Commutator subgroup G’ is the unique smallest normal subgroup N such that G/N is abelian.

Commutator subgroup G' is the unique smallest normal subgroup N such that G/N is abelian.

If G is a group, then G' is a normal subgroup of G and G/G' is abelian. If N is a normal subgroup of G, then G/N is abelian iff N contains G'.

Proof

Let f:G\to G be any automorphism. Then \displaystyle f(aba^{-1}b^{-1})=f(a)f(b)f(a)^{-1}f(b)^{-1}\in G'.

It follows that f(G')\leq G'. In particular, if f is the automorphism given by conjugation by a\in G, then aG'a^{-1}=f(G')\leq G', so G'\unlhd G.

Since (ab)(ba)^{-1}=aba^{-1}b^{-1}\in G', abG'=baG' and hence G/G' is abelian.

(\implies) If G/N is abelian, then abN=baN for all a,b\in G. Hence ab(ba)^{-1}=aba^{-1}b^{-1}\in N. Therefore, N contains all commutators and G'\leq N.

(\impliedby) If G'\subseteq N, then ab(ba)^{-1}=aba^{-1}b^{-1}\in G'\subseteq N. Thus abN=baN for all a,b\in G. Hence G/N is abelian.

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