Past exam of the algebra course of the University of Oxford / 2009 / B / 4

For the Abelian case of Cauchy theorem for groups, use induction on $|G|$. Choose $a\ne e$. If $p$ divides $m=|a|$, then $a^{m/p}$ has order $p$. Otherwise $⟨a⟩$ is a nontrivial normal subgroup of order prime to $p$, and $G/⟨a⟩$ has smaller order still divisible by $p$. Induction gives a coset of order $p$; any lift $b$ has order divisible by $p$, because the order of its image divides its own order. A suitable power of $b$ then has order $p$. This uses no classification of finite Abelian groups.

The map $g\mapsto g^{-1}$ partitions $S$ into disjoint two-element sets $\{g,g^{-1}\}$, so $|S|$ is even. Since $|G|$ is even, $|G\setminus S|$ is even as well. This complement contains the identity and consists of elements with $g^2=e$. It therefore contains at least one further element $h\ne e$, and that element has order two. Thus every finite group of even order contains an involution, without using Cauchy theorem for groups.

By Lagrange's theorem, a nonidentity element in a group of order four has order two or four. In the latter case the group is cyclic. Otherwise every nonidentity element has order two, and $(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$ while $(ab{)}^{-1}=ab$, so the group is Abelian. Choose distinct nonidentity elements $a,b$. Their cyclic subgroups meet trivially and $ab$ is the fourth element, so $(i,j)\mapsto a^i{b}^j$ gives an isomorphism $C_2\times C_2\to G$. HenceNo classification of finite Abelian groups is used.

By Cauchy theorem for groups, choose $a$ of order $p$ and $b$ of order two. The subgroup $P=⟨a⟩$ has index two, hence is normal, and $b\notin P$. Thus $a,b$ generate $G$. Conjugation by $b$ has $ba{b}^{-1}=a^r$ with $r\not\equiv 0(\mathrm{mod}p)$ and $r^2\equiv 1(\mathrm{mod}p)$. Since $p$ is prime, $r\equiv 1$ or $-1$.

If $r=1$, the two generators commute and $ab$ has order $2p$, so $G$ is cyclic. If $r=-1$, then $(ab{)}^2=e$ and $ab\ne b$. The two involutions $b,ab$ generate $G$, because $(ab)b=a$. Hence $G$ is dihedral in the stated convention. Thuswhere $D_{2p}$ denotes the dihedral group of order $2p$. In both cases $P$ is a proper nontrivial normal subgroup, so $G$ is not simple.