Equivalence of conjugacy and commutator definitions of normality

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This article gives a proof/explanation of the equivalence of multiple definitions for the term normal subgroup
View a complete list of pages giving proofs of equivalence of definitions


Suppose H is a subgroup of a group G and g \in G is an element. Then the following are equivalent:

  1. gHg^{-1} \subseteq H.
  2. The commutator [g,H] \subseteq H, where [g,H] is the subgroup generated by [g,h] = ghg^{-1}h^{-1}, h \in H.

If these equivalent conditions hold for all g \in G, then H is a normal subgroup of G. Note that the second condition on all g \in G would translate to saying that [G,H] := \langle [g,h] \mid g \in G, h \in H\rangle is contained in H.

Note that gHg^{-1} \subseteq H is a weaker condition than gHg^{-1} = H for a particular g \in G.

More on the proof techniques

The survey article manipulating equations in groups discusses similar proof techniques involving equations that deal with elements and subsets of groups.


(1) implies (2)

Given: Group G, subgroup H, g \in G such that gHg^{-1} \subseteq H.

To prove: [g,H] \subseteq H.

Proof: Since H is a subgroup, it suffices to show that all the generators of [g,H] are in H, i.e., that [g,h] \in H for every h \in H. For this, note that:

\! [g,h] = ghg^{-1}h^{-1} = (ghg^{-1})h^{-1}

By assumption, since h \in H ghg^{-1} \in H, and so the quotient (ghg^{-1})h^{-1} is also in H. This completes the proof.

(2) implies (1)

Given: Group G, subgroup H, g \in G such that [g,H] \subseteq H.

To prove: gHg^{-1} \subseteq H.

Proof: For any h \in H, we have:

\! ghg^{-1} = (ghg^{-1}h^{-1}h = [g,h]h

By assumption, both [g,h] and h are in H, hence ghg^{-1} \in H.