Equivalence of conjugacy and coset 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
This article gives a proof/explanation of the equivalence of multiple definitions for the term normalizer
View a complete list of pages giving proofs of equivalence of definitions

Statement

The following are equivalent for a subgroup H of a group G and an element g \in G:

  1. gHg^{-1} = H. Here, gHg^{-1} = \{ ghg^{-1} \mid h \in H \} is the conjugate subgroup of H by g.
  2. gH = Hg. Here, gH = \{ gh \mid h \in H\} is the Left coset (?) of g for the subgroup H and Hg = \{ hg \mid h \in H \} is the right coset of g for the subgroup H.

The set of g \in G such that the equivalent conditions (1) and (2) hold is termed the Normalizer (?) of H in G. H is a Normal subgroup (?) of G if the above two conditions hold for all g \in G.

The techniques used in this proof

For a survey article describing these techniques in more detail, see manipulating equations in groups.

Proof

(1) implies (2)

Given: H a subgroup of G, g \in G, and gHg^{-1} = H.

To prove: gH = Hg.

Proof: Informally, we start with:

gHg^{-1} = H

and multiply both on the right by g. We get:

gHg^{-1}g = Hg

which simplifies to:

gH = Hg.

A clearer justification of the manipulation done on the left side can be obtained by looking at things elementwise. We have gHg^{-1} = \{ ghg^{-1}\mid h \in H \}. Thus, (gHg^{-1})g = \{ (ghg^{-1})g \mid h \in H \} = \{ hg : h \in H\}.

(2) implies (1)

Given: H a subgroup of G, g \in G, and gH = Hg.

To prove: gHg^{-1} = H.

Prof: We start with:

gH = Hg

We multiply both sides on the right by g^{-1}, and obtain:

gHg^{-1} = Hgg^{-1}

which simplifies to:

gHg^{-1} = H.

A clearer justification of the manipulation done on the right side can be obtained by looking at things elementwise. We have Hg = \{ hg \mid h \in H \} and (Hg)g^{-1} = \{ (hg)g^{-1} \mid h \in H \} = \{ h \mid h \in H \} = H.