# Algebra. center. prime numbers

## Homework Statement

Suppose $p$ is some prime number, and $G$ a group such that $\# G = p^n$ with some $n\in\{1,2,3,\ldots\}$. Prove that the center

$$Z(G) = \{g\in G\;|\; gg'=g'g\;\forall g'\in G\}$$

contains more than a one element.

## Homework Equations

Obviously $1\in Z(G)$, so the task is to find some other element from there too.

A hint is given, that conjugacy classes

$$[x]=\{x'\in G\;|\; \exists y\in G,\; x'=yxy^{-1}\}$$

are supposed to be examined.

## The Attempt at a Solution

Nothing to be done in sight.

I have some results concerning Sylow p-subgroups, but I don't see how they could be used.

## Answers and Replies

jbunniii
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Hint 1: you don't need anything about Sylow p-subgroups, or anything related to the Sylow theorems.

Hint 2: The word "class" is a hint. Can an element of G be in more than one conjugacy class? Can it be in none?

Hint 3: If g is in Z(G), does that tell you anything about what conjugacy class(es) it lives in?

$x'\in [x]$ is an equivalence relation $x'\sim x$, and always $x\in [x]$. So $G$ is a disjoint union of all conjugacy classes.

$g\in Z(G)$ is equivalent with $[g]=\{g\}$.

So I must prove that there is at least one element $g\neq 1$ so that $[g]=\{g\}$?

jbunniii
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$x'\in [x]$ is an equivalence relation $x'\sim x$, and always $x\in [x]$. So $G$ is a disjoint union of all conjugacy classes.

$g\in Z(G)$ is equivalent with $[g]=\{g\}$.

So I must prove that there is at least one element $g\neq 1$ so that $[g]=\{g\}$?

Yes, that's right. It's basically a counting argument. Add up the sizes of all the conjugacy classes and see what you can conclude.

The remaining thing that you'll need to answer is: what sizes do the conjugacy classes have? For those conjugacy classes that contain more than one element, can you relate them somehow to certain subgroups of G, thereby constraining their sizes?

A natural subset of $G$, related to a given conjugate class $[x]$, would be

$$H([x]) = \{y\in G\;|\;\exists x'\in [x],\; yx'y^{-1}\in [x]\},$$

but I was unable to verify that this is a subgroup. If $y,z\in H([x])$, is it true that $yz\in H([x])$?

Is there some other natural subsets of $G$, related to a given conjugate class $[x]$?

jbunniii
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A natural subset of $G$, related to a given conjugate class $[x]$, would be

$$H([x]) = \{y\in G\;|\;\exists x'\in [x],\; yx'y^{-1}\in [x]\},$$

but I was unable to verify that this is a subgroup. If $y,z\in H([x])$, is it true that $yz\in H([x])$?

Is there some other natural subsets of $G$, related to a given conjugate class $[x]$?

Hint: if you pick two arbitrary elements from the same conjugacy class, what can you say about their centralizers? Can you find a relationship between centralizers and conjugacy classes?

So the correct natural subset of $G$, related to a given conjugate class $[x]$, is

$$Z([x]) = \{g\in G\;|\; gx'=x'g,\;\;\forall x'\in [x]\}?$$

(update: This was the fastest reply from me in this thread so far. It could be that I didn't yet think enough. I'm merely thinking by posting now)

Centralizer of a single element is

$$Z(x) = \{g\in G\;|\; gx=xg\}.$$

If $x,x'$ are from the same conjugate class, then $Z(x)$ and $Z(x')$ have the same size. Is this the relevant remark?

update:

So for all conjugate classes $[x]$ we get a collection of centralizers $(Z(x'))_{x'\in [x]}$, which all share the same size. Is there supposed to be some explicit relation between the sizes of $Z(x)$ and $[x]$?

update:

If $\#Z(x)=p^m$, where $0\leq m\leq n$, then $\# [x] \leq p^n - p^m$.

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jbunniii
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If $x,x'$ are from the same conjugate class, then $Z(x)$ and $Z(x')$ have the same size. Is this the relevant remark?

