# Trivial subgroup is normal

This article gives the statement, and possibly proof, of a subgroup property (i.e., normal subgroup) satisfying a subgroup metaproperty (i.e., trivially true subgroup property)

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## Contents

## Statement

Let be any group. The trivial subgroup of , which is the one-element subgroup comprising the identity element, is a normal subgroup of .

## Related facts

## Proof

### Conjugation definition of normality

**Given**: A group with identity element . An element .

**To prove**: .

**Proof**: We have , completing the proof. `Further information: manipulating equations in groups`

### Kernel of homomorphism definition of normality

**Given**: A group with identity element .

**To prove**: There exists a homomorphism for some group such that the kernel of is precisely the subgroup .

**Proof**: Take to be the identity map (i.e., for all ). This clearly satisfies the conditions for a homomorphism: , , and .

The kernel of is defined as the set of elements that map to . But since this is the identity map, the set is precisely , completing the proof.

### Coset definition of normality

**Given**: A group with identity element .

**To prove**: For any , .

**Proof**: Both are clearly the same as the singleton set , because .

### Union of conjugacy classes definition of normality

**Given**: A group with identity element

**To prove**: is a union of conjugacy classes.

**Proof**: In fact, is a single conjugacy class, because for any , .

### Commutator definition of normality

**Given**: A group , with identity element .

**To prove**: If , then is contained in .

**Proof**: is the subgroup generated by elements of the form , where . But . Thus, all elements of the form equal , so the subgroup generated is equal to .