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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?

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    $\begingroup$ You are right, the problem statement should really ask you to show that $F^{\{a,b,c\}}$ is isomorphic to a subgroup of $F^{\{a,b\}}$ $\endgroup$ Feb 22 '19 at 21:34
  • $\begingroup$ You need to find three words in $F^{\{a,b\}}$ that satisfy no relations. $ab, ba,$ and $b^2$ look like they work to me, but I don't know how to prove it. $\endgroup$
    – saulspatz
    Feb 22 '19 at 21:44
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    $\begingroup$ I'm picky, but "is a subgroup" should be "is isomorphic to a subgroup" $\endgroup$
    – YCor
    Feb 22 '19 at 22:11
  • $\begingroup$ @YCor But in that case, you should also object to "the free group on two generators". $\endgroup$
    – Derek Holt
    Feb 23 '19 at 8:29
  • $\begingroup$ @DerekHolt I objected at many occasions on the word "generator". Still I see no relation between the two. For my previous remark, the "is a subgroup" terminology implies that $\mathbf{C}$ is a subgroup of $\mathbf{R}$. $\endgroup$
    – YCor
    Feb 23 '19 at 9:12
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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 $a^n b^m aba^{-1}b^{-1} a^{-n}b^{-m}$.

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    $\begingroup$ Thank you for your proof. It was must illuminating. As you have shown that the free group on any number of generators is isomorphic to a subgroup of the free group on two generators, I wonder, as the free group on two generators is clearly isomorphic to a subgroup of the free group on more than two generators, does this mean that the free group on any number of generators is isomorphic to the free group on two generators? A consequence of this would be that all free groups are isomorphic, so surely this cannot be. Where is the flaw in my reasoning? $\endgroup$ Feb 27 '19 at 15:48
  • $\begingroup$ Using the universal property one can show a free group has a well defined rank, i.e. two free groups are isomorphic if and only if they are free on a basis of the same cardinality. This is in contrast to the case of sets where if two sets both inject into each other they are isomorphic. $\endgroup$ Feb 27 '19 at 16:24
  • $\begingroup$ The flaw in your reasoning is that the conclusion doesn’t follow from the premises. You know that a subgroup of G is isomorphic to H and a subgroup of H is isomorphic to G. These don’t imply G is isomorphic to H. $\endgroup$ Feb 27 '19 at 19:57
  • $\begingroup$ @ConnorMalin Apparently all free group of all possible ranks is contained in $F_2$, including as you proved $F_\infty$. I just encountered this fact today, and I'm so surprised that it's true. Is there a way to make this surprising result more intuitive? $\endgroup$
    – chhro
    Feb 26 at 17:02
  • $\begingroup$ @chhro I suppose you can view rank as a measure of complexity, and I don’t think it should be very surprising that an object can contain a subobject that is more complex. For example, a disk is homotopically very simple, but it has a subspace, the circle, that is more complex. As you probably know, nonabelian groups are much more complicated than abelian groups, and the complexity of the subgroups of the free group is directly tied to the nonabelian nature of the free group. $\endgroup$ Feb 26 at 17:37
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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 $F(a,b)\to G$.

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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.

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