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Centre of a group Z(G).

  1. Feb 27, 2008 #1
    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 [itex]Z(G) = g \epsilon G: \forall x \epsilon G, xg = gx[/itex]

    (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 [itex] |G| = p^n[/itex] ( p prime) then [itex] Z(G) \neq {1} [/itex]
    (iii) If G is non abelian show that G/Z(G) is not cyclic.
    (iv) Decude that any group of order [itex] p^2[/itex] is abelian.
    (V) Deduce that a gorup of oder [itex] p^2 [/itex] is isomorhpic either to [itex] C_{p^2}[/itex] or to [itex] C_p \times C_p[/itex]
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    WHAT I HAVE SO FAR: PLEASE VERIFY THEM
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    (i) is not too hard: we let [itex]x\in \text{Z}(G)[/itex] and so we prove that [itex]gxg^{-1} \in \text{Z}(G)[/itex] for any [itex]g\in G[/itex].

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

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

    please verify these and help me out with the rest. very many thanks :)
     
    Last edited: Feb 27, 2008
  2. jcsd
  3. Feb 27, 2008 #2

    morphism

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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.
     
  4. Feb 28, 2008 #3
    (iii) but wouldnt proving the contrapositive be the OPPOSITE of what is required for the question?
     
  5. Feb 28, 2008 #4

    CompuChip

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    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
  6. Feb 29, 2008 #5
    (iii)
    ok here is the contrapositive.

    we let [itex]H=\text{Z}(G)[/itex].
    now if [itex]G/H[/itex] is cyclic then there is [itex]aH[/itex] which generates the group [itex]G/H[/itex]. We Let [itex]x,y\in G[/itex]. Also, Note [itex]xH,yH\in G/H[/itex] thus [itex]xH=a^nH[/itex] and [itex]yH=a^mH[/itex]. This means [itex]x = a^n z_1[/itex] and [itex]y=a^mz_2[/itex] where [itex]z_1,z_2\in H[/itex]. But then [itex]xy = a^n z_1 a^mz_2 = a^{n+m}z_1z_2[/itex] and [itex]yx = a^m z_2 a^n z_1 = a^{n+m}z_1z_2[/itex] because [itex]z_1,z_2[/itex] commute with everything. So,[itex]G[/itex] is abelian.

    where would you go from here? how would i prove the original statement by contradiction?
     
  7. Feb 29, 2008 #6

    morphism

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    Isn't it obvious? If G is nonabelian, then there is no way G/Z(G) is cyclic...
     
  8. Feb 29, 2008 #7

    NateTG

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    It's just a generic proof by contradiction - start with:
    Consider a non-abelian group [itex]G[/itex], and assume that [itex]G/Z(G)[/itex] is cyclic by contradiction. (And try to find a contradiction...)
     
  9. Feb 29, 2008 #8
    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 :)
     
  10. Mar 1, 2008 #9

    CompuChip

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    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.
     
  11. Mar 1, 2008 #10
    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

    how about question (v)? any hints on that please?
     
  12. Mar 1, 2008 #11

    morphism

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    Didn't you already solve it?
     
  13. Mar 1, 2008 #12
    sorry i meant (iv), any suggestions there?
    i guess it follows on from the previous question...
     
  14. Mar 1, 2008 #13

    morphism

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    Pretty much. Just combine (ii) and (iii) together and it falls through.
     
  15. Mar 2, 2008 #14
    is this correct?
    (iv)
    Burnside's Lemma states that the center is non-trivial. (using part 2) (i.e. Z(G) [itex]\neq[/itex] {1}), forming the factor group we have a cyclic group. Thus the original group needs to be abelian, including [itex]p^2[/itex].
     
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