= Solution
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 \b[$G$ is the cyclic group $C_{10}$]. Otherwise choose $a$ of order five. Its <cyclic subgroup> $H=\langle a\rangle$ 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 $a^jb$. <Conjugation> by $b$ restricts to an <automorphism> of $H$, say $bab^{-1}=a^r$. Since $b^2=1$, $r^2\equiv1\pmod5$, 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 \b[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.
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