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Let $G$ be a p-group $|G| = p^n$. Then G is nilpotent in the following sense:

Let $C^1(G) = [G,G] = G'$ And $C^n(G) = [G,C^{n-1}]$.

G is nilpotent iff there is a $n_0$ such that $C^n = \{e\}$.

What I proved in previous exercises is:

If H is a proper subgroup of a nilpotent group then H is a normal proper subgroup of its normalizer. i.e. if $H \lneq G$ then $H \lneq N_G(H)$.

I use induction: If $|G| = p$, then G is isomorphic to $\mathbb{Z}/p\mathbb{Z}$ which is cyclic. therefore $C^1(G) = G' = \{e\}$. Suppose that the following assertion is true:

if $|L| = p^i$, for $i \lt n$ then L is nilpotent.

Let G be a group of order $p^n$. If I can just prove $G' \neq G$, then we'd have $|G'| = p^i, i \lt n$ and in such a case G' would be nilpotent. So that G would be nilpotent.

Is it? If G is a p-group, then $G'\neq G$?

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Yes: if $G$ is a $p$-group, then $G'\neq G$. –  Arturo Magidin May 3 '12 at 13:26
    
how to prove it? –  Henrique Tyrrell May 3 '12 at 13:36
3  
Assuming that you could prove that $G' \ne G$ and hence assume inductively that $G'$ is nilpotent, how would you then prove that $G$ is nilpotent? It is not true for all groups that $G'$ nilpotent implies $G$ nilpotent. –  Derek Holt May 3 '12 at 20:46
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1 Answer

up vote 2 down vote accepted

A very easy way to prove that $G'\neq G$ is to note that $G'$ is the smallest normal subgroup of $G$ such that $G/G'$ is abelian. If you can find any proper normal subgroup of $G$, $N$, such that $G/N$ is abelian, then it will follow that $G'\subseteq N$, and so $G'\neq G$.

For example: have you proven that a maximal subgroup of a $p$-group must be normal? This can be done easily by induction, using the fact that $Z(G)$ (the center) is nontrivial. With that in hand, it follows that $G'$ is always contained in a maximal subgroup of $G$, and you'd be done.

(The previous exercise you quote can't help you because it assumes your group is nilpotent, and here you are trying to prove that a group is nilpotent).

Added. Another way of proving this:

Show that if $f\colon G\to K$ is an onto group homomorphism, then $f([G,G])=[K,K]$. In particular, if $[K,K]\neq K$, then $[G,G]\neq G$. Then use the fact that if $G$ is a $p$-group then $Z(G)\neq \{1\}$ and induction to show that either $Z(G)=G$ (so $[G,G]=\{1\}$) or else $[G/Z(G),G/Z(G)]\neq G/Z(G)$ and get the result.

Note: Your notation for the lower central series is uncommon. It is more common to use $G_1=G$ and $G_{n+1}=[G_n,G](=[G,G_n])$.


Added. I wasn't paying enough attention when I answered this, and what Derek Holt points out in comments is quite true: it is not enough to prove that $[G,G]$ is nilpotent in order to prove that $G$ is nilpotent, even if $[G,G]\neq G$: there are groups where $[G,G]$ is nilpotent, but $G$ is not. For example, $S_3$ has $[G,G]\cong A_3$, which is abelian, but $S_3$ is not nilpotent.

Instead, we can proceed by induction on $n$, where $|G|=p^n$.

If $n=1$ or $n=2$, then $G$ is abelian, hence $G$ is nilpotent.

Assume that $G$ has order $p^n$. If $Z(G)=G$, then $G$ is abelian and hence is nilpotent. Otherwise, consider $\mathfrak{G}=G/Z(G)$. This is a $p$-group of order strictly less than $p^n$ (since the center of a finite $p$-group is always nontrivial), so by the induction hypothesis, $\mathfrak{G}$ is nilpotent. Now we use the lemma mentioned above, slightly strengethened (I will use $H_n$ to denote the $n$th term of the lower central series of $H$, that is, $H_n$ is what the original post calls $C^n(H)$):

Lemma. Let $f\colon G\to K$ be a group homomorphism. Then for every $m$ we have $f(G_m) \subseteq K_m$. If $f$ is onto, then $f(G_m)=K_m$ for all $m$.

Proof. The result holds for $m=1$. Assume $f(G_m)\subseteq K_m$. A generator for $G_{m+1}$ is of the form $[g,g']$, where $g'\in G_m$, hence $f([g,g']) = [f(g),f(g')]\in [K,K_m]=K_{m+1}$, so $f(G_{m+1})\subseteq K_{m+1}$, as claimed. Assume now that $f$ is onto and $f(G_m)=K_m$. A generator for $K_{m+1}$ is of the form $[k,k']$ with $k\in K$ and $k'\in K_{m+1}$; since $f$ is onto there exists $g\in G$ with $f(g)=k$; since $f(G_m)=K_m$, there exists $g'\in G_m$ with $f(g')=k'$. Hence $f([g,g'])=[f(g),f(g')] = [k,k']$, and since $[g,g']\in G_{m+1}$, it follows that $K_{m+1}\subseteq f(G_{m+1})$, as desired. $\Box$

Now, since $\mathfrak{G}$ is nilpotent, there exists $r$ such that $\mathfrak{G}_r$ is trivial; therefore, by the lemma, $f(G_r)$ is trivial, so $G_rZ(G) = Z(G)$. Hence $G_r\subseteq Z(G)$, so $G_{r+1}=[G,G_r]\subseteq [G,Z(G)] = \{1\}$, so $G$ is nilpotent, as claimed. $\Box$

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But G is a general p-group. I can't answer with certainty that there is such a subgroup. –  Henrique Tyrrell May 3 '12 at 13:36
    
@Henrique: Every nontrivial finite group has maximal subgroups (even if they are trivial). All you need to do is prove that a maximal subgroup of a $p$-group is normal. –  Arturo Magidin May 3 '12 at 13:37
    
I see, I'll try that. –  Henrique Tyrrell May 3 '12 at 13:37
    
But if this maximal subgroup is trivial, as in, the whole group, then I can't conclude that. The result I was thinking about using is this: Let G be a finite group. $|G| = p^n m$ then there is an ascending chain of subgroups $0 = H_0 < H_1 < .. < H_n$ and $|H_i| = p^i$ and each $H_i$ is normal in $H_j$. –  Henrique Tyrrell May 3 '12 at 13:45
    
@HenriqueTyrrell: Maximal subgroups are never the whole group. "Trivial" means the trivial subgroup, $\{e\}$, which can only be maximal in the case of cyclic groups of prime order. The result you want to use works, because $H_n=G$, so that means that $H_{n-1}$ is normal in $G$. What is the order of $G/H_{n-1}$? It's $p$. So $G/H_{n-1}$ must be abelian. So $G'\subseteq H_{n-1}\neq G$. –  Arturo Magidin May 3 '12 at 13:58
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