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How to show that for $p$ prime, every group of order $p^5$ is metabelian? A group is metabelian if and only if its commutator subgroup $G'$ is abelian.

If $G$ is a group of order $p^5$, then $G$ has a normal subgroup $N$ of order $p^3$ (in $p$-groups, there is a normal subgroup of every possible order). Then $G/N$ is abelian and thus $G' \leq N$. Therefore the commutator subgroup has order $1$, $p$, $p^2$ or $p^3$. The only case where $G'$ could be nonabelian is when it has order $p^3$. This actually happens when $G = D_{32}$, the dihedral group of order $32$, so proving that $|G'| = p^3$ is impossible is not the way to go.

Any ideas where to go from here?

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4 Answers 4

If $G'$ has order less than $p^3$, then it is abelian. So we may assume that $G'$ has order exactly $p^3$, in which case $G/G'$ is of order $p^2$, and thus $2$-generated. let $x$ and $y$ be the two generators. Moreover, $G_3=[G,G']$ has order at most $p^2$, $G_4$ has order at most $p$, and so $G_5$ will certainly be trivial; that is, $G$ has class at most $4$.

Since $G$ is a finite $p$-group, $G'$ is generated by the basic commutators in $x$ and $y$. We know that $G_n/G_{n+1}$ is generated by the basic commutators of weight $n$. So $G_2/G_3$ is cyclic, generated by $[y,x]$; $G_3/G_4$ is generated by $[y,x,x]$ and $[y,x,y]$. And $G_4/G_5 = G_4$ is generated by the basic commutators of weight $4$; but the only basic commutators of weight $4$ that are nontrivial are $[y,x,x,x]$, $[y,x,x,y]$, and $[y,x,y,y]$, since there is no nontrivial basic commutator obtained as $[c,c']$ with $c$ and $c'$ of weight $2$.

Thus, $G'$ is generated by $[y,x]$, $[y,x,x]$, $[y,x,y]$, $[y,x,x,x]$, $[y,x,x,y]$, and $[y,x,y,y]$. Since $G'$ is of class at most $4$, they commute pairwise (the only commutator not of weight greater than $4$ is $[[y,x],[y,x]]$, which is trivial).

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Being metabelian is the same as having derived length $d \le 2$. A $p$-group of order $p^5$ has nilpotency class $c\le 4$. Now using the standard inequality $$ d < 1+\log_2(c+1),$$

one sees what you said is true as long as $G$ is not of maximal class. I don't know how to handle the maximal class case "easily". What you say is true, because if $G$ is maximal class it has a positive degree of commutativity, implying $\gamma_2(G)$ is central in $\gamma_1(G)$. This is enough to conclude $G$ is metabelian. Hopefully someone will come along and provide a more elementary answer.

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Thanks to you and @ArturoMagidin for answering. I'm afraid both answers are over my head right now, so I'll leave this question open for a while. I have no idea about nilpotency and I've never heard of "commutator weight", for example. I am wondering too about if there is an elementary solution (whatever that means here).. I started thinking about this question after proving the statement for groups of order $p^4$. I can't solve this even for special cases like $p = 2$, $p = 3$ (although I guess they might not be any easier than the general case) –  spin Mar 25 '12 at 19:33
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@spin: In fact, smaller primes are harder than larger primes. In general, it is easier to understand groups of order $p^n$ with $n\leq p$ than with $n\gt p$, so for a fixed $n$, larger primes makes things easier. This is a consequence of "regularity" (a property that groups of prime power order may or may not have; when they have it, they behave, in some ways, much like abelian groups). When $n\leq p$, then you fall into the so-called "small class case", which is by far easier. I suspect you would find it easier to work with, say, $p=7$ than $p=2$ or $p=3$. –  Arturo Magidin Mar 25 '12 at 22:17
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@spin: I'll think about whether I can express my answer without invoking (or re-proving, for that matter) results on the lower central series and basic commutators. By the way: I did not get "pinged" by your comment, because I had neither commented nor edited this answer. You can only "ping" one other person that is already "involved" in the question. –  Arturo Magidin Mar 25 '12 at 22:18
    
