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Let $F$ be the free group on the generators $x,y$, let $n > 1$ an integer, and let $M := F/F''F^n$, so that $M$ is the rank 2 free metabelian group of exponent $n$. Let $\overline{x},\overline{y}$ be the images of $x,y$ in $M$.

Certainly, $M^{ab} = C_n\times C_n$, and so $M$ is an extension: $$1\rightarrow M'\rightarrow M\rightarrow C_n\times C_n\rightarrow 1$$ Is it possible to determine the structure of $M'$?

This is certainly an abelian group, generated by the conjugates of $[\overline{x},\overline{y}]$, all of which have the same order. I would expect $M'$ to be a product of copies of $C_n$, but I'm not sure what the rank would be, or even if this is true.

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    It is just as usual to define $[x,y]:=x^{-1}y^{-1}xy$. I tried your presentation with $n=3$ on the computer and got a group of order $729$ and exponent $9$. For $n=3$ your group $M$ should have order $27$. – 2017-02-01
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    @Derek Are you using GAP? If so, how do you obtain the group M for $n=3$? – 2017-02-01
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    The largest $2$-generator group of exponent $3$ of any kind is $B(2,3)$, which is well-known to have order $27$. No I wasn't using GAP. – 2017-02-01
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    @DerekHolt Ah, interesting! Good to know. I've changed the question to be perhaps more tractable. – 2017-02-01

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