Let $G$ be a $p$-group of nilpotency class at most 2, where $p$ is an odd prime.Then $A=\{{x^p| x \in G}\}$ is a subgroup of $G$.

As $G$ is a group of nilpotency class at most 2, if the nilpotency class of $G$ equal 1 then $G$ is an abelian group and the result is obvious.

Now we consider the nilpotency class equal 2. Then we have $G^\prime \leq Z(G)$.

Also as a result for every natural number like $n$ and every $x,y$ in $G$ we have $$(xy)^n= x^ny^n[y,x]^{n\frac{n-1}{2}}.$$

Now I consider $x^p, y^p$ in $A$ where $x,y$ are in $G$. I want to show $x^py^p$ in $A$. According to the above fact we have $$ (xy)^p= x^py^p[y,x]^{p\frac{p-1}{2}}.$$

I want to show $[y,x]^{p\frac{p-1}{2}}=1$, but I don't know how.

  • 2
    $\begingroup$ You don't need to show it is trivial! You need to show that the product of two $p$th powers is a $p$th power, and what you have shows that $x^py^p = (xy)^p([x,y]^{(p-1)/2})^p \in G^p$. Or you can note that $[y,x]^{p(p-1)/2} = [y^p,x^{(p-1)/2}] = y^{-p} (y^p)^{x^{(p-1)/2}}$, so it is a product of $p$-th powers. Thus, the product of two elements in $G^p$ is in $G^p$. $\endgroup$ – Arturo Magidin Mar 22 at 18:32
  • $\begingroup$ @ArturoMagidin We define $G^p =\langle x^p|x \in G\rangle$ and it is obviously a subgroup of $G$. But here $A=\{x^p|x \in G\}$. We want to show the procut of two $p$-power elements is in $A$ not in $G^p$. Thanks for your complete answer. $\endgroup$ – Yasmin Mar 23 at 4:05
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    $\begingroup$ Well, as you see in the answer, the comment is incomplete; the point is that one of the factors is central, and a product of two $p$-th powers, one of which is central, is a $p$th power. P.S. Use \langle rather than <, and \rangle rather than >. $\endgroup$ – Arturo Magidin Mar 23 at 4:07
  • $\begingroup$ @ArturoMagidin yes, I see the answer, I only wanted to say that the way you said in comment dosen't complete. Thanks for your help. $\endgroup$ – Yasmin Mar 23 at 4:14

You won’t be able to prove the commutator is trivial, because it doesn’t have to be.

Consider the multiplicative group of upper triangular $3\times 3$ matrices with entries in $\mathbb{Z}/p^n\mathbb{Z}$ with $1$s in the diagonal. Multiplication is given by $$\left(\begin{array}{ccc} 1&a&c\\ 0 & 1 & b\\ 0 & 0 & 1\end{array}\right) \left(\begin{array}{ccc} 1 & \alpha & \gamma\\ 0 & 1 & \beta\\ 0 & 0 & 1 \end{array}\right) = \left(\begin{array}{ccc} 1 & a+\alpha & c + \gamma + a\beta\\ 0 &1 & b+\beta\\ 0 & 0 & 1\end{array}\right).$$ This is a nilpotent group of order $p^{3n}$ and class $2$. The commutator subgroup is generated by the matricx with $a=b=0$, $c=1$.

If you let $x$ be the matrix corresponding to $a=1$, $b=c=0$; and $y$ the matrix corresponding to $a=c=0$, $b=1$, then $[x,y]=x^{-1}y^{-1}xy$ is the matrix with $a=b=0$, $c=1$, which has order $p^n$. In particular, if $n\gt 1$, then $[x,y]^{p(p-1)/2}\neq e$.

However, it is not necessary to show the commutator is trivial. As you already computed, in a nilpotent group of class at most $2$, we have $$ (xy)^n = x^n y^n [y,x]^{n(n-1)/2}.$$

So, to show that $A$ is a subgroup, we note that $e^p=e\in A$; that if $x^p\in $, then $(x^p)^{-1} = (x^{-1})^p\in A$. And then the hard one: showing that the product of two $p$th powers is a $p$th power. Suppose $x^p,y^p\in A$. Then, as previously computed we have: $$\begin{align*} (xy)^p &= x^py^p[y,x]^{p(p-1)/2}\\ (xy)^p[x,y]^{p(p-1)/2} &= x^py^p\\ (xy)^p\left([x,y]^{(p-1)/2}\right) &= x^py^p. \end{align*}$$ To finish off, we note that since $[x,y]^{(p-1)/2}$ commutes with $xy$ (in fact, it is central), then $$ \left( (xy)[x,y]^{(p-1)/2}\right)^p = (xy)^p[x,y]^{p(p-1)/2} = x^py^p.$$ Thus, $x^py^p$ is equal to a $p$th power, and hence lies in $A$. Thus, $A$ is closed under products, and we are done.

  • $\begingroup$ Thank you, I understand. $\endgroup$ – Yasmin Mar 23 at 4:01

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