= Groups of order at most six
{title2=$C_1,C_2,C_3,C_4,C_2\times C_2,C_5,C_6,S_3$}
Up to <group isomorphism>, the <finite groups> with at most six elements are the <trivial group>, the <cyclic groups> $C_2,C_3,C_4,C_5,C_6$, the <Klein four-group> $C_2\times C_2$, and the <symmetric group> $S_3$.
A <group> of prime <cardinality> is a <cyclic group>, since <Lagrange's theorem> forces any nonidentity element to generate it. For four elements, either an element has <order of a group element> four, or every nonidentity element is an <involution>. In the latter case the <group of exponent two is abelian> identity shows that two distinct nonidentity elements $a,b$ generate $\{e,a,b,ab\}\cong C_2\times C_2$.
For a <group of order six>, either an element generates it or some element $a$ has <order of a group element> three: otherwise all nonidentity elements would be <involutions>, producing a four-element <subgroup> in contradiction with <Lagrange's theorem>. Pairing each element with its <inverse element> gives an <involution> $b$ outside $\langle a\rangle$. The <subgroup> $\langle a\rangle$ has <index of a subgroup> two and is a <normal subgroup>, so $bab^{-1}=a$ or $a^{-1}$. The commuting case makes $ab$ have <order of a group element> six, whereas the second case gives the multiplication rules of $S_3$ on the six distinct elements $a^i,a^ib$, $0\leq i<3$.
All these <groups> embed in $S_4$ except $C_5$ and $C_6$. The former is excluded by <Lagrange's theorem>; the latter is excluded because the <disjoint cycle decomposition> of a <permutation> of four letters cannot have <least common multiple> of cycle lengths equal to six.
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