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

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:

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

• It looks as if you are asked to show there is an element other than the identity in the centre, that the centre has more than one element. Aug 17, 2015 at 2:25
• @AndréNicolas nice! Any hints in how this is related to the element having order multiple of a prime? Aug 17, 2015 at 2:28
• @GuerlandoOCs Power, not multiple. Aug 17, 2015 at 2:41
• This is a standard theorem proved in almost any text on group theory (in the one standard, textbook way). If you really want an intuitive understanding of this fact, you absolutely need to have an understanding of group actions and the orbit-stabilizer theorem. Aug 23, 2015 at 6:16

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.

• I should note the equation $(\ast)$ is the famous class equation, which is the way I've usually seen this result derived. Aug 23, 2015 at 6:06
• why $|Z(G)|$ is the sum of the orders of all singletons? Aug 23, 2015 at 19:04
• @GuerlandoOCs Those aren't just any singletons, those are the singleton conjugacy classes, which are precisely the elements of $Z(G)$ (see the second paragraph of the answer). So adding them up is the same as counting the elements of $Z(G)$. Aug 23, 2015 at 19:09
• I'm 90% at understanding this. Could you please be more clear in the part that you say $Gx_i=C_i$? Aug 23, 2015 at 19:40
• I'm just confused because my book never mentioned the orbit stabilizer theorem, he just mentions this exercise: math.stackexchange.com/questions/1400267/… and then asks for this one. Is there any relation between these two that can take a shortcut and not use the orbit stabilizer theorem? Aug 23, 2015 at 19:47

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 . . .)

• Maybe not a typo, could be unidiomatically translated from Spanish. Aug 17, 2015 at 2:41
• Noah, somehow your answer looks like the C.E. in words . . . or am I missing something? Cheers! Aug 17, 2015 at 2:42
• I would still call that a typo, but maybe I'm defining "typo" too broadly. Aug 17, 2015 at 2:42
• @RobertLewis Sorry, what is "the C.E."? Aug 17, 2015 at 2:43
• @GuerlandoOCs That is a routine thing to do that you should really try to work out for yourself. Just find other equivalent ways of writing that identity and interpret them. Aug 23, 2015 at 21:03

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 $|Z(G)|\geq p$