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Let $A\leq F_2$ where $F_2$ is the free group on $\{a,b\}$.

Assume $A$ is generated by words on $a,b$ such that the total powers of $a$ are $3x$ for some $x\in \mathbb{Z}$ and the total powers of $b$ are $-2x$ for the same $x$.

Explicitly, for instance, a generator of $A$ would be $a^3b^{-2}$ or $a^2b^{-1}ab^{-1}$.

Besides writing them all out, is there a way to concisely provide a presentation for $A$ using generator notation? Of course being able to write $A=\langle a^{3x}b^{-2x} \rangle$ would be nice, but $F_2$ is non-abelian.

It seems like it should be close to something like, $A = \langle a^{3x} b^{-2x} : x\in\mathbb{Z}, ab=ba \rangle$, but I'm not convinced throwing in the commutivity relation makes too much sense.

Maybe utilizing $\prod$?

pyridoxal_trigeminus
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  • You could define $\varphi\colon F_2\to\mathbb Z^2$ as the abelianization. Then $A$ is generated by $u\in F_2$ such that $\varphi(u)=(3x,-2x)$ for some $x$. – Cheerful Parsnip Mar 20 '22 at 05:16

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Define a group morphism $\pi:F_2 \to \Bbb Z$ by setting $\pi(a) = 2$ and $\pi(b) = 3$. Thus for instance $\pi(a^3b^{-2}) = 0$ and $\pi(a^2b^{-1}ab^{-1}) = 0$, etc. Consequently, your group is generated by $\ker \pi$, and since $\ker \pi$ is a group, it is actually equal to $\ker \pi$.

I claim that $\ker \pi$ is a free group generated by $a^3b^{-2}$ and by the commutator subgroup $[F_2, F_2]$ of $F_2$. Let $u \to |u|_a$ be the group morphism from $F_2$ onto $\Bbb Z$ which maps $a$ to $1$ and $b$ to zero. Intuitively, $|u|_a$ counts the number of occurrences of $a$ in $u$, with $a^{-1}$ counting as $-1$. The map $u \to |u|_b$ is defined in a similar way. Note that $$[F_2, F_2] = \{u \in F_2 \mid |u|_a = |u|_b = 0\}.$$

Let now $u \in \ker \pi$. Since $\pi(u) = 2 |u|_a + 3|u|_b$, one has $2|u|_a + 3 |u|_b = 0$ and thus $|u|_b$ is even. Now $(a^3b^{-2})^{|u|_b/2}u \in [F_2, F_2]$ since $$ \bigl(|(a^3b^{-2})^{|u|_b/2}u|_a, |(a^3b^{-2})^{|u|_b/2}u|_b\bigr) = \bigl(3|u|_b/2 + |u|_a, -|u|_b + |u|_b\bigr) = (0, 0) $$ which proves the claim. Finally, according to this question, $[F_2, F_2]$ is generated by the commutators $[a^n, b^n]$, for $n \in \Bbb Z \setminus \{0\}$. This gives you the following set of generators for your group: $$ \{a^3b^{-2}\} \cup \{[a^n, b^n] \mid n \in \Bbb Z \setminus \{0\}\} $$

J.-E. Pin
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  • Indeed, but is there any explicit, concise way to present $\ker \pi$? – pyridoxal_trigeminus Mar 20 '22 at 06:45
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    @pyridoxal_trigeminus Unfortunately, the commutator subgroup of $F_2$ is free of infinite rank, so it would be difficult to describe via a presentation. The best you can do is almost certainly what Pin does: say it is generated by $[F_2,F_2]$ and $a^3b^{-2}$. – Arturo Magidin Mar 20 '22 at 16:39