Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 3 / 6D (Groups)

Lagrange's theorem for finite groups states that if $H\le G$ and $G$ is finite, thenso $|H|$ divides $|G|$. Indeed, the left cosets of $H$ partition $G$, and multiplication by a coset representative is a bijection from $H$ to each coset.

Applying the theorem to the cyclic subgroup $⟨g⟩$ gives

In $C_3\times C_9$, an element has order dividing three exactly when its first coordinate is arbitrary and its second belongs to the unique subgroup of order three. There are $3\cdot 3=9$ such elements, and after removing the identity the answer is $\boxed{8}$.

Write the dihedral group as $D_{2n}=⟨r,s:r^n=s^2=e,srs=r^{-1}⟩$. Every reflection has order two, so an element of order three must be a rotation. Such rotations exist exactly when $3\mid n$, and then they are $r^{n/3}$ and $r^{2n/3}$. The answer is therefore

An element of order three in $S_7$ has cycle type $31^4$ or $3^{2}1$. There aresingle three-cycles. For two disjoint three-cycles, choose the two unordered three-element supports and an orientation on each:Hence $S_7$ contains $\boxed{350}$ elements of order three.

A three-cycle is an even permutation, as is a product of two disjoint three-cycles. Thus every element counted in part (c) lies in $A_7$, and every order-three element of $A_7$ was already counted there. The answer remains $\boxed{350}$.