Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics I / 6

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

The order of a group element $g\in G$ is the least positive integer $k$ such that $g^k=e$. It exists because $G$ is finite. The cyclic subgroup $⟨g⟩$ consists of exactly $e,g,\dots ,g^{k-1}$, so has order $k$. Applying Lagrange's theorem to this subgroup gives

A map $\varphi :G\to G$ is a group isomorphism if it is a bijection and a group homomorphism, meaningSuch a map from a group to itself is an group automorphism. Its inverse is also a group homomorphism: if $u=\varphi (x)$ and $v=\varphi (y)$, then ${\varphi }^{-1}(uv)=xy={\varphi }^{-1}(u){\varphi }^{-1}(v)$.

Write $C_p=⟨g⟩$. A group homomorphism from this cyclic group is determined by the image of $g$, because $\varphi (g^j)=\varphi (g{)}^j$. Each choice $\varphi (g)=g^a$, with $a=0,\dots ,p-1$, gives a well-defined group homomorphism. It is an group automorphism exactly when $g^a$ generates $C_p$.

For a prime number $p$, every $a=1,\dots ,p-1$ is coprime to $p$, so the powers of $g^a$ run through all of $C_p$. The choice $a=0$ gives the trivial group homomorphism and fails to be bijective. Consequently

For $h\in H$, define $\pi (h)(n)=hn{h}^{-1}$. Since $N$ is a normal subgroup, this sends $N$ into itself. It preserves multiplication, and its inverse is conjugation by $h^{-1}$, so it is an group automorphism of $N$. MoreoverThus $\pi :H\to \mathrm{Aut}(N)$ is a group homomorphism, induced by the group action of conjugation.

If $\pi$ is trivial, every element of $H$ commutes with every element of $N$. Henceis a group homomorphism, since $(n_1{h}_1)(n_2{h}_2)=n_1{n}_2{h}_1{h}_2$. Its image of a group homomorphism is a subgroup containing both $N$ and $H$, and every product $nh$ lies in $⟨N,H⟩$, so its image is exactly $⟨N,H⟩$. If $nh=e$, then $n=h^{-1}\in N\cap H=\{e\}$, so $n=h=e$. Thus $F$ is injective and gives

The stabilizer subgroup of $H$ for this conjugation group action is its normalizerIf it is all of $G$, the equality $gH{g}^{-1}=H$ holds for every $g\in G$, which is precisely the definition of a normal subgroup.

If instead $\mathrm{Stab}(H)=H$, the orbit-stabilizer theorem says that the group orbit of $H$ has size $[G:H]$. Every conjugate $gH{g}^{-1}$ is isomorphic to $H$, with group isomorphism $h\mapsto gh{g}^{-1}$, so this group orbit is a subset of $\Omega$. ThereforeThe group action need not be transitive on all of $\Omega$; this one group orbit supplies the lower bound.

By Lagrange's theorem, every nonidentity group element has order $5$, $7$, or $35$. If some element has order $35$, its seventh and fifth powers have orders $5$ and $7$, respectively, and we are done.

Otherwise every nonidentity element has order $5$ or $7$. Distinct subgroups of a given prime number order intersect only in the identity: a shared nonidentity element generates both. If there were no element of order $5$, the $34$ nonidentity elements would therefore partition into sets of six, the nonidentity elements of the cyclic subgroups of order seven. This would force $6\mid 34$, a contradiction. If there were no element of order $7$, the analogous partition into sets of four would force $4\mid 34$, again a contradiction. Thus elements of both orders exist.

Let $P$ and $Q$ be subgroups of orders five and seven, which exist by i. Their normalizers contain $P$ and $Q$, respectively. By Lagrange's theorem, $N_G(P)$ has order $5$ or $35$, and $N_G(Q)$ has order $7$ or $35$. If either has order $35$, its subgroup is normal by iv.

Suppose neither is normal. Then iv gives at least seven distinct subgroups of order five and at least five distinct subgroups of order seven. Distinct subgroups of the same prime order intersect only in the identity, and a subgroup of order five and one of order seven also intersect only in the identity by Lagrange's theorem. Thus $G$ would have at leastnonidentity elements, contradicting $|G|-1=34$. Hence at least one of $P,Q$ is a normal subgroup.

Choose a normal subgroup $N$ of prime order $p\in \{5,7\}$ as in ii, and choose a subgroup $H$ of the other prime order $q$. Their intersection is trivial by Lagrange's theorem. Both are cyclic groups, and ii gives $|\mathrm{Aut}(N)|=p-1$.

The image of a group homomorphism $\pi :H\to \mathrm{Aut}(N)$ from iii has order dividing $q$, because its kernel of a group homomorphism partitions $H$ into equally sized cosets. Its order also divides $p-1$ by Lagrange's theorem. Butso $\pi$ is trivial in either case. By iii, $⟨N,H⟩\cong C_p\times C_q$ has order $pq=35$ and therefore equals $G$.

If $n$ and $h$ generate $N$ and $H$, the element $(n,h)$ has order $\mathrm{lcm}(p,q)=pq=35$, since its $k$th power is the identity exactly when both $p$ and $q$ divide $k$. Thus this direct product of groups is cyclic, and