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

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.