Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 9

Act on the four labelled vertices. A proper rotation gives an even permutation, identifying the rotational tetrahedral symmetry group with $A_4$, an alternating group. Geometrically there are one identity, three half-turns about opposite-edge midpoint axes, and two nontrivial rotations about each of four vertex-face axes. ThereforeThe tetrahedral rotation conjugacy classes areTo justify the splitting, the centralizer of a 3-cycle within $A_4$ is its three-element cyclic subgroup, so its conjugacy class has size $12/3=4$. Its inverse is not conjugate to it in $A_4$: the permutations that reverse that cycle while preserving its fixed vertex are odd. Conjugating $(123)$ by the even permutations produces the displayed $C_{+}$, and taking inverses gives $C_{-}$. The three half-turns are conjugate, since rotations move their axes into one another. Their centralizer has order four and their class therefore has size three.

LetThe product of any two distinct nonidentity elements is the third, and each squares to $e$, so $K$ is a Klein four-group. Conjugating a double transposition simply relabels its letters, keeping it in the set. Hence $K$ is a normal subgroup of $A_4$.

Choose $r=(123)$. The cosets $K,rK,r^{2}K$ are distinct and cover the twelve elements. With $\zeta =e^{2\pi i/3}$, defineThe coset decomposition makes this well-defined. Normality givesso $\Phi$ preserves products. It is onto $\{1,\zeta ,{\zeta }^2\}\cong C_3$, andThis is the tetrahedral quotient onto the cyclic group of order three. In terms of (a), it sends $C_0\cup C_2$ to $1$, $C_{+}$ to $\zeta$, and $C_{-}$ to ${\zeta }^2$. The normality check above also follows abstractly because a group homomorphism kernel is normal.

The group is the direct product of groups ${\mathbb{Z}}_p\times {\mathbb{Z}}_q$. Repeatedly adding $(1,1)$ returns to $(0,0)$ for the first time when both $p$ and $q$ divide the step number. Its order of a group element is therefore $\mathrm{lcm}(p,q)$. If $p,q$ are coprime, this is $pq$, so it generates every element:Explicitly, $[t{]}_{pq}\mapsto ([t{]}_p,[t{]}_q)$ is the Chinese remainder theorem group isomorphism.

If $d=\gcd (p,q)>1$, every pair satisfiesThus every element order divides $pq/d<pq$, and no element can generate the whole group. HenceThis is the cyclicity of a product of two finite cyclic groups.