Show that the free group on three generators is a subgroup of the free group on two generators

Question

I have been asked to show that the free group on three generators is a subgroup of the free group on two generators.

The following definition has been taken from the appendix to Armstrong's $\textit{Basic Topology}$:

The free subgroup $F^X$on $r$ generators $X=\{x_1,...,x_r\}$ has been defined as the infinite set of words obtained by concatenating the generators $x_i$ and their inverses $x_i^{-1}$ into words, where the inverse relation is $x_ix_i^{-1}=e$, the empty word, which is the identity element of the group, and naturally satisfies the relation that $e$ concatenated with any word $w\in F^X$ produces the same word $w$.

So the free group on three generators would be $F^X$, where $X=\{a,b,c\}$, and the free group on two generators would be $F^Y$, where $Y=\{a,b\}$.

We want to show that $F^X$ is a subgroup of $F^Y$. Now a requisite for a group being a subgroup of another, is that it is a subset of the group. But I cannot seem to see how $F^X$ can be a subset of $F^Y$ seeing as $c\not\in F^Y$. Even when one considers relabelling, one cannot discount the fact that the longest word we can create using distinct letters in $F^X$ is $ac^{-1}b^{-1}a^{-1}bc$, or some valid rearrangement of those letters. This word is of length 6. On the other hand, the longest word one can create using distinct letters in $F^Y$ is $a^{-1}bab^{-1}$, which has length 4. So $F^X$ must contain elements that are not in $F^Y$.

The notion that a group generated by a greater number of free elements should be a subgroup of one generated by a lesser number seems absurd to me, and I have almost convinced myself that the statement must be false.

All help and input would be highly appreciated.

Follow-up: The comments below have clarified the matter by indicating that the free group on three generators is $\textit{isomorphic}$ to a subgroup of the free group on two generators, which is what the two answers given below have proven. In response to this, I ask the following: As the free group on two generators clearly is a subgroup of the free group on three generators, does this imply that they are isomorphic groups?

Answer 1

It is no coincidence this was in an algebraic topology textbook. I will show the stronger statement "The free group on countably infinite generators is a subgroup of the free group on two generators."

Take the usual universal cover of $S^1 \times S^1$ which is given by the product of the universal covers for $S^1$, notably it's domain is $\mathbb{R}^2$. Now if we restrict this cover to the grid lines through $\mathbb{Z}^2$ we get a cover for $S^1 \vee S^1$. Pick your favorite spanning tree for the grid and contract it to see that it has the homotopy type of a wedge of countably many circles. Such a thing has fundamental group a free group on the inclusion of each circle, so is free on countably infinite generators. $S^1 \vee S^1$ has fundamental group that is free on two generators. The last thing you need is that any covering map induces an injection on fundamental groups.

If you actually pick a spanning tree, you can use it to write exactly what the basis of this group is. I think a basis element looks like $a^n b^m aba^{-1}b^{-1} a^{-n}b^{-m}$.

Answer 2

In the free group $F(a,b)$ on 2 generators $a,b$, the family $(a^nba^{-n})_{n\in\mathbf{Z}}$ is free.

One way to prove this is to consider the free group $F(\mathbf{Z})$ on the generators $(b_n)_{n\in\mathbf{Z}}$, its automorphism $f$ induced by the assignment $b_n\mapsto b_{n+1}$, and consider the semidirect product $G=\mathbf{Z}\ltimes_fF(\mathbf{Z})$, where the positive generator $t$ of $\mathbf{Z}$ acts by $f$. Then there is a unique homomorphism $u:F(a,b)\to G$ mapping $a\mapsto t$, $b\mapsto b_0$. Then $u$ maps $a^nba^{-n}$ to $b_n$. Since $(b_n)_{n\in\mathbf{Z}}$ is free, it follows that $(a^nba^{-n})_{n\in\mathbf{Z}}$ is free too.

Remark (not used above): one can show that $u$ is an isomorphism $F(a,b)\to G$.

Answer 3

If you order the words on $\langle a,b\ |\quad \rangle$ as $$1<a<a^{-1}<b<b^{-1}<a^2<ab<ab^{-1}< a^{-2}<a^{-1}b<a^{-1}b^{-1}< ba<ba^{-1}<b^2< b^{-1}a<b^{-1}a^{-1}<b^{-2}< a^3<a^2b<\ ...$$ you'll discover that the 1st three words of length two generate all the remaining nine and hence all the even's length words, this means that the set $$\{a^2\ ,\ ab\ ,\ ab^{-1}\}$$ generate a subgroup on three generators and is free by the Schreier's Subgroup Theorem.