Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last updated: 24 September 2014

Lecture 26: Centres and p-groups

Let p be a prime in >0.

A p-group is a group G such that there exists a>0 with Card(G)=pa.

Let G be a p-group, Card(G)=pa.

(a) G contains an element of order p.
(b) 𝒵(G){1}.


(a) To show: There exists yG with order(y)=1.
Let xG with x1.
Then order(x) divides Card(G) and order(x)1.
So order(x)=pb with 0<ba.
Let y=xpb-1.
Then y1 and yp= (xpb-1)p =xpb=1. So order(y)=p.
(b) To show: 𝒵(G){1}.
We know that 𝒵(G) is the union of the conjugacy classes of size 1.
We know that, if 𝒞s is a conjugacy class in G then Card(𝒞s)= Card(GStab(s)) dividesCard(G)=pa. So, either Card(𝒞s)=1 or Card(𝒞s) is divisible by p.
Then pa=Card(G)= 1+1++1 number of conj.classes of size 1 + conj. classes𝒞swithCard(𝒞s)>1 Card(𝒞s). So (number of conjugacy classes of size 1) is divisible by p.
So Card(𝒵(G)) is divisible by p.
So 𝒵(G){1}.

Let G be a group with Card(G)=p2. Then G is abelian.


To show: 𝒵=𝒵(G) is all of G.
We know, from (b) of the last Proposition, that 𝒵{1}.
We know, Card(𝒵) divides Card(G)=p2.
So Card(𝒵)=p or Card(𝒵)=p2.
Case 1: Card(𝒵)=p.
Let xG with x𝒵.
Then x𝒵 generates G𝒵= { 𝒵,x𝒵,x2𝒵,, xp-1𝒵 } (𝒵 is a normal subgroup of G and G𝒵p).
Let gG. Then there exists 0k<p and z𝒵 with g=xkz. So gx = xkzx= xkxz,sincez𝒵 = xk+1z= x(xkz) =xg. So x𝒵.
This is a contradiction to x𝒵.
So Card(𝒵)p.
Case 2: Card(𝒵)=p2.
Since Card(G)=p2, then 𝒵=G.
So G is abelian.

About conjugacy classes, normal subgroups and centres

(1) Let N be a normal subgroup of G. Then N is a union of conjugacy classes of G.


To show: If nN then 𝒞nN.
To show: If nN and gG then gng-1N.
This is true since N is normal.

(2) Let G be a group and 𝒵=𝒵(G). Then 𝒵 is a normal subgroup of G and if z𝒵 then 𝒞z={z}.


To show: If z𝒵 then 𝒞z={z}.
Assume z𝒵.
To show: 𝒞z={z}. 𝒞z= {gzg-1|gG}= {gg-1z|gG}= {z}.

If 𝒞z={z} then z𝒵.


Assume zG and 𝒞z={z}.
To show: z𝒵.
To show: If gG then gz=zg.
Assume gG.
Then gz=(gzg-1)g=zg, since gzg-1𝒞z={z}.

Notes and References

These are a typed copy of Lecture 26 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on October 5, 2011.

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