Commutator

As always, [ilmath]1[/ilmath] and [ilmath]e[/ilmath] will be used to denote the identity of a group.

Definition

Given a group [ilmath](G,\times)[/ilmath] we define the commutator of two elements, [ilmath]g,h\in G[/ilmath] as:

• [math][g,h]=ghg^{-1}h^{-1}[/math] (I use this definition, as does Serge Lang)

Although some people use:

• [math][g,h]=g^{-1}h^{-1}gh[/math]

I prefer and use the version given by Serge Lang, just because it better aligns with alphabetical order, that is to say that [ilmath]g,h[/ilmath] commute is to say [ilmath]gh=hg[/ilmath] (which leads to [ilmath]ghg^{-1}h^{-1}=e[/ilmath]) and [ilmath]hg=gh[/ilmath] while logically equivalent, seems a little bit nastier to write (and leads to [ilmath]hgh^{-1}g^{-1}=e[/ilmath])

Important property

Theorem: The commutator [ilmath][g,h]=e[/ilmath] if and only if the elements [ilmath]g[/ilmath] and [ilmath]h[/ilmath] commute

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To say [ilmath]g,h[/ilmath] commute is to say [ilmath]gh=hg[/ilmath].

Proof:

Proof that [ilmath][g,h]=e\implies gh=hg[/ilmath]

Suppose [ilmath][g,h]=e[/ilmath] then [ilmath]ghg^{-1}h^{-1}=e[/ilmath]

[ilmath]\implies ghg^{-1}=h[/ilmath]

[ilmath]\implies gh=hg[/ilmath]

It is shown that if [ilmath][g,h]=e[/ilmath] then [ilmath]gh=hg[/ilmath] as required

Proof that [ilmath]gh=hg\implies [g,h]=e[/ilmath]

Suppose [ilmath]gh=hg[/ilmath] this

[ilmath]\implies ghg^{-1}=h[/ilmath]

[ilmath]\implies ghg^{-1}h^{-1}=e[/ilmath]

But this is the very definition of the commutator, so:

[ilmath][g,h]=e[/ilmath], as required.

This completes the proof.

See also

• Subgroup
• Group
• Coset
• Normal subgroup