Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 1 / 1B (Algebra and Geometry)

Compose permutations from right to left. Tracking the four symbols through the two permutation cycles gives $1\mapsto 2\mapsto 1$ and $3\mapsto 4\mapsto 3$. Thus the cycle decomposition of a permutation isThe disjoint transpositions commute, and squaring their product gives the identity element, while the product itself is nonidentity. Its order of a group element is therefore two. Each transposition has negative sign of a permutation, so the product has sign $(-1{)}^2=\boxed{+1}$: it is an even permutation.

If $y=gx{g}^{-1}$, cancellation of successive $g^{-1}g$ factors gives $y^k=g{x}^k{g}^{-1}$ for every positive integer $k$. Hence $y^k$ is the identity element exactly when $x^k$ is, proving that conjugate group elements have the same order of a group element. For permutations, the sign group homomorphism similarly givesThus conjugate permutations have the same sign of a permutation.

The converse is false: permutation order and sign do not determine conjugacy. In the symmetric group $S_6$, take $x=(123)$ and $y=(123)(456)$. Both have order three and positive sign. However, $x$ fixes three symbols whereas $y$ fixes none. If $x$ fixes a symbol $j$, then $gx{g}^{-1}$ fixes $g(j)$, and applying the inverse conjugation gives a bijection between the fixed-symbol sets. Their different fixed-point counts therefore prove that $x,y$ are not conjugate. Equivalently their cycle types $31^3$ and $3^2$ differ. Equal order and sign are necessary, but not sufficient, for conjugacy.