Commutator

As always, $1$ and $e$ will be used to denote the identity of a group.

Definition

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

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

Although some people use:

• $$[g,h]=g^{-1}h^{-1}gh$$

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

Important property

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

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To say $g,h$ commute is to say $gh=hg$.

Proof:

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

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

$\implies ghg^{-1}=h$

$\implies gh=hg$

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

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

Suppose $gh=hg$ this

$\implies ghg^{-1}=h$

$\implies ghg^{-1}h^{-1}=e$

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

$[g,h]=e$, as required.

This completes the proof.

See also

• Commutator subgroup
• Subgroup
• Group
• Coset
• Normal subgroup