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Math Help - p-Groups Question

  1. #1
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    p-Groups Question

    I solved the 1st half of this question and I am stuck on the second part

    Let G be a group and let Z_{1} be the center of G. For every n\geq 1, we'll define Z_{n+1}=\{ x\in G|\forall y\in G, [x,y]\in Z_{n}\}.

    1. Show that for every n\geq 1, Z_{n}\unlhd G.

    2. If [G]=p^{n} ( p a prime number), show that Z_{n}=G

    I showed part 1 by induction on n and by showing that the center of G/Z_{n} is isomorphic to Z_{n+1} therfore it is normal in G.

    For part 2, I'm thinking that since Z_{n}\subseteq G then all that is left to show is that [Z_{n}]=p^{n}, but I can't figure out a way to do it.

    Appreciate any ideas

    SK
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by skyking View Post
    I solved the 1st half of this question and I am stuck on the second part

    Let G be a group and let Z_{1} be the center of G. For every n\geq 1, we'll define Z_{n+1}=\{ x\in G|\forall y\in G, [x,y]\in Z_{n}\}.

    1. Show that for every n\geq 1, Z_{n}\unlhd G.

    2. If [G]=p^{n} ( p a prime number), show that Z_{n}=G

    I showed part 1 by induction on n and by showing that the center of G/Z_{n} is isomorphic to Z_{n+1} therfore it is normal in G.

    For part 2, I'm thinking that since Z_{n}\subseteq G then all that is left to show is that [Z_{n}]=p^{n}, but I can't figure out a way to do it.

    Appreciate any ideas

    SK
    This is really rough, but this should work...double check the details though.

    Prove that Z_1\unlhd Z_2\unlhd \cdots so that it's an ascending chain. Note then that if Z_k\triangleleft Z_{k+1} then 1<\left[Z_{k+1}:Z_k\right] and thus \left[Z_{k+1}:Z_k\right]=p^a,\text{ }a>0. Assume then that Z_k\ne G for any k\in[n-1] (otherwise we're done) then note that since G is a p-group we have that |Z_1|=p^b,\text{ }b>0. It follows then that |Z_n|=|Z_{n-1}|\left|Z_n:Z_{n+1}\right|\geqslant p|Z_{n-1}|=p\left|Z_{n-1}\right|\left[Z_{n-1}:Z_{n-2}\right]\geqslant p^2|Z_{n-2}|\geqslant\cdots
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by skyking View Post
    I solved the 1st half of this question and I am stuck on the second part

    Let G be a group and let Z_{1} be the center of G. For every n\geq 1, we'll define Z_{n+1}=\{ x\in G|\forall y\in G, [x,y]\in Z_{n}\}.

    1. Show that for every n\geq 1, Z_{n}\unlhd G.

    2. If [G]=p^{n} ( p a prime number), show that Z_{n}=G

    I showed part 1 by induction on n and by showing that the center of G/Z_{n} is isomorphic to Z_{n+1} therfore it is normal in G.

    For part 2, I'm thinking that since Z_{n}\subseteq G then all that is left to show is that [Z_{n}]=p^{n}, but I can't figure out a way to do it.

    Appreciate any ideas

    SK
    This sequence is called the upper central series. A group where Z_n=G is called nilpotent. It is, essentially, identical to the lower central series,

    \gamma_{m+1}(G) = [\gamma_{m}(G), G] (see * at bottom for explanation of notation)

    where \gamma_0(G)=G (and \gamma_1(G)=G^{\prime}, the derived subgroup).

    Now, Drexel has given a proof using the upper central series. I've never liked the U.C.S., I much prefer the L.C.S. I also quite like a proof of this result using the L.C.S. which I've written below. It is stolen (appologetically) from a course a did during my undergrad.

    So, note that if |G|=p^n and if G is nilpotent then it must happen in at most n-steps. So basically you are wanting to prove that every p-group is nilpotent.

    So, induct on n.

    If n=0 G is trivial and we're done (alternatively, n=1 so G is cyclic and so abelian).

    Now suppose n>0. Then Z(G) \neq 1, as G is a non-trivial p-group. By induction, G/Z(G) is nilpotent, so \gamma_c(G/Z(G)) = 1 for some c.

    Let \phi: G \rightarrow G/Z(G) be the natural homomorphism.

    Now, it is relatively easy to prove that if \theta: H \rightarrow K then \gamma_i(H) \mapsto \gamma_i(K) (induct on i). Thus, \gamma_c(G) \mapsto \gamma_c(G/Z(G))=1, and so \gamma_c(G) \leq ker(\pi) = Z(G).

    Therefore, \gamma_{c+1} = [\gamma_c(G), G] \leq [Z(G), G]=1, as required. ( [H, Z(G)] = 1 for all H \leq G, by the definition of [-, -] and because everything in the center commutes with everything else).

    *We write [H, K] = \langle[h, k]:h \in H, k \in K\rangle (with [h, k] = h^{-1}k^{-1}hk). Note that [h, k]=1 \Leftrightarrow hk=kh. [g,h] is called the commutator of g and h.
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  4. #4
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    Quote Originally Posted by skyking View Post
    I solved the 1st half of this question and I am stuck on the second part

    Let G be a group and let Z_{1} be the center of G. For every n\geq 1, we'll define Z_{n+1}=\{ x\in G|\forall y\in G, [x,y]\in Z_{n}\}.

    1. Show that for every n\geq 1, Z_{n}\unlhd G.
    (1) Normality is preserved upon surjective homomorphisms, and is also preserved upon taking inverse images (link). i.e.,

    if f:G \rightarrow G^\prime is a group homomorphism and N is a normal subgroup of G, then f(N) is a normal subgroup of f(G). If N' is a normal subgroup of f(G), then f^{-1}(N^\prime) is a normal subgroup of G.


    Let h:G \rightarrow G/Z_{i}(G) be the canonical projection. We know that if G is a group, then Z(G) is normal in G. Since Z_{i+1}(G)=h^{-1}(Z(G/Z_{i}(G)), it follows that Z_{i}(G) for i \geq 0 is normal in G by (1) inductively (verify this), where Z_0(G)=\{e\}.

    2. If [G]=p^{n} ( p a prime number), show that Z_{n}=G

    I showed part 1 by induction on n and by showing that the center of G/Z_{n} is isomorphic to Z_{n+1} therfore it is normal in G.

    For part 2, I'm thinking that since Z_{n}\subseteq G then all that is left to show is that [Z_{n}]=p^{n}, but I can't figure out a way to do it.

    Appreciate any ideas

    SK
    As said above, it is a standard textbook problem. If G is a group of order p^n, then G is nilpotent of nilpotence class at most n. For each i \geq 0, G/Z_i(G) is a p-group. Note that a nontrivial p-group has a non-trivial center. Thus, if Z_i(G) \neq G,then |Z_{i+1}(G)| \geq p|Z_i(G)|. It follows that |Z_{i+1}(G)| \geq p^{i+1}. Now we see that |Z_n(G)|=p^n.
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  5. #5
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    Thanks to all. We havn't covered nilpotence yet so I am afraid I can't use that, but I was able to show it using Drexel's approach. Thanks again.

    SK
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