Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 9

A group homomorphism $f:G\to H$ preserves the operation: $f(ab)=f(a)f(b)$ for all $a,b\in G$. A group isomorphism is a bijective group homomorphism; its inverse also preserves the operation. The kernel of a group homomorphism is $\ker f=\{g\in G:f(g)=e_H\}$, a normal subgroup of $G$.

Choose $g\ne e$. By Lagrange's theorem, its cyclic subgroup has order dividing the prime 101. It is not of order one, so $⟨g⟩=G$. Thus the group is cyclic, and hence abelian. In particular, every pair of elements is $g^i,g^j$, and $g^i{g}^j=g^{i+j}=g^{j+i}=g^j{g}^i$.

A group homomorphism from the additive cyclic group ${\mathbb{Z}}_6$ is determined by the image $t$ of its generator $1$, with the consistency condition $6t=0$ in the target. Every element of either target satisfies this condition. ThereforeandThese formulas are well-defined because changing $n$ by six changes $tn$ by a target multiple of its modulus. They list respectively two and three maps.

The symmetric group $S_3$ is generated by transpositions, each of order two. A group homomorphism to ${\mathbb{Z}}_3$ must map every transposition to zero, because the only target element annihilated by two is zero. Hence the map to ${\mathbb{Z}}_3$ is necessarily trivial.

For $S_3\to {\mathbb{Z}}_2$, the 3-cycles map to zero because their order divides three. All transpositions are conjugate group elements, and a map to an abelian group takes conjugate elements to the same value. That common value is either zero or one. Both choices work: they give the trivial map and the sign homomorphism, with an odd permutation sent to $1$ and an even permutation sent to $0$.