# Central implies normal

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., central subgroup) must also satisfy the second subgroup property (i.e., normal subgroup)

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## Statement

Suppose is a subgroup of that is a central subgroup of , i.e., is contained in the center of . Then, is a normal subgroup of .

## Related properties

### Intermediate properties

Property | Proof that central subgroup implies it | Proof that it does not imply central | Proof that it implies normal subgroup | Proof that normal subgroup does not imply it |
---|---|---|---|---|

central factor | central implies central factor | central factor not implies central | central factor implies normal | normal not implies central factor |

hereditarily normal subgroup | central implies hereditarily normal | hereditarily normal not implies central | (immediate) | (via transitively normal) |

transitively normal subgroup | (via hereditarily normal) | (via hereditarily normal) | (immediate) | normality is not transitive |

abelian normal subgroup | central implies abelian and central implies normal | abelian normal not implies central | (immediate) | every group is normal in itself, combined with existence of non-abelian groups. |

Here's a more complete list: Abelian normal subgroup, Amalgam-characteristic subgroup, Amalgam-strictly characteristic subgroup, Center-fixing automorphism-invariant subgroup, Central factor, Class two normal subgroup, Commutator-in-center subgroup, Conjugacy-closed normal subgroup, Dedekind normal subgroup, Direct factor over central subgroup, Hereditarily normal subgroup, Join-transitively central factor, Nilpotent normal subgroup, SCAB-subgroup, Transitively normal subgroup|FULL LIST, MORE INFO

## Related facts

### Converse

The converse is not true in general: normal not implies central.

However, some versions are true:

- Normal of least prime order implies central
- Totally disconnected and normal in connected implies central
- Cartan-Brauer-Hua theorem

## Proof

### Proof using coset definition of normality

**Given**: A group , a central subgroup of .

**To prove**: for all (the cosets definition of normality).

**Proof**: Since is central, this means that for all , . Thus, for a fixed , the sets and are equal, because each equals the corresponding .

### Proof using conjugation definition of normality

**Given**: A group , a central subgroup of .

**To prove**: For all and , we have .

**Proof**: Since is central, we have, by definition, that for all . Multiplying both sides on the right by , we obtain that for all . Since by assumption, and , we obtain that .

### Proof using commutator definition of normality

**Given**: A group , a central subgroup of .

**To prove**: For all , , we have .

**Proof**: Since we have by definition. Multiplying both sides by on the right, we get (i.e., it is the identity element). Since any subgroup contains the identity element, , so **Failed to parse (syntax error): ghg^{-1}{h^{-1} \in H**
.

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

**Given**: A group , a central subgroup of .

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

**Proof**: Every element of the center of forms a conjugacy class of size 1. Since comprises only central elements, it is the union of these singleton conjugacy classes.