# 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]
^{[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:**

- 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

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

- 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.

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

This completes the proof.

## Identities

## See also

## References

- ↑ Serge Lang - Algebra - Revised Third Edition - GTM
- ↑ http://en.wikipedia.org/w/index.php?title=Commutator&oldid=660112221#Group_theory