$G$ has order $p^a$, then the center of $G$ counts more than the identity

Question

This question makes no sense to me. I don't know what it means by "counts more than the identity". Then, the exercise gives me a hint: "break G into equivalence classes of conjugacy elements and see that an element belongs to the center iff it is its own conjugate"

I understand that, by the definition, an element $x$ belongs to the center of $G$ is, for all $g\in G$, we have that $xg = gx$. This implies that $x = gxg^{-1}$. I think the exercise is asking me to break each $x$ like this. But then I don't know what it is asking for, and I know if you guys could understand it by the title, but maybe the hint tells about what's expected to be the answer. Could somebody help me?

UPDATE:

With a GREAT help of Ben West in the answers, I was able to synthetize this proof by using a result previously asked by my book, so I didn't need to use the orbit stabilizer theorem as he did, so I'll leave this proof here:

Lets take all the conjugacy classes of $G$, that is:

$$Gx = \{g\cdot x: g \in G\} = \{gxg^{-1}: g\in G\}$$

Now, observe that $x\in Z(G)$ if and only if $gx=xg$, that is, $gxg^{−1} =x$ for all $g∈G$ . So $x\in Z(G)$ if and only if its conjugacy class is the singleton $\{x\}$. Then, there will be two cases: the conjugacy classes that have only $1$ element, here called $g_k$, and the ones that have more than $1$ element, here called $C_G(x)$. Since the conjugacy classes partition the group, we must have:

$$|G| = \sum |g_k| + \sum |C_k|$$

but $\sum |g_k|$ is exactly |$\text{Center}(G)$|, and $C_k(x)$ is the $k$-$th$ conjugacy class with more than $1$ element.

By this question (which were presented for me before this exercise, as a hint), we have that the cardinality of the conjugacy class is

$|C_k|=|G|/|N_G(x)| \implies |C_k||N_G(x)| = |G|$

Therefore, we have that:

$$p^a = |G| = |Z(G)| + \sum |C_k|$$ Since $|C_k|$ divides the order of the group, looking at the equation $\text{mod $p$}$ gives us:

$$0 = |Z(G)| \ \ mod \ p \implies Z(G) \text{ contains more than the identity}$$

Curiosity:

For just one element, the normalizer is the equivalence class:

$N_G(x)=\{g\in G:gxg^{-1}\subseteq\{x\}\}=\{g\in G:gxg^{-1}=x\}=C_G(x)$

Answer 1

Note that we can define an equivalence relation on $G$ by $g\sim h$ if $g$ and $h$ are conjugate. This implies that the conjugacy classes partition the group $G$.

Observe that $x\in Z(G)$ if and only if $gx=xg$, that is, $gxg^{-1}=x$ for all $g\in G$. So $x\in Z(G)$ if and only if its conjugacy class is the singleton $\{x\}$.

Let $G$ act on itself by conjugation, that is, we define the action $g\cdot x=gxg^{-1}$. For $x\in G$, the orbit $$ Gx=\{g\cdot x:g\in G\}=\{gxg^{-1}:g\in G\} $$ is precisely the conjugacy class containing $x$, by definition more or less.

For good measure, the stabilizer $G_x$ of $x$ is $$ G_x=\{g\in G:g\cdot x=x\}=\{g\in G:gxg^{-1}=x\}=C_G(x), $$ so the stabilizer of $x$ is just the centralizer in this case.

By the Orbit=Stabilizer Theorem, we have $$ |Gx|=\frac{|G|}{|G_x|}=\frac{|G|}{|C_G(x)|}. $$

Now write $Z(G)=\{e,g_2,\dots,g_k\}$, it is precisely the union of all conjugacy classes which are singletons, as observed above. Let $C_1,\dots,C_m$ be the remaining conjugacy classes which are not singletons, so that $\{e\},\{g_2\},\dots,\{g_k\},C_1,\dots,C_m$ are all the conjugacy classes of $G$, hence partition $G$.

This means $$ |G|=|\{e\}|+|\{g_2\}|+\cdots+|\{g_k\}|+\sum_{i=1}^m |C_i|=|Z(G)|+\sum_{i=1}^m|C_i|\qquad(\ast). $$

Pick $x_i\in C_i$. So $Gx_i=C_i$, since the conjugacy class of $x_i$ is just its orbit under the conjugation action. So by Orbit-Stabilizer, $$ |C_i|=|Gx_i|=\frac{|G|}{|C_G(x_i)|} $$ implying $|C_i||C_G(x_i)|=|G|$, which means $|C_i|$ divides $|G|$. Note that nothing here has assumed $|G|=p^a$ yet. Now assume $|G|=p^a$, hence each $|C_i|$ is a power of $p$ itself, not equal to $1$ since we have $|C_i|>1$ by definition.

Reducing the equation $(\ast)$ modulo $p$ yields $$ 0\equiv |Z(G)|\pmod{p}. $$

It follows that $Z(G)\neq\{e\}$, because $0\not\equiv 1\pmod{p}$. So $Z(G)$ must contain more than the identity.

Answer 2

I imagine "counts" is a typo for "contains."

As to the problem itself: if $G$ is a group, a conjugacy class in $G$ is a set of the form $C_z=\{x: \exists y(x=yzy^{-1})\}$ for some $z\in G$. You know that the following are equivalent:

• $z$ is in the center of $G$;

• the conjugacy class $C_z$ of $z$ has exactly one element.

Moreover, you know that the center has at least one element, the identity - that is, there is at least one singleton conjugacy class. And the group itself is a disjoint union of conjugacy classes.

Finally, you should also know that the size of a conjugacy class divides the order of the whole group.

So the question is: can you put these things together? Suppos $G$ has order $p^a$ for some prime $p$, and $a\not=0$. Since the size of each conjugacy class divides $p^a$, we know that each conjugacy class either has size a multiple of $p$, or has exactly one element. Now, what happens if there is only one conjugacy class with exactly one element? (HINT: add the sizes of the conjugacy classes mod p . . .)

Answer 3

Proof: $G$ acts on $\Omega := N$ by conjugation, where $N$ is a normal subgroup of $G$, and $C_{\Omega}(P) = Z(P) \cap N$. Since $N$ is a $p$-group:

${|C_{\Omega}(P)| \equiv |Ω| \equiv 0} {\mod p}$

Now $1\in C_{\Omega}(P)$ gives $| C_{\Omega}(P)| \geq p$. In particuar, let $N=G$ means $|Z(G)|\geq p$