Past exam of the mathematics course of the University of Cambridge / 2004 / ia / Paper 3 / 1D (Algebra and Geometry)

For a subgroup $H$ of a finite group $G$, Lagrange's theorem states $|G|=[G:H]|H|$. Thus a subgroup's order, and in particular the order of a group element, divides $|G|$.

For $|G|=10$, if every element had order one or two, the order of a finite group of exponent two would make $|G|$ a power of two, a contradiction. Hence an element has order five or ten. In the latter case $G$ is the cyclic group $C_{10}$. Otherwise choose $a$ of order five. Its cyclic subgroup $H=⟨a⟩$ has index two and is therefore a normal subgroup.

In the noncyclic case an element $b$ outside $H$ has order two: its order divides ten, it cannot have order ten, and an element of order five has trivial image in the order-two quotient group $G/H$. Every element of $G$ is then $a^j$ or $a^{j}b$. Conjugation by $b$ restricts to an automorphism of $H$, say $ba{b}^{-1}=a^r$. Since $b^2=1$, $r^2\equiv 1(\mathrm{mod}5)$, so $r=1$ or $-1$ modulo five.

For $r=1$, the generators commute and $ab$ has order ten, yielding the cyclic case. For $r=-1$, these are the relations of the dihedral group of the pentagon. The classification of groups of order ten is therefore exactly $C_{10}$ and $D_{10}$, where the subscript on $D_{10}$ denotes its order. They are not isomorphic because one is Abelian and the other is not.