pi -1$, the character of the permutation representation with its constant line removed. For every nontrivial irreducible character $\theta$ of $H$, form the virtual characterIts values are $\theta (1)$ at the identity and at every fixed-point-free element. At an element with one fixed point, conjugate it to $h\in H\setminus \{1\}$; induction gives ${\phi }_{\theta }(g)=\theta (h)$, because there is exactly one fixed coset.

The identity and the $n-1$ fixed-point-free elements together contribute $n\theta (1)\overline{\eta (1)}$ to the inner product. The remaining elements are partitioned into the nonidentity parts of the $n$ stabilizer subgroups. Hence character orthogonality givesA virtual character of norm one is plus or minus an irreducible character: its coefficients in the irreducible-character basis are integers whose squares sum to one. Its positive degree $\theta (1)$ selects the plus sign. Thus each ${\phi }_{\theta }$ is an actual irreducible character.

For a group representation, $\chi (g)=\chi (1)$ holds exactly on its kernel: make the representation unitary and compare the sum of its unit-modulus eigenvalues with its dimension. All of $N$ therefore lies in the intersection of the kernels of the ${\phi }_{\theta }$. Conversely, a nonidentity element fixing a point gives $h\ne 1$ in $H$. Some nontrivial irreducible character of $H$ has $\theta (h)\ne \theta (1)$, since otherwise the regular representation of $H$ would not vanish at $h$. ConsequentlyThis proves normality and subgroup closure. It has order $n$ and no nonidentity element fixing a point, so it is a regular permutation subgroup.

Now $H$ acts transitively by conjugation on $N\setminus \{1\}$: identify an element of $N$ with its image of $\alpha$ and use transitivity of $H$ on the remaining points. Thus all nonidentity elements of $N$ have the same order. Taking a suitable power of one element shows this common order is a prime $p$. By Cauchy's theorem no other prime divides $|N|$, so $N$ is a $p$-group. Its nontrivial center is $H$-invariant, so transitivity forces the center to be all of $N$. Therefore $N$ is elementary abelian of order $p^d=n$.

Since $|H|=p^d-1$ is prime to $p$, $N$ is the unique Sylow $p$-subgroup of $G$. Uniqueness makes it characteristic under every group automorphism. We have provedThe character argument supplies the regular kernel of a finite sharply two-transitive group; the final Sylow argument establishes the stronger characteristic assertion.

Iwasawa's simplicity lemma states the following. Suppose $G$ acts faithfully and primitively on a set, and a stabilizer subgroup $H$ has an abelian normal subgroup $A$ whose $G$-conjugates generate $G$. Then every nontrivial normal subgroup $N$ contains $G^{\prime }$. In particular, if $G$ is nontrivial and perfect, then $G$ is simple.

Indeed a nontrivial normal subgroup in a faithful primitive action is transitive, so $G=NH$. Since $H$ normalizes $A$, all conjugates of $A$ have the same image in $G/N$. Those images generate the quotient, which is therefore abelian. This gives $G^{\prime }\le N$ and proves the stated conclusion.

Use the field $F={\mathbb{F}}_2[\zeta ]/({\zeta }^3+\zeta +1)$. The defining polynomial has no root in ${\mathbb{F}}_2$, hence is irreducible. Label its elements bySince $F^{\ast }$ has prime order seven, $\zeta$ has order seven. Multiplication by $\zeta$ is exactly $b$. The identities ${\zeta }^3=\zeta +1$, ${\zeta }^6={\zeta }^2+1$ and ${\zeta }^5={\zeta }^4+1$ show that translation by one is exactly $a$.

Conjugating $a$ by powers of $b$ gives the translations $T_{{\zeta }^i}:z\mapsto z+{\zeta }^i$. The translations by $1,\zeta ,{\zeta }^2$ generate all eight translations. ConsequentlyFor distinct $u,v$ and distinct target points $u^{\prime },v^{\prime }$, the unique affine map has $s=(v^{\prime }-u^{\prime })/(v-u)$ and $t=u^{\prime }-su$. Thus the action is sharply two-transitive.

Its regular characteristic subgroup isTranslations act regularly, $K$ is normal, and its order eight makes it the unique Sylow two-subgroup, hence characteristic. This identifies the abstract subgroup and explicit generators in the original permutation notation.

Adjoin the label $9=\infty$. Inversion on the projective line, with zero and infinity interchanged, isThe label $2=1\in F$ is fixed. Set $H=G_1=\{M_s:z\mapsto sz\}$, where the subscript one denotes the original point label, namely field zero. Then $x^2=1\in H$ and $x{M}_{s}x=M_{s^{-1}}$, so $x$ normalizes $H$.

For $g(z)=sz+t$ outside $H$, we have $t\ne 0$. On the projective line,The equality uses characteristic two and is valid as an equality of fractional linear transformations, including poles and infinity. ThusAll conditions of the double-coset criterion for a one-point extension hold. Therefore $G^{+}=⟨G,x⟩$ is a one-point extension, with $|G^{+}|=9\cdot 56=504$. Its stabilizer subgroup is sharply two-transitive, so its action on nine points is sharply three-transitive.

Use the projective-line realization from part (ii). The subgroup $K$ of translations is abelian and normal in the stabilizer subgroup $G$ of infinity. We verify both remaining conditions of Iwasawa's simplicity lemma, instead of concluding simplicity from transitivity alone.

Let $R$ be generated by all conjugates of $K$ in $G^{+}$. It contains the matricesHere matrices act by fractional linear transformations. For $t\ne 0$,In particular $w(1)=x\in R$, and $w(t)w(1)=\mathrm{diag}(t,t^{-1})$ acts as multiplication by $t^2$. Squaring is a bijection of $F^{\ast }$, so every $M_s$ belongs to $R$. Hence $R$ contains $G$ and $x$, and $\boxed{R=G^{+}}$.

Choose $s\ne 1$. The commutator, with convention $[u,v]=uv{u}^{-1}{v}^{-1}$, isAs $t$ varies, this gives all translations. Thus $(G^{+}{)}^{\prime }$ contains $K$, and normality makes it contain every conjugate of $K$. Since those generate $G^{+}$, the group is a perfect group.

A sharply three-transitive action on nine points is a primitive group action, and this permutation action is a faithful group action. All Iwasawa hypotheses now hold, soIt may be identified with ${\mathrm{PSL}}_2(8)={\mathrm{PGL}}_2(8)$: the generators are fractional linear transformations and $|{\mathrm{PGL}}_2(8)|=8(8^2-1)=504$, while all nonzero field elements are squares. The simplicity proof above does not rely on assuming simplicity of that named family.