Past exam of the mathematics course of the University of Oxford / 2014 / Prelims / Mathematics I / 5 / b

Let $n=o(g)$. A group homomorphism preserves powers and the identity element, soThe preceding divisibility property givesNow assume $\varphi$ is surjective and choose $h\in H$ of order $m$. Take a preimage $g$ of $h$, and write $o(g)=mr$, as the divisibility result permits. The element $g^r$ has order exactly $m$: its $m$th power is $e$, while $(g^r{)}^s=e$ implies $mr\mid rs$, hence $m\mid s$. Therefore $G$ contains an element of order $m$, namely $g^r$.

The preimage $g$ itself need not have order $m$; taking the suitable power is the essential step. The constructed element need not map to the original $h$. This proves that element orders in a finite quotient occur in the original group.

A group homomorphism from an additive cyclic group ${\mathbb{Z}}_n$ is determined by the image $a$ of $1$, subject exactly to $a^n=e$ in multiplicative notation, or $na=0$ in an additive target.

For ${\mathbb{Z}}_{18}\to {\mathbb{Z}}_{15}$, the relation is $18a\equiv 0(\mathrm{mod}15)$, equivalently $3a\equiv 0(\mathrm{mod}15)$. The possibilities are $a=0,5,10$, and each defines a homomorphism between cyclic groups. Thus there are three such maps, an instance of the number of homomorphisms between finite cyclic groups.

For ${\mathbb{Z}}_9\to S_4$, the image of $1$ must be a permutation whose order of a group element divides $9$. A permutation on four symbols can have order $1,2,3$ or $4$, so the allowed images are the identity and the three-cycles. There are $\left(\genfrac{}{}{0ex}{}{4}{3}\right)(3-1)!=8$ three-cycles: choose their three symbols and one of the two cyclic orientations. Hence there are nine maps.

Finally, each transposition in $S_5$ has order $2$, so its image in ${\mathbb{Z}}_5$ must have order dividing both $2$ and $5$, and is therefore the identity. Every permutation is a product of transpositions, so every element must map to the identity. There is one map. In the order asked, the counts are