What Are the Properties of the Center of a Group in Group Theory?

[mathusers]

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

QUESTION:
--------------------------------------
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
--------------------------------------

WHAT I HAVE SO FAR: PLEASE VERIFY THEM
--------------------------------------
(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 :)

 

[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.

 

[mathusers]

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

 

[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).

 

[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?

 

[morphism]

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

 

[NateTG]

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

It's just a generic proof by contradiction - start with:
Consider a non-abelian group G, and assume that G/Z(G) is cyclic by contradiction. (And try to find a contradiction...)

 

[mathusers]

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 :)

 

[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.

 

[mathusers]

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

how about question (v)? any hints on that please?

 

[morphism]

how about question (v)? any hints on that please?

Didn't you already solve it?

 

[mathusers]

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

 

[morphism]

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

 

[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.