A group G of order 2p, with p prime, contains an element a of order p and an involution (element of order 2) b. The subgroup <a> is normal in G. If the element b commutes with a, then G is cyclic and hence isomorphic to Z/2pZ. If b does not commute with a, then bab^-1 = a^k for some k. But then a = b(bab^-1)b^-1 = b(a^k)b^-1 = (a^k)^k = a^k2. But that means that k² = 1 mod p and hence k = -1 mod p. In particular, bab^-1 = a^-1 and G is isomorphic to D_2p.

We have already seen some groups of order 12. Namely, a cyclic group Z/12Z, a direct product of cyclic groups Z/2Z×Z/2Z×Z/3Z, the dihedral group D₆, the product Z/2Z×S₃ and the alternating group A₄. But there are more groups of order 12.

Let G be the subgroup of SL₂(C) generated by the following two matrices.

A =

x 0 0 x-1

,

B =

0 -1 1 0

,

where x equals (-1 + i3)/2.

The element A is of order 3 and the element B is of order 4. Furthermore, BA = A²B. Hence, every element in G can be written as A^kB^l, where k is in {0,1,2} and l in {0,1,2,3}. In particular, G has order 12.

The group G, denoted by Q₁₂ is not commutative and contains elements of order 4. Hence, it is not isomorphic to one of the examples from the table.

Every group of order 12 is isomorphic to one of the examples described above. We will not prove this, but leave it as a (rather nontrivial) exercise to the reader.