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Math Help - centre of a group Z(G)

  1. #1
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    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
    Last edited by joanne_q; February 27th 2008 at 06:23 AM.
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  2. #2
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    Quote Originally Posted by joanne_q View Post
    (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.
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  3. #3
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    Quote Originally Posted by joanne_q View Post
    (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.
    Last edited by ThePerfectHacker; February 25th 2008 at 09:35 PM.
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  4. #4
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    Quote Originally Posted by joanne_q View Post
    (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.
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  5. #5
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    Thanks a lot for the help really appreciate it. makes it easier to understand the subject aswell.

    For part (ii), your solution was:
    Quote Originally Posted by ThePerfectHacker View Post
    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.
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  6. #6
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    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..?
    Quote Originally Posted by ThePerfectHacker View Post
    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.
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  7. #7
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    Quote Originally Posted by joanne_q View Post
    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.
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  8. #8
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    Quote Originally Posted by joanne_q View Post
    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.
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