Thread: centre of a group Z(G)

1. centre of a group Z(G)

Hey there, i have a question on the center of a group.

The centre Z(G) of a group G is defined by $Z(G) = g \epsilon G: \forall x \epsilon G, xg = gx$

(i) Show that Z(G) is normal subgroup of G
(ii) By considering the Class Equation of G acting on itself by conjugation show that if $|G| = p^n$ ( p prime) then $Z(G) \neq {1}$

(iii) If G is non abelian show that G/Z(G) is not cyclic.
(iv) Decude that any group of order $p^2$ is abelian.
(V) Deduce that a gorup of oder $p^2$ is isomorhpic either to $C_{p2}$ or to $C_p \times C_p$

any hints would be greatly appreicated and ill atempt the question once i have a clear idea of how to do them. consequently, i will post them online once i have worked them out. cheers guyz

2. Originally Posted by joanne_q
(i) Show that Z(G) is normal subgroup of G
Let $x\in \text{Z}(G)$ prove that $gxg^{-1} \in \text{Z}(G)$ for any $g\in G$.

(ii) By considering the Class Equation of G acting on itself by conjugation show that if $|G| = p^n$ ( p prime) then $Z(G) \neq {1}$
The conjugacy class equation says,
$|G| = |\text{Z}(G)| + \sum [G:\text{C}(x)]$.
Where the $x$'s are taken from distinct conjugacy classes of more than one element.
We know that the left hand side is divisible by $p$ so the right hand side is divisible by $p$. Now $\text{C}(x)$, the centralizer, is not whole $G$ because we are picking those $x$, this means $1\leq |\text{C}(x)| \leq p^{n-1}$. This means $p$ divides the index of $\text{C}(x)$ under $G$. Thus, this proves that $\text{Z}(G)$ is divisible by $p$. Thus, the center is non-trivial.

3. Originally Posted by joanne_q
(iii) If G is non abelian show that G/Z(G) is not cyclic.
Contrapositive is easier. Let $H=\text{Z}(G)$. If $G/H$ is cyclic then there is $aH$ which generates the group $G/H$. Let $x,y\in G$. Note $xH,yH\in G/H$ thus $xH=a^nH$ and $yH=a^mH$. This means $x = a^n z_1$ and $y=a^mz_2$ where $z_1,z_2\in H$. But then $xy = a^n z_1 a^mz_2 = a^{n+m}z_1z_2$ and $yx = a^m z_2 a^n z_1 = a^{n+m}z_1z_2$ because $z_1,z_2$ commute with everything. Thus $G$ is abelian.

(iv) Decude that any group of order $p^2$ is abelian.
By Burnside's lemma (that is (ii)) we have that the center is non-trivial, forming the factor group we have a cyclic group. Thus the original needs to be abelian.

4. Originally Posted by joanne_q
(V) Deduce that a gorup of oder $p^2$ is isomorhpic either to $C_{p2}$ or to $C_p \times C_p$
Let $|G|=p^2$. Pick $a\not = 1$. Form the subgroup $H=\left< a \right>$ if $H = G$ then the group is cyclic and proof is complete. Otherwise choose $b\in G\setminus H$ and form $K=\left< b\right>$. Now $H\cap K = \{ 1\}$ this means $HK = G$*. Also $H,K\triangleleft G$ because the group is abelian. This means $G\simeq H\times K \simeq \mathbb{Z}_p \times \mathbb{Z}_p$.**

*)Because $|HK| = |G|$ by using the fact $|HK||H\cap K| = |H||K|$.

**)Theorem: If $H,K$ are normal subgroups with $H\cap K = \{ 1 \}$ and $HK = G$ then $G\simeq H\times K$.

5. Thanks a lot for the help really appreciate it. makes it easier to understand the subject aswell.

For part (ii), your solution was:
Originally Posted by ThePerfectHacker
The conjugacy class equation says,
$|G| = |\text{Z}(G)| + \sum [G:\text{C}(x)]$.
Where the $x$'s are taken from distinct conjugacy classes of more than one element.
We know that the left hand side is divisible by $p$ so the right hand side is divisible by $p$. Now $\text{C}(x)$, the centralizer, is not whole $G$ because we are picking those $x$, this means $1\leq |\text{C}(x)| \leq p^{n-1}$. This means $p$ divides the index of $\text{C}(x)$ under $G$. Thus, this proves that $\text{Z}(G)$ is divisible by $p$. Thus, the center is non-trivial.

here is my solution, is it correct aswell?

$G \equiv |Z(G)| (mod p)$ since Z(G) is a fixed point set.
Now $|Z(G)| \equiv p^n(mod p)$, |Z(G)|=0.
So Z(G) has atleast p elements.

6. for part (iii) i believe you have proved the opposite of the question? i.e. G/Z(G) is cyclic and thus abelian...

to prove that it is non cyclic, do all the points you stated have to be contradicted..?
Originally Posted by ThePerfectHacker
Contrapositive is easier. Let $H=\text{Z}(G)$. If $G/H$ is cyclic then there is $aH$ which generates the group $G/H$. Let $x,y\in G$. Note $xH,yH\in G/H$ thus $xH=a^nH$ and $yH=a^mH$. This means $x = a^n z_1$ and $y=a^mz_2$ where $z_1,z_2\in H$. But then $xy = a^n z_1 a^mz_2 = a^{n+m}z_1z_2$ and $yx = a^m z_2 a^n z_1 = a^{n+m}z_1z_2$ because $z_1,z_2$ commute with everything. Thus $G$ is abelian.

7. Originally Posted by joanne_q
For part (ii), your solution was
Have you ever done the conjugacy class equation? There is a way around it if you never done it that way, I think I have an idea of what you might have done.

Let $G$ be a finite $p$-group. Let $G$ act on a non-empty finite set $X$. Then $|X|\equiv |X^G|(\bmod p)$ where $X^G$ is the invariant subset fixed by $G$.

here is my solution, is it correct aswell?

$G \equiv |Z(G)| (mod p)$ since Z(G) is a fixed point set.
Now $|Z(G)| \equiv p^n(mod p)$, |Z(G)|=0.
So Z(G) has atleast p elements.
Let $G$ act on itself by conjugation (i.e. $X=G$ and $g*x = gxg^{-1}$). Also $G$ is a finite $p$-group which fits the above result thus $|G| \equiv |G^G| (\bmod p)$ but $G^G = \text{Z}(G)$ because that is the subset left fixed under conjugation. Thus, $|G| \equiv |\text{Z}(G)|(\bmod p)$ which means the center needs to be divisible by $p$, i.e. it cannot be trivial.

8. Originally Posted by joanne_q
for part (iii) i believe you have proved the opposite of the question? i.e. G/Z(G) is cyclic and thus abelian...

to prove that it is non cyclic, do all the points you stated have to be contradicted..?
No, there is no contradiction argument. I proved the contrapositive statement.