centralizer (original) (raw)

To illustrate an application of this concept we prove the following lemma.

Proof:
If x,y∈G are in the same right coset, then y=c⁢x for some c∈C⁢(a). Thus y-1⁢a⁢y=x-1⁢c-1⁢a⁢c⁢x=x-1⁢c-1⁢c⁢a⁢x=x-1⁢a⁢x.Conversely, if y-1⁢a⁢y=x-1⁢a⁢x then x⁢y-1⁢a=a⁢x⁢y-1 and x⁢y-1∈C⁢(a) giving x,y are in the same right coset. Let [a] denote the conjugacy class of a. It follows that |[a]|=[G:C(a)] and |[a]|∣|G|.

We remark that a∈Z⁢(G)⇔C⁢(a)=G⇔|[a]|=1, where Z⁢(G) denotes the center of G.

Now let G be a p-group, i.e. a finite group of order pn, where p is a prime and n is a positive integer. Let z=|Z⁢(G)|.Summing over elements in distinct conjugacy classes, we have pn=∑|[a]|=z+∑a∉Z⁢(G)|[a]|since the center consists precisely of the conjugacy classes ofcardinality 1. But |[a]|∣pn, so p∣z. However, Z⁢(G) is certainly non-empty, so we conclude that everyp-group has a non-trivial center.