# p-Groups Question

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• Jan 20th 2011, 06:15 PM
skyking
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
• Jan 20th 2011, 06:36 PM
Drexel28
Quote:

Originally Posted by skyking
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 $|Z_n|=|Z_{n-1}|\left|Z_n:Z_{n+1}\right|\geqslant p|Z_{n-1}|=p\left|Z_{n-1}\right|\left[Z_{n-1}:Z_{n-2}\right]\geqslant p^2|Z_{n-2}|\geqslant\cdots$
• Jan 21st 2011, 12:23 AM
Swlabr
Quote:

Originally Posted by skyking
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 $h$.
• Jan 21st 2011, 01:39 AM
TheArtofSymmetry
Quote:

Originally Posted by skyking
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 $Z_0(G)=\{e\}$.

Quote:

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 $|Z_n(G)|=p^n$.
• Jan 21st 2011, 06:28 AM
skyking
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