Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 9

Write $H=\{e,h\}$ with $h\ne e$. For any $g\in G$, the normal subgroup property implies $gh{g}^{-1}\in H$. This conjugate cannot be $e$, since multiplication by $g^{-1}$ and $g$ would then give $h=e$. There is only one remaining choice, soThus $h$ commutes with every element of $G$; the identity does too. HenceThis proves that a normal subgroup of order two is central. The proof does not require $G$ to be finite: conjugation permutes $H$ and fixes its identity, so it must fix its only other element.

A group homomorphism $\varphi :H\to G$ is a map preserving multiplication:It necessarily sends $e_H$ to $e_G$ and sends inverses to inverses, by applying this relation to $e_H^2$ and $h{h}^{-1}$. Its kernel of a group homomorphism isThe kernel is a normal subgroup of the domain: if $k\in \ker \varphi$ and $h\in H$, then $\varphi (hk{h}^{-1})=\varphi (h)e_G\varphi (h{)}^{-1}=e_G$. The kernel’s identity condition refers to the identity in the target group.

Let $H$ be a proper subgroup. By Lagrange's theorem, $|H|$ divides $21=3\cdot 7$. Since $H\ne G$, its possibilities areThe subgroup of order one is generated by the identity and is cyclic. For either prime order $p=3$ or $7$, choose $h\in H$ other than the identity. Its order of a group element divides $p$ and is not one, so it is $p$. Thus its powers exhaust $H$, proving $H=⟨h⟩$.

ThereforeThis uses the fact that a group of prime order is cyclic. It does not require the order-21 group itself to be Abelian or cyclic.