Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 24 September 2014

Lecture 26: Centres and $p -groups$

Let $p$ be a prime in $ℤ_{> 0} .$

A $p -group$ is a group $G$ such that there exists $a \in ℤ_{> 0}$ with $Card (G) = p^{a} .$

Let $G$ be a $p -group,$ $Card (G) = p^{a} .$

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

Proof.

(a)

To show: There exists

y \in G

with

order (y) = 1 .

Let

x \in G

with

x \neq 1 .

Then

order (x)

divides

Card (G)

and

order (x) \neq 1 .

order (x) = p^{b}

with

0 < b \leq a .

Let

y = x^{p^{b - 1}} .

Then

y \neq 1

and

y^{p} = {(x^{p^{b - 1}})}^{p} = x^{p^{b}} = 1 .

order (y) = p .

(b)

To show:

𝒵 (G) \neq {1} .

We know that

𝒵 (G)

is the union of the conjugacy classes of size

1 .

We know that, if

𝒞_{s}

is a conjugacy class in

G

then

Card (𝒞_{s}) = Card (\frac{G}{Stab (s)}) divides Card (G) = p^{a} .

So, either

Card (𝒞_{s}) = 1

Card (𝒞_{s})

is divisible by

p .

Then

p^{a} = Card (G) = \underset{\underset{\underset{classes of size 1}{number of conj.}}{⏟}}{1 + 1 + \dots + 1} + \sum_{\underset{with Card (𝒞_{s}) > 1}{conj. classes 𝒞_{s}}} Card (𝒞_{s}) .

So (number of conjugacy classes of size 1) is divisible by

p .

Card (𝒵 (G))

is divisible by

p .

𝒵 (G) \neq {1} .

$□$

Let $G$ be a group with $Card (G) = p^{2} .$ Then $G$ is abelian.

Proof.

To show: $𝒵 = 𝒵 (G)$ is all of $G .$
We know, from (b) of the last Proposition, that $𝒵 \neq {1} .$
We know, $Card (𝒵)$ divides $Card (G) = p^{2} .$
So $Card (𝒵) = p$ or $Card (𝒵) = p^{2} .$
Case 1: $Card (𝒵) = p .$
Let $x \in G$ with $x \notin 𝒵 .$
Then $x 𝒵$ generates $\frac{G}{𝒵} = {𝒵, x 𝒵, x^{2} 𝒵, \dots, x^{p - 1} 𝒵}$ $(𝒵$ is a normal subgroup of $G$ and $\frac{G}{𝒵} ≃ \frac{ℤ}{p ℤ}).$
Let $g \in G .$ Then there exists $0 \leq k < p$ and $z \in 𝒵$ with $g = x^{k} z .$ So $\begin{matrix} g x & = & x^{k} z x = x^{k} x z, since z \in 𝒵 \\ = & x^{k + 1} z = x (x^{k} z) = x g . \end{matrix}$ So $x \in 𝒵 .$
This is a contradiction to $x \notin 𝒵 .$
So $Card (𝒵) \neq p .$
Case 2: $Card (𝒵) = p^{2} .$
Since $Card (G) = p^{2},$ 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 $G .$

Proof.

To show: If $n \in N$ then $𝒞_{n} \leq N .$
To show: If $n \in N$ and $g \in G$ then $g n g^{- 1} \in N .$
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 $z \in 𝒵$ then $𝒞_{z} = {z} .$

Proof.

To show: If $z \in 𝒵$ then $𝒞_{z} = {z} .$
Assume $z \in 𝒵 .$
To show: $𝒞_{z} = {z} .$ $𝒞_{z} = {g z g^{- 1} | g \in G} = {g g^{- 1} z | g \in G} = {z} .$

$□$

If $𝒞_{z} = {z}$ then $z \in 𝒵 .$

Proof.

Assume $z \in G$ and $𝒞_{z} = {z} .$
To show: $z \in 𝒵 .$
To show: If $g \in G$ then $g z = z g .$
Assume $g \in G .$
Then $g z = (g z g^{- 1}) g = z g,$ since $g z g^{- 1} \in 𝒞_{z} = {z} .$

$□$

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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