# Commutator (group)

*This article refers to commutators of groups, not to be confused with commutator groups.*

In a group, the **commutator** of two elements and , denoted or , is the element . If and commute, then . More generally, , or
It then follows that
We also have
where denote the image of under the inner automorphism , as usual.

## Relations with Commutators

**Proposition 1.** For all in a group, the following relations hold:

- ;
- ;
- ;
- ;
- .

*Proof.* For the first equation, we note that
From the earlier relations,
hence the relation. The second equation follows from the first by passing to inverses.

For the third equation, we define . We then note that By cyclic permutation of variables, we thus find

For the fourth equation, we have The fifth follows similarly.

## Commutators and Subgroups

If and are subgroups of a group , denotes the subgroup generated by the set of commutators of the form , for and .

The group is trivial if and only if centralizes . Also, if and only if normalizes . If and are both normal (or characteristic), then so is , for if is an (inner) automorphism, then Note also that since , .

**Lemma.** Let be a closed subsets of (not necessarily a subgroups); denote by the subgroup of generated by elements of the form , for , . Then .

*Proof.* Let be elements of and be an element of . Then

**Proposition 2.** Let be three subgroups of .

- The group normalizes the group .
- If the group normalizes , then the set of elements , for , , , generates the group .
- If , , and are normal subgroups of , then is a subgroup of the group .

*Proof.* For the first, we note that for any and any ,
by Proposition 1.

For the second, we have for any , , , , Since normalizes , the element lies in . It then follows from induction on that for all , , the element lies in the subgroup generated by elements of the form . Similarly, lies in the subgroup generated by elements of the form ; it then follows that does. Then using the observation we prove by induction on that the element lies in the subgroup generated by elements of the form . This proves the second result.

For the third part, we first note that since are normal subgroups, so are , , and ; in particular, is a group. From the previous result of this proposition, it suffices to show that lies in this group, for all , , and . To that end, we note that since is normal, there exists such that . Then by the third result of Proposition 1, or

\[(a,(b,c)) = \bigl[ (b^c, (c,a')) \cdot (c^{a'}} , (a',b)) \bigr]^{-1} \in (B,(C,A)) \cdot (C, (B,A)) ,\] (Error compiling LaTeX. ! Extra }, or forgotten $.)

as desired.