# Equivalence of conjugacy and commutator definitions of normality

This article gives a proof/explanation of the equivalence of multiple definitions for the term normal subgroup

View a complete list of pages giving proofs of equivalence of definitions

## Statement

Suppose is a subgroup of a group and is an element. Then the following are equivalent:

- .
- The commutator , where is the subgroup generated by .

If these equivalent conditions hold for all , then is a normal subgroup of . Note that the second condition on all would translate to saying that is contained in .

Note that is a *weaker* condition than for a *particular* .

## More on the proof techniques

The survey article manipulating equations in groups discusses similar proof techniques involving equations that deal with elements and subsets of groups.

## Proof

### (1) implies (2)

**Given**: Group , subgroup , such that .

**To prove**: .

**Proof**: Since is a subgroup, it suffices to show that all the generators of are in , i.e., that for every . For this, note that:

By assumption, since , and so the quotient is also in . This completes the proof.

### (2) implies (1)

**Given**: Group , subgroup , such that .

**To prove**: .

**Proof**: For any , we have:

By assumption, both and are in , hence .