= Solution
<Lagrange's theorem for finite groups> states that for a <subgroup> $H$ of a finite <group> $G$, $|G|=[G:H]|H|$. For $x\in G$, two powers among $1,x,\ldots,x^{|G|}$ coincide, giving $x^m=1$ for some positive $m$. The least such exponent $d$ is the <order of a group element>, and the <cyclic subgroup> generated by $x$ consists of the $d$ distinct powers $1,x,\ldots,x^{d-1}$. Apply <Lagrange's theorem for finite groups> to obtain $d\mid|G|$.
<Cauchy's theorem for finite groups> states that if a prime $p$ divides the order of a finite <group>, the <group> contains an element of order $p$.
The classification of a <group of order eight> is
$$
\boxed{C_8,\quad C_4\times C_2,\quad C_2\times C_2\times C_2,\quad D_8,\quad Q_8.}
$$
Here $D_8$ denotes the <dihedral group> of order eight, not a <group> of order sixteen. To prove exhaustiveness, every element order is $1,2,4$ or $8$. An element of order eight generates the whole <group>, giving $C_8$.
If no element has order four or eight, every nonidentity element is an <involution>. Since $(ab)^{-1}=ab$ while $a^{-1}=a$ and $b^{-1}=b$, we have $ab=ba$. Choose an <involution> $a$, another $b$ outside $\langle a\rangle$, and then $c$ outside the four-element <subgroup> $\langle a,b\rangle$. The factors commute and each added factor has trivial intersection with the preceding <subgroup>. The <internal direct product theorem> gives $G\cong C_2^3$, an <elementary abelian group>.
Otherwise choose $a$ of order four and set $A=\langle a\rangle$. It has index two, so is normal. Choose $b\notin A$. The two <cosets> show that $a,b$ generate $G$ and that $b^2\in A$. There is no element of order eight, so $b^2$ is either $1$ or $a^2$.
If $G$ is abelian and $b^2=1$, the <internal direct product theorem> yields $G=A\times\langle b\rangle\cong C_4\times C_2$. If instead $b^2=a^2$, let $c=a^{-1}b$; commutativity gives $c^2=1$ and $c\notin A$, producing the same direct product.
If $G$ is nonabelian, <conjugation> by $b$ is a nonidentity <group automorphism> of $A$: otherwise $a,b$ would commute and hence so would all of $G$. Thus $bab^{-1}=a^{-1}$. With $b^2=1$ these are the relations of $D_8$; with $b^2=a^2$ they are the relations of the <quaternion group> $Q_8$. In either presentation every word reduces to $a^ib^j$, $0\leq i<4$, $0\leq j<2$. The standard eight-element model maps onto $G$, and equal orders make that <group homomorphism> an isomorphism.
Finally <group isomorphisms> preserve element orders. The following numbers of elements distinguish all five possibilities:
|| <Group>
|| Order 1
|| Order 2
|| Order 4
|| Order 8
| $C_8$
| 1
| 1
| 2
| 4
| $C_4\times C_2$
| 1
| 3
| 4
| 0
| $C_2^3$
| 1
| 7
| 0
| 0
| $D_8$
| 1
| 5
| 2
| 0
| $Q_8$
| 1
| 1
| 6
| 0
Thus the list is both exhaustive and pairwise non-isomorphic.
Back to article page
