# Centre of a group Z(G).

1. Feb 27, 2008

### mathusers

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

QUESTION:
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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 $C_{p^2}$ or to $C_p \times C_p$
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WHAT I HAVE SO FAR: PLEASE VERIFY THEM
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(i) is not too hard: we let $x\in \text{Z}(G)$ and so we prove that $gxg^{-1} \in \text{Z}(G)$ for any $g\in G$.

(ii) $G \equiv |Z(G)| (mod p)$ 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.

(v) We let $|G|=p^2$. We choose $a\not = 1$. We form a subgroup $H=\left< a \right>$ if $H = G$. This implies that the group is cyclic and so the proof is complete. If this is not the case then we pick $b\in G\setminus H$ and form $K=\left< b\right>$. This means $H\cap K = \{ 1\}$ which further implies $HK = G$. Also, since the group is abelian $H,K\triangleleft G$. So $G\simeq H\times K \simeq \mathbb{Z}_p \times \mathbb{Z}_p$.

please verify these and help me out with the rest. very many thanks :)

Last edited: Feb 27, 2008
2. Feb 27, 2008

### morphism

(iii) Try proving that if G/Z(G) is cyclic, then G is abelian.

(iv) Show that G/Z(G) is cyclic in this case.

The rest looks good.

3. Feb 28, 2008

### mathusers

(iii) but wouldnt proving the contrapositive be the OPPOSITE of what is required for the question?

4. Feb 28, 2008

### CompuChip

Sort of. Assume you have proven it, then try to prove the original statement by contradiction (suppose it is non-abelian but G/Z(G) is cyclic).

Last edited: Feb 28, 2008
5. Feb 29, 2008

### mathusers

(iii)
ok here is the contrapositive.

we let $H=\text{Z}(G)$.
now if $G/H$ is cyclic then there is $aH$ which generates the group $G/H$. We Let $x,y\in G$. Also, 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. So,$G$ is abelian.

where would you go from here? how would i prove the original statement by contradiction?

6. Feb 29, 2008

### morphism

Isn't it obvious? If G is nonabelian, then there is no way G/Z(G) is cyclic...

7. Feb 29, 2008

### NateTG

Consider a non-abelian group $G$, and assume that $G/Z(G)$ is cyclic by contradiction. (And try to find a contradiction...)

8. Feb 29, 2008

### mathusers

that is merely restating the question..

im fairly new to algebra topics.. ive searched here and there but its still a little hazy..

why cant G/Z(G) be cyclic if G is non abelian? are there any hints or explanations behind the dynamics?? ill attempt the proof myself if i get a clearer backgroup picture of how nonabelian and cycles work? please explain . thnx :)

9. Mar 1, 2008

### CompuChip

You are really overlooking something quite easy, so let me write it out for you.

You have proven what I will call lemma 1: if G/Z(G) is cyclic, then G is abelian.

You want to prove theorem 1: if G is non abelian show that G/Z(G) is not cyclic.

Proof: Let G be a non-abelian group. Suppose that G/Z(G) is cyclic. Then by Lemma 1, G is abelian. So G is both abelian and non-abelian. This is a contradiction, therefore the assumption (namely, that G/Z(G) is cyclic) must have been false. Thus, G/Z(G) must be non-cyclic.

10. Mar 1, 2008

### mathusers

thanks, didnt really expect it to be that easy. i thought there must have been something else to it, but i guess thats maths for us lol

11. Mar 1, 2008

### morphism

12. Mar 1, 2008

### mathusers

sorry i meant (iv), any suggestions there?
i guess it follows on from the previous question...

13. Mar 1, 2008

### morphism

Pretty much. Just combine (ii) and (iii) together and it falls through.

14. Mar 2, 2008

### mathusers

is this correct?
(iv)
Burnside's Lemma states that the center is non-trivial. (using part 2) (i.e. Z(G) $\neq$ {1}), forming the factor group we have a cyclic group. Thus the original group needs to be abelian, including $p^2$.