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

Among the $|G|+1$ powers $e,g,g^2,\dots ,g^{|G|}$ in the finite group $G$, the pigeonhole principle forces two to be equal: $g^i=g^j$ with $0\le i<j\le |G|$. Multiply by $g^{-i}$ to obtainThis proves the existence of a positive power equal to the identity element, without presupposing the order of a group element.

The order of a group element $o(g)$ is the least positive integer $n$ for which $g^n=e$, whose existence follows from part (i). If $g^k=e$, use the Euclidean division to write $k=qn+r$, where $0\le r<n$. ThenMinimality of $n$ forces $r=0$. ThusThe argument also works for negative $k$, using integer powers and the same remainder convention.

Lagrange's theorem states that for a subgroup $H$ of a finite group $G$,where $[G:H]$ is the index of a subgroup, the number of its left cosets. In particular, $|H|$ divides $|G|$.

The cyclic subgroup $⟨g⟩$ consists of the distinct elements $e,g,\dots ,g^{n-1}$, where $n=o(g)$. Every power reduces to one of these by the Euclidean division, and a repetition among them would contradict minimality of $n$. Thus $|⟨g⟩|=o(g)$. Applying Lagrange's theorem gives

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