@ArturoMagidin: Thanks. This problem turned out to be harder than I thought. –  spin Mar 26 '12 at 13:51
    
I think that the natural way to handle such problems is to separate two cases, as Arturo mentioned: when $n\le p$ and when $n > p$. The first is usually the easier case. The second involves more "heavy lifting", either by resorting to more advanced arguments, or a tedious case-by-case verification. –  user641 Mar 26 '12 at 21:19

The question is to show that for $p$ prime, every group $P$ with order $p^5$ is metabelian.

As usual, let $\Phi(P)$ denote the Frattini subgroup of $P.$ Also let $W$ be the centre of $\Phi(P).$ If $x\in\Phi(P)$ is not in $W,$ then $x$ cannot centralise any maximal subgroup of $P,$ so the conjugacy class of $x$ in $P$ has size divisible by $p^2.$ It follows that the orders of $W$ and $\Phi(P)$ are equal mod $p^2.$

Now in the present case $P$ has order $p^5,$ so the order of $\Phi(P)$ is at most $p^3.$ The only possibilities consistent with the above are $W=\Phi(P)$ or that the orders of $W$ and $\Phi(P)$ are $p^2$ and $p^3$ respectively. The second case is impossible however, since centre-by-cyclic implies abelian. Hence $\Phi(P)$ is abelian, and $P$ is metabelian as claimed.

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Let $G$ be a group of order $p^5$.

Claim: $G$ has abelian normal subgroup of order $p^3$.

Since any finite $p$-group has normal series (series of normal subgroups); consider a normal series $1\triangleleft H \triangleleft K\triangleleft \cdots \triangleleft G$, where $|H|=p$, $|K|=p^2$. Then $K$ is an abelian normal subgroup of order $p^2$. The normality of $K$ implies that $G$ acts on $K$ by conjugation: $(g,k)\mapsto gkg^{-1}$. This gives a homomorphism $\varphi\colon G\rightarrow Aut(K)$, $g\mapsto \{\varphi_g\colon k\mapsto gkg^{-1}\}$,

and $ker(\varphi)$ continas those elements of $G$ which commute with all elements of $K$.

Now, $K\cong C_p\times C_p$ or $C_{p^2}$; hence $|Aut(K)|\in \{p(p-1)(p^2-1), p(p-1)\}$. We see that $|Im(\varphi)|\in \{1,p\}$ (since $G$ is $p$-group), i.e. $[G\colon ker(\varphi)]\leq p$. Therefore, $|ker(\varphi)|\geq p^{4}> p^3$. Consider the normal series $1 \triangleleft H\triangleleft K\triangleleft ker(\varphi)\triangleleft G$, and take a refinement $1 \triangleleft H\triangleleft K\triangleleft L\triangleleft \cdots \triangleleft ker(\varphi)\triangleleft G$, where $|L|=p^3$. Then $L$ is abelian (why?) normal subgroup of order $p^3$. Can you see that $G$ is meta-abelian?

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Thanks, this works! The subgroup $L$ is abelian because $K \leq Z(L)$ and thus $Z(L)$ has order at least $p^2$, and there are no groups of order $p^3$ with center of order $p^2$. Also, there are groups of order $p^4$ with center of order $p^2$, so this same proof does not work out with the case $|G| = p^6$. But that's also because there exist groups of order $p^6$ that are not metabelian, so maybe this center thing is what it comes down to? –  spin Sep 14 '12 at 12:37
    
Hmm, actually I looked it up and every group of order $2^6$ and $3^6$ is metabelian. But there are non-metabelian groups of order $p^6$ when $p > 3$ is prime. So maybe it's not so simple. –  spin Sep 14 '12 at 12:41

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