Yes, that's right.

So for all conjugate classes $[x]$ we get a collection of centralizers $(Z(x'))_{x'\in [x]}$, which all share the same size. Is there supposed to be some explicit relation between the sizes of $Z(x)$ and $[x]$?

Yes. Hint: Can you show that the number of cosets of $Z(x)$ equals the number of elements of $[x]$? Then can you use this fact to constrain the size of $[x]$?

jbunniii
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Not yet. I'm not sure why $(Z(x'))_{x'\in [x]}$ would be disjoint. Are they, and is it relevant?

Well, you can't have disjoint subgroups at all - they all have to contain the identity.

But what I meant was, consider the cosets of $Z(x)$, i.e., $\{gZ(x)\}_{g\in G}$. They are all the same size and they partition $G$. So in particular, the number of cosets must divide $|G|$, which is useful for this problem if you can argue that the number of cosets of $Z(x)$ equals the size of $[x]$. You can do this by trying to find a bijection between the two, or by arguing directly.

You were fast. I don't think that that message was here for more than couple of seconds.

jbunniii
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You were fast. I don't think that that message was here for more than couple of seconds.

Ha, lucky (or unlucky?) timing on my part!

I'm still making no significant progress. Two cosets $gZ(x), g'Z(x)$ are the same if $(g')^{-1}g x = x(g')^{-1}g$.

There seems to be two potentially good looking mappings. One is

$$[x]\to G/Z(x),\quad x'\mapsto x'Z(x),$$

and other one is

$$F\to G/Z(x),\quad y_{x'}\mapsto y_{x'}Z(x),$$

where $F\subset G$ is such set that for all $x'\in [x]$ there is a unique $y_{x'}\in F$ such that $x'=y_{x'}xy_{x'}^{-1}$.

I don't see any reason why either one of these would be injective.

update:

I had a mistake in the first equation originally, namely having $g$ and $(g')^{-1}$ in wrong order. Now it seems, that in fact $F\to G/Z(x)$ is injective.

And surjectivity is clear. I'll continue...

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final strike attempt

So for all $x\in G$ there is $0\leq m(x) \leq n$ such that $\# [x]=p^{m(x)}$.

If $J\subset G$ is collection of representative elements of the conjugate classes, excluding $1\in G$, then

$$\sum_{x\in J} p^{m(x)} = p^n - 1$$

If $m(x)>0$ for all $x\in J$, then the left side is divisible by $p$, but the right side is not. Hence there must exist $x\in J$ such that $m(x)=0$, and then $\#[x] = 1$.

jbunniii
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So for all $x\in G$ there is $0\leq m(x) \leq n$ such that $\# [x]=p^{m(x)}$.

If $J\subset G$ is collection of representative elements of the conjugate classes, excluding $1\in G$, then

$$\sum_{x\in J} p^{m(x)} = p^n - 1$$

If $m(x)>0$ for all $x\in J$, then the left side is divisible by $p$, but the right side is not. Hence there must exist $x\in J$ such that $m(x)=0$, and then $\#[x] = 1$.

Looks good to me!

One last thing you can observe is that $$|Z(G)|$$ equals the number of conjugacy classes of size 1. Then you can neatly summarize the results in the following equation, called the class equation, which often comes in handy when dealing with finite groups:

$$|G| = |Z(G)| + \sum_j |C_j|$$

where the $$C_j$$ are the conjugacy classes with 2 or more elements.

And, as you just showed, $$|C_j|$$ = |G|/|Z(x_j)| for some element $$x_j$$, and so in particular each $$|C_j|$$ divides $$|G|$$.

An easier way to prove the relation between the size of the conjugacy class $$[x]$$ and the centralizer $$Z(x)$$ is to use the orbit-stabilizer theorem. Indeed, $$G$$ defines a group action on itself by conjugation: $$g \cdot x \equiv g x g^{-1}$$ (it's trivial to verify that this actually is a group action). The stabilizer of $$x$$ under this action is $$Z(x)$$ and its orbit is $$[x]$$. Thus, the theorem gives $$|[x]| = [G : Z(G)]$$.