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Let $F$ be a free group, $g,h \in F$ with $g^5 = h^7$. Then I want to show these are in a cyclic subgroup.

The strategy I'm trying and failing with is to show that $g$ and $h$ commute, then they are in an abelian subgroup which must be cyclic (only free abelian groups are $\mathbb{Z}$ or trivial).

There must be some clever manipulation of the relation $g^5 h^{-7}$ that will give us $ghg^{-1}h^{-1}$, right? I can't see it though.

Another idea is to use the fact that $F$ is residually finite. If $gh \neq hg$ then there is a $f : F \to G$ with $G$ finite and $f(gh) \neq f(hg)$. I can't see how this helps, although we are now in a finite group $G$, so maybe we can do some trick with finite orders (we couldn't do this before since all nontrivial elems of $F$ have infinite order).

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  • $\begingroup$ Are you able to use the result that subgroups of free groups are free? $\endgroup$ – Derek Holt Apr 23 at 19:12
  • $\begingroup$ Yes, sure. In fact that is how I know the only abelian subgroup of $F$ is $\mathbb{Z}$, since it is the only abelian free group. $\endgroup$ – pizzaroll Apr 23 at 19:33
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    $\begingroup$ So $H = \langle g,h \rangle$ is a free group. But since $g^5 = h^7$ commutes with both $g$ and $h$, we have $g^5 \in Z(H)$. Nonabelian free groups have trivial centre, so $H$ must be free of rank $1$ - i.e. cyclic. $\endgroup$ – Derek Holt Apr 23 at 21:08
  • $\begingroup$ Hm, I don't know how I didn't see that, so obvious in hindsight. There are alternate solutions that don't use this theorem that are still of interest, though, so I don't think asking this question was a complete waste :) $\endgroup$ – pizzaroll Apr 23 at 22:32
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If $g=1$, the claim is clear. So assume $g\ne1$. Let $I$ be the generators of $F$ and write $g,h$ as word over the alphabet $I\cup I^{-1}$. We may assume wlog that among all conjugates of $(g,h)$, our pair is one where the word for $g$ is of minimal length. Then $g$ cannot be of the form $x\alpha x^{-1}$ or $x^{-1}\alpha x$ as that would allow us to find a shorter conjugate. Therefore, in the multiplication of $g^5$, no cancellation occurs. Also, the first symbol of $h$ must be the first symbol of $g$ and similarly for the last symbols. We conclude that no cancellation occurs in the multiplication of $h^7$, either. Therefore, $g^5=h^7$ is a word that can be split into 5 equal parts as well as into 7 equal parts, in particular has a length divisible by $35$. Split it into $35$ parts $\alpha_1,\ldots,\alpha_{35}$ of equal length. Then $$\alpha_1\alpha_2\alpha_3\alpha_4\alpha_5=\alpha_6\alpha_7\alpha_8\alpha_9\alpha_{10}=\ldots =\alpha_{31}\alpha_{32}\alpha_{33}\alpha_{34}\alpha_{35}=h$$ and we conclude $$\alpha_i=\alpha_j\qquad\text{if }i\equiv j\pmod 5.$$ The same way, we conclude from $$ \alpha_1\alpha_2\alpha_3\alpha_4\alpha_5\alpha_6\alpha_7=\ldots =g$$ that $$\alpha_i=\alpha_j\qquad\text{if }i\equiv j\pmod 7.$$ Together, these imply $$ \alpha_i=\alpha_j,$$ $$ g=\alpha_1 7,\quad h=\alpha_1^5.$$

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