# Centre of a group Z(G).

• mathusers
In summary, the centre of a group is defined by the equation Z(G) = g \epsilon G. If G is non-abelian, then Z(G) is not cyclic.
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 :)

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(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.

(iii) but wouldn't proving the contrapositive be the OPPOSITE of what is required for the question?

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).

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(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?

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

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

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

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

that is merely restating the question..

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

why can't 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 :)

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.

CompuChip said:
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.

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

mathusers said:

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

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

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 $p^2$.

## 1. What is the Centre of a group Z(G)?

The Centre of a group Z(G) is the set of all elements in the group that commute with every other element in the group. In other words, it is the set of elements that can be switched with any other element in the group without changing the result.

## 2. How is the Centre of a group Z(G) related to the identity element?

The identity element is always included in the Centre of a group Z(G) because it commutes with every other element in the group. However, the Centre may also include other elements depending on the group's structure.

## 3. Can the Centre of a group Z(G) be empty?

Yes, the Centre of a group Z(G) can be empty. This occurs when no elements in the group commute with every other element, resulting in an empty set.

## 4. How does the Centre of a group Z(G) relate to the group's normal subgroups?

The Centre of a group Z(G) is always a subgroup of the original group G. However, it may not always be a normal subgroup. In fact, the Centre is a normal subgroup if and only if the group is abelian.

## 5. How can the Centre of a group Z(G) be used in group theory?

The Centre of a group Z(G) has many applications in group theory. For example, it can be used to determine if a group is abelian, to find normal subgroups, and to classify finite groups. It also plays an important role in the concept of group isomorphism.

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