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


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][1] (I use this definition, as does Serge Lang)

Although some people use:

  • [math][g,h]=g^{-1}h^{-1}gh[/math][2]

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

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


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


  1. Serge Lang - Algebra - Revised Third Edition - GTM