Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 3 / 4

Fix ordered branch positions $P=(p_1,\dots ,p_r)$ and conjugacy classes $C_i$ in $G$. Form the generating product-one setThe rigid ramification type condition is that this set is nonempty and consists of one orbit under simultaneous inner conjugation by $G$. Rationality of classes is a further arithmetic condition and is not needed for this complex uniqueness statement.

Let $U={\mathbb{P}}^1(\mathbb{C})\setminus P$. Choose a base point and positively oriented peripheral loops ${\gamma }_i$ with sole relation ${\gamma }_1\cdots {\gamma }_r=1$. A connected Galois cover over $U$ gives a surjective monodromy map ${\pi }_1(U)\to G$, and changing the labeling of one fiber conjugates it. Its peripheral images are exactly a tuple in the set above.

For two extensions of the same rigid type, rigidity conjugates their tuples to agree. The loops generate ${\pi }_1(U)$, so their entire monodromy maps then agree. This constructs an isomorphism of the unramified covers directly: label fibers at the base point identically, and transport that matching along paths. Independence of path follows from equal monodromy. The resulting map is holomorphic because the cover maps are local biholomorphisms.

Near each missing point, the cover has local form $u={\pi }^{e_i}$. The isomorphism therefore extends over the added points, equivalently by the removable-singularity theorem in these local parameters. Pullback of meromorphic functions gives an isomorphism of the function fields fixing $\mathbb{C}(t)$. Hence there is at most one extension of the prescribed rigid type. This is uniqueness for fixed branch positions and peripheral conventions, not uniqueness after arbitrarily moving the branch points.

For $n\ge 3$, take $C_1$ to be the class of $n$-cycles, $C_2$ the class of transpositions, and $C_3$ the class of an $(n-1)$-cycle with one fixed point. With rightmost permutations applied first, a witness isMultiplying an $n$-cycle by a transposition splits it at the two exchanged letters. The resulting cycles have the lengths of the two arcs between those letters in the cyclic ordering. Thus $\alpha \beta$ has type $(n-1,1)$ exactly when the exchanged letters are cyclically adjacent.

Any product-one tuple can first be conjugated so that its first entry is $\alpha$. Its second entry is then one of the $n$ adjacent transpositions. The centralizer $⟨\alpha ⟩$ acts transitively on these choices; the third entry is forced to be the inverse product. Therefore every product-one tuple lies in a single simultaneous-conjugation orbit.

The conjugates ${\alpha }^j\beta {\alpha }^{-j}$ exchange all adjacent letters around the cycle. Their connected-edge graph generates $S_n$, so the witness, and every tuple in that orbit, generates $S_n$. HenceFor $n=3$ the last two classes coincide, which causes no difficulty: the same three adjacent swaps form one orbit and generate $S_3$.

The relevant cycle types are $(5,1)$, $(2,2,2)$ and $(3,2,1)$. Conjugate the first entry to $\alpha =(12345)$, fixing six. An entry $\beta$ of type $(2,2,2)$ pairs six with one of the five other letters. Powers of $\alpha$ move that partner to one. With the pair $(16)$ fixed, only three matchings remain:Thus there is exactly one centralizer orbit of eligible second entries, consisting of five matchings before normalization. The third entry is forced to be $(\alpha \beta {)}^{-1}$, so there is at most one inner orbit of product-one tuples of these types.

For the actual displayed $g_2$, direct multiplication giveswhich belongs to the class of the displayed $g_3$. This is a product-one witness. The three displayed permutations themselves need not multiply to one; the problem concerns their conjugacy classes.

The group $⟨\alpha ,g_2⟩$ is transitive, because $\alpha$ moves the first five points transitively and $g_2$ connects six to them. Its stabilizer of six contains the five-cycle $\alpha$ and is transitive on the other points. Hence the group is two-transitive and therefore primitive. Moreover $(\alpha g_2{)}^3=(13)$ is a transposition. The primitive transposition lemma proves that the group is $S_6$. Consequently all tuples in the sole inner orbit generate $S_6$, andAs a count check, there are $6\cdot 4!=144$ first entries and five eligible second entries for each, giving $720$ tuples. A generating tuple has simultaneous stabilizer equal to the trivial center of $S_6$, so its inner orbit also has size $720$.