# Every group is normal in itself

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

View all subgroup metaproperty satisfactions | View all subgroup metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for subgroup properties

Get more facts about normal subgroup |Get facts that use property satisfaction of normal subgroup | Get facts that use property satisfaction of normal subgroup|Get more facts about identity-true subgroup property

## Contents

## Statement

### Statement with symbols

Suppose is a group. Consider as a subgroup of itself -- is a normal subgroup of itself.

## Related facts

### Opposite facts

### Similar facts

- Invariance implies identity-true
- Balanced implies identity-true
- Every group is characteristic in itself
- Trivial subgroup is normal

## Proof

### Proof using the conjugation definition of normality

**Given**: A group .

**To prove**: For any and , .

**Proof**: This is obvious from the fact that is closed under multiplication.

### Proof using the kernel of homomorphism definition of normality

**Given**: A group .

**To prove**: There exists a homomorphism for some group such that every element of maps to the identity element.

**Proof**: Let be the trivial group and be the map sending every element of to the identity element of . Clearly, satisfies the conditions for being a homomorphism: for any , both and equal the identity element of . Moreover, every element of is sent to the identity element of under .

### Proof using the cosets definition of normality

**Given**: A group .

**To prove**: For every element , .

**Proof**: Note that

- : Clearly, . Also, for any , , so . Thus, .
- : Clearly, . Also, for any , , so . Thus, .

Combining the two steps, we obtain that .

### Proof using the union of conjugacy classes definition of normality

**Given**: A group .

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

**Proof**: The conjugacy classes form a partition of (arising from the equivalence relation of being conjugate), so is their union.

### Proof using the commutator definition of normality

**Given**: A group .

**To prove**: For every , .

**Proof**: This is direct from the fact that is closed under multiplication and inverses, and the commutator is defined in terms of these operations.