# Linearly primitive implies cyclic-center

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property must also satisfy the second group property
View all group property implications | View all group property non-implications
|
This article describes a fact or result that is not basic but it still well-established and standard. The fact may involve terms that are themselves non-basic
View other semi-basic facts in group theory
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|

## Statement

### Verbal statement

If a group has a faithful irreducible representation over complex numbers, then the center of the group is cyclic.

### Property-theoretic statement

The group property of having a faithful irreducible representation over complex numbers (viz the property of being a linearly primitive group) implies the group property of having a cyclic center (viz the property of being a cyclic-center group).

### Symbolic statement

Let $G$ be a group with a faithful irreducible representation $\rho$ over complex numbers. Then, $Z(G)$ is a cyclic group.

## Generalizations

It is in fact true that if a group has a faithful irreducible representation over any field of characteristic zero, then the center of the group is cyclic.

## Proof

The proof follows from two facts.

### Center goes to scalars

Schur's lemma states that for an irreducible representation $\rho:G \to GL(V)$, the only elements of $GL(V)$ that commute with everything in $\rho(G)$ are the scalar matrices.

Since the image of any element in $Z(G)$ must commute with everything in $\rho(G)$, we conclude that the image of any element in $Z(G)$ is scalar.

In particular, thus, we have a homomorphism from $Z(G)$ to the multiplicative group of complex numbers ( $\mathbb{C}^*$).

In particular, if $\rho$ is faithful, $Z(G)$ is embedded as a subgroup of $\mathbb{C}^*$.

### Any finite subgroup of multiplicative group of complex numbers is cyclic

We know that the finite subgroups of the multiplicative group of complex numbers are precisely the cyclic groups of $d^{th}$ roots of unity. These are always cyclic groups.

Combining these two facts, we obtain that the center is always cyclic.