# If $F(a, b)=\langle a, B\rangle$ then $B=a^ib^{\epsilon}a^j$: a neat proof?

If you take a generating pair for $F(a, b)$, $(a, B)$, then it is intuitively obvious that $B=a^ib^{\epsilon}a^j$ for $i, j\in \mathbb{Z}$, $\epsilon=\pm 1$. However, I cannot come up with a neat proof of this, and so was wondering if someone could either provide a reference or come up with a more elegant approach.

My proof is basically,

Assume $B\neq a^ib^{\epsilon}a^j$, so $B=a^{i_1}b^{i_2}a^{i_3}\bar{B}a^{j_3}b^{j_2}a^{j_1}$, so you end up with $\langle a, \hat{B}\rangle=F(a, b)$ where $\hat{B}=b^{i_2}a^{i_3}\bar{B}a^{j_3}b^{j_2}$ and no free cancellation happens between $a$ and $\hat{B}$, and $|\hat{B}|>1$. This means that $a\not\in \langle a, \hat{B}\rangle$, a contradiction.

I do not like this proof, but it seems to be the best I can come up with...so any ideas would be appreciated!

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You don't specify, but I assume that $F(a,b)$ is the free group on two generators, and by a generating pair, you mean two elements of the group which generate the entire group? Also, what would constitute an elegant approach? We know exactly what the elements of $F(a,b)$ are, and your idea relies on nothing more than that. Would you like something that uses covering space theory? Or something that characterizes the automorphisms of $F(a,b)$? –  Aaron Jul 16 '11 at 18:20
This is related to an old result of Magnus: if two words $r$ and $s$ in a free group have the same normal closure, then $r$ is conjugate to $s$ or $s^{-1}$. –  user641 Jul 16 '11 at 20:21
@Steve D: I can see how that would make my approach neater - I end up with B=$B_0^{-1}b^{\epsilon}B_0$ freely reduced and just need to prove that $b$ is not a subword of $B_0$. Was this what you were getting at? –  user1729 Jul 17 '11 at 11:45
@Aaron: I suppose my question is bad because "elegant" is subjective. I perhaps meant "elementary and short" or "can be obtained from another result in an elementary and short way". –  user1729 Jul 17 '11 at 11:47

For any nonempty word $w\in F(a,b)$ and any integer $p$, we have $\lvert w^p\rvert\ge \lvert p\rvert$. This follows by writing $w=cqc^{-1}$, where the first and last letters of $q$ don't cancel. Since $w$ is nonempty, $\lvert q\rvert\ge1$ and therefore $\lvert w^p\rvert=2\lvert c\rvert+\lvert q^p\rvert\ge \lvert p\rvert$.

We use this fact to prove your statement.

If $F(a,b)=\langle a,B\rangle$ then $F(a,b)=\langle a,\tilde B\rangle$, where $\tilde B=a^iBa^j$ for any $i,j\in\mathbf Z$. Choose $i$ and $j$ so that $\tilde B$ has no leading or trailing powers of $a$. Then $b=\tilde B^p$ for some $p\in\mathbf Z$ and so $1=\lvert \tilde B^p\rvert\ge\lvert p\rvert$. This implies $\lvert p\rvert=1$, and the statement follows.

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You should clarify that $p$ is an integer, not a word. –  Arturo Magidin Jul 26 '11 at 20:25
Thanks! I've added that. –  Will Orrick Jul 26 '11 at 20:57
Very nice proof. –  Grumpy Parsnip Jul 30 '11 at 23:23
@WillO Can you explain the step where you write $b$ as $\tilde B^p$? Why cannot $b$ be, for instance, $\tilde B^p a \tilde B^q$ instead? Thanks. –  Srivatsan Jul 31 '11 at 4:22
@Swlabr : The essence is undoubtedly the same, although I wouldn't say that one version is more formal than the other. I was trying to emphasize and clarify in my own mind what seemed to be the key point - that powers of words do not compress to less than a certain length. I suppose I misunderstood what it was that dissatisfied you about your own proof. I was expecting, and still hope, that others would provide interesting alternatives. –  Will Orrick Aug 2 '11 at 4:48