# Thread: group of order 64 cannot have a trivial center

1. ## group of order 64 cannot have a trivial center

I have to prove that a group G of order 64 cannot have a trivial center.

In the case that G is Abelian, this is trivial, as the center of Abelian group is the group itself.

What to do in the case that G is non-Abelian?

I've found this Center of Group of Prime Power Order is Non-Trivial, - ProofWiki but the proof uses many things we have not discussed in class so there has to be another way.

2. For each $x\in G$, the centralizer of $x$ in $G$ is the subgroup $C_G(x)=\{g\in G:gxg^{-1}=x\}.$

Define a relation $\sim$ on $G$ by $x\sim y\ \Leftrightarrow\ \exists\,g\in G$ such that $gxg^{-1}=y.$ Then $\sim$ is an equivalence relation, and so partitions $G$ into equivalence classes, called conjugacy classes. The conjugacy class containing $x$ is the set $\{gxg^{-1}:g\in G\}.$ The number of elements in this conjugacy class is equal to the number of left cosets of $C_G(x)$ in $G$ because the mapping $gxg^{-1}\mapsto gC_G(x)$ is a bijection.

Note that the conjugacy class containing the identity $1_G$ is always $\{1_G\}.$ If there are $r$ distinct conjugacy classes not containing the identity, let $x_1,\ldots,x_r$ be representatives from each of them. Then

$64\ =\ |G|\ =\ 1+\sum_{i\,=\,1}^r\left|G:C_G(x_i)\right|$

If there is no $x_i$ such that $\left|G:C_G(x_i)\right|=1$ then the RHS would be odd since $\left|G:C_G(x_i)\right|$ divides 64 and so must be a power of 2. It follows that there must be some $x_i$ such that $\left|G:C_G(x_i)\right|=1.$

Hence $C_G(x_i)=G$

$\Rightarrow\ gx_ig^{-1}=x_i$ for all $g\in G$

$\Rightarrow\ x_i\in Z(G)$

$\Rightarrow\ Z(G)$ is not trivial

3. I would have never figured this one out myself!
Thank you!!!

4. You’re welcome. But, believe it or not, the proof I gave is more or less the same as the proof given on Wikipedia. I merely rewrote it in what was probably an easier way to follow (I also prefer the term “centralizer” to “normalizer”).

5. Originally Posted by georgel
I have to prove that a group G of order 64 cannot have a trivial center.

In the case that G is Abelian, this is trivial, as the center of Abelian group is the group itself.

What to do in the case that G is non-Abelian?

I've found this Center of Group of Prime Power Order is Non-Trivial, - ProofWiki but the proof uses many things we have not discussed in class so there has to be another way.

As JaneBennet mentioned, the class equation is
$64\ =\ |G|\ =\ |Z(G)|+\sum_{i\,=\,1}^r\left|G:C_G(x_i)\right|$
Since $C_G(x_i)$ is a subgroup of G, $|G:C_G(x_i)| > 1$ divides $2^6$ (by Lagrange's theorem).
We know that |Z(G)| >= 1 and 2 divides each $|G:C_G(x_i)|$. That implies |Z(G)| has at least 2 elements.

6. Originally Posted by JaneBennet
You’re welcome. But, believe it or not, the proof I gave is more or less the same as the proof given on Wikipedia. I merely rewrote it in what was probably an easier way to follow (I also prefer the term “centralizer” to “normalizer”).
It was definitely easier to follow! I got lost on Wikpedia, with all the clicking around to se what was what, because, as I said, we didn't mention a lot of the terms and theorems used there. Not yet, at least.

Thanks again!

Now, off to another problem.

7. Originally Posted by aliceinwonderland
As JaneBennet mentioned, the class equation is
$64\ =\ |G|\ =\ |Z(G)|+\sum_{i\,=\,1}^r\left|G:C_G(x_i)\right|$
Since $C_G(x_i)$ is a subgroup of G, $|G:C_G(x_i)| > 1$ divides $2^6$ (by Lagrange's theorem).
We know that |Z(G)| >= 1 and 2 divides each $|G:C_G(x_i)|$. That implies |Z(G)| has at least 2 elements.
People here are just great!
Thank you!