# If $g^5 = h^7$ in a free group, then $g$ and $h$ are in a cyclic subgroup

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).

• Are you able to use the result that subgroups of free groups are free? – Derek Holt Apr 23 at 19:12
• 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. – pizzaroll Apr 23 at 19:33
• 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. – Derek Holt Apr 23 at 21:08
• 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 :) – pizzaroll Apr 23 at 22:32

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 $$\alpha_i=\alpha_j,$$ $$g=\alpha_1 7,\quad h=\alpha_1^5.$$