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

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

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

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

    For part 2, I'm thinking that since $\displaystyle Z_{n}\subseteq G$ then all that is left to show is that $\displaystyle [Z_{n}]=p^{n}$, 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 $\displaystyle G$ be a group and let $\displaystyle Z_{1}$ be the center of $\displaystyle G$. For every $\displaystyle n\geq 1$, we'll define $\displaystyle Z_{n+1}=\{ x\in G|\forall y\in G, [x,y]\in Z_{n}\}$.

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

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

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

    For part 2, I'm thinking that since $\displaystyle Z_{n}\subseteq G$ then all that is left to show is that $\displaystyle [Z_{n}]=p^{n}$, 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 $\displaystyle Z_1\unlhd Z_2\unlhd \cdots$ so that it's an ascending chain. Note then that if $\displaystyle Z_k\triangleleft Z_{k+1}$ then $\displaystyle 1<\left[Z_{k+1}:Z_k\right]$ and thus $\displaystyle \left[Z_{k+1}:Z_k\right]=p^a,\text{ }a>0$. Assume then that $\displaystyle Z_k\ne G$ for any $\displaystyle k\in[n-1]$ (otherwise we're done) then note that since $\displaystyle G$ is a $\displaystyle p$-group we have that $\displaystyle |Z_1|=p^b,\text{ }b>0$. It follows then that $\displaystyle |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 $\displaystyle G$ be a group and let $\displaystyle Z_{1}$ be the center of $\displaystyle G$. For every $\displaystyle n\geq 1$, we'll define $\displaystyle Z_{n+1}=\{ x\in G|\forall y\in G, [x,y]\in Z_{n}\}$.

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

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

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

    For part 2, I'm thinking that since $\displaystyle Z_{n}\subseteq G$ then all that is left to show is that $\displaystyle [Z_{n}]=p^{n}$, 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 $\displaystyle Z_n=G$ is called nilpotent. It is, essentially, identical to the lower central series,

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

    where $\displaystyle \gamma_0(G)=G$ (and $\displaystyle \gamma_1(G)=G^{\prime}$, 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 $\displaystyle |G|=p^n$ 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 $\displaystyle n$.

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

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

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

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

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

    *We write $\displaystyle [H, K] = \langle[h, k]:h \in H, k \in K\rangle$ (with $\displaystyle [h, k] = h^{-1}k^{-1}hk$). Note that $\displaystyle [h, k]=1 \Leftrightarrow hk=kh$. $\displaystyle [g,h]$ is called the commutator of $\displaystyle g$ and $\displaystyle h$.
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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 $\displaystyle G$ be a group and let $\displaystyle Z_{1}$ be the center of $\displaystyle G$. For every $\displaystyle n\geq 1$, we'll define $\displaystyle Z_{n+1}=\{ x\in G|\forall y\in G, [x,y]\in Z_{n}\}$.

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

    if $\displaystyle f:G \rightarrow G^\prime$ 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 $\displaystyle f^{-1}(N^\prime)$ is a normal subgroup of G.


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

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

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

    For part 2, I'm thinking that since $\displaystyle Z_{n}\subseteq G$ then all that is left to show is that $\displaystyle [Z_{n}]=p^{n}$, 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 $\displaystyle p^n$, then G is nilpotent of nilpotence class at most n. For each $\displaystyle i \geq 0$, $\displaystyle G/Z_i(G)$ is a p-group. Note that a nontrivial p-group has a non-trivial center. Thus, if $\displaystyle Z_i(G) \neq G$,then $\displaystyle |Z_{i+1}(G)| \geq p|Z_i(G)|$. It follows that $\displaystyle |Z_{i+1}(G)| \geq p^{i+1}$. Now we see that $\displaystyle |Z_n(G)|=p^n$.
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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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