There is no group G such that G’=D2n, for n≥3.

Let G be a group. We denote with G’ the commutator subgroup of G and with D2n the diedral group of order 2n, with n a natural number ≥3.

Let us prove that there is no group G such that G’=D2n. Suppose that such a group exists, then since G’=<r,s> we get:

<r> char G’ char G —> <r>◅G. Then there is an action of G on <r> given by

φ: G—->Aut(<r>),
g |—>(n|—>g⁻¹ng)

since <r> is cyclic, Aut(<r>) is abelian, and therefore G/Ker(φ) is abelian, that is Ker(φ)>G’ but then G’ is contained in the centralizer of <r> in G, hence <r>≤Z(G’)=Z(D2n) which imply <r> has order ≤ 2, contradiction. □