Determining the number of conjugacy classes in a p-group

[jstrunk]

TL;DR Summary

I can never do it

The book I am using has a number of exercises like this:
If G is a p-group, show that G cannot have exactly p+1 conjugacy classes.
If G is a p-group with p+2 conjugacy classes, show that the order of G is 4.
I can never solve any problem that involves connecting the number of conjugacy classes to the order of a p-group.
The author seems to think he showed how to do it, but if so, it went right over my head.
Please don't make me show my work.
I have spent a month on just these two problems and gotten nowhere and given myself mental block.
Is there any place I can find a clear explanation of this at a introductory level?
At the point I am in my book, Sylow subgroups have been introduced but we haven't done the Sylow Theorems yet,
so that should give you an idea of what I mean by an introductory level.

 

[fresh_42]

Did you understand the orbit-stabilizer-formula?
https://en.wikipedia.org/wiki/Group_action#Orbit-stabilizer_theorem

Or how equivalence relations divide a group into orbits of equal size?

 

[jstrunk]

Orbit-Stabilizer Theorem:
For an action of group G on set S |orb(s)|=[G:stab(s)]=|G|/|stab(s)| for any s element of S.
For the action of conjugation, this becomes |cls(s)|=[G:CG(s)]|G|/|cls(s)|, CG(s) being the centralizer of s in G.

Equivalence relations divide a group into orbits of equal size:
Well, this part I don't get.
I know that any group action such as conjugation is an equivalence relation.
I know that cosets partition a group into sets of equal size.
But I didn't think that was a universal property of equivalence relations.
I thought conjugacy classes partitioned the group but that the partitions could be of different sizes.
In fact, the conjugacy class of the Identity is of order 1, so unless all conjugacy classes are of order 1,
I don't see how that could be true.

 

[Stephen Tashi]

But I didn't think that was a universal property of equivalence relations.

Considering a group merely as a set, you could define an equivalence relation on it that partitions it in an arbitrary way, even in a way that divides it into subsets of unequal size. However, to say " '$\equiv$' is an equivalence relation on a group" implies the equivalence relation also satisfies: (for each $a,b,x$ in the group) if $a \equiv b$ then $ax \equiv bx$. Sometimes an equivalence relation with such added conditions is called a "congruence".

 

[fresh_42]

Right. Arbitrary equivalence relations do not partition sets into classes of equal size. But in a group we can jump from class to class via multiplication. The concept for groups involves the structure of groups.

Let's start with a finite group $G$ and a subgroup $U$. Then we consider $G/U$ by defining $a\sim b$ iff $a^{-1}b\in U$ or $aU=bU.$ Then all classes have the size of $U$. In case $U=\{e\}$ we have as many classes as we have elements of $G$. In general we have the entire subgroup $U$ as neutral class in $G/U$, which is a group again, if and only if $U$ is a normal subgroup. But we do not need a group structure on $G/U$ for the consideration of equivalence classes.

Now if we start with conjugation and not with a subgroup. Then we define $x\sim y$ iff there is an element $g\in G$ such that $g^{-1}xg=y$. The class of $e$ is thus
$$
\{x\sim e\} = \{x\in G\,|\,\exists \,g\in G: g^{-1}xg=e \Longleftrightarrow x=e\}=\{e\}
$$
So you are right. We get different possible sizes for $x=e$ and $x\neq e.$ The stabilizer-orbit-theorem quantifies this relation. If $x=e$ then $|G/G_e|=|G.\{e\}|=|\{e\}|=1$ and $|G_e|=|G|.$ But $x=e$ is the exception. All other orbits $G.x=\{g^{-1}xg\,|\,g\in G\}$ are of equal size, namely the size of the factor group $G/G_x$ along the stabilizer subgroup of $x$, where $U=G_x$ as above in that case.

 

[jstrunk]

OK, thank you. I had it in my head that the conjugacy classes could be all different sizes and I had to look for clues work out the size of each one individually.

You said that x=e is the exception. Wouldn't it be better to say that any element of Z(G) is the exception, since
if x is an element of Z(G) then cls(x)={x}? There will be one conjugacy class for every element of Z(G).
And if I can find the center for any x not an element of Z(G), G/CG(x) will give me size of all the other conjugacy classes

Anyway, I will now go try these exercises again and see if this helps.

 

[fresh_42]

If you look at conjugation alone, then you should see it as a group homomorphism:
$$
\varphi \, : \,G\longrightarrow \operatorname{Inn}(G)\, , \,g\longmapsto (x\longmapsto g^{-1}xg)
$$
Then $Z(G)=\operatorname{ker}\varphi$ and $G/Z(G)$ is of interest, i.e. the equivalence classes on the set of conjugations rather than a single one. If we have only a single conjugation $\iota_g$ of $G$, then we look at orbits and stabilizers. From this point of view the group is merely a set which $\iota_g$ operates on. If we look at the set $\operatorname{Inn}(G)=\{\iota_g\,|\,g\in G\}$ of all such operators, then $G$ is still only a set where they act upon: $\operatorname{Inn}(G)\times G \longrightarrow G$. The group structure of $G$ is irrelevant here, since we consider the group $\operatorname{Inn}(G)$ instead.

It is confusing because $\operatorname{Inn}(G)$ and $G$ are of course related, so it is their different roles as set on which we operate and as pool of available operators via conjugation which makes it messy.