Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Show that the even subgroup of ${\bf F}_2$ is generated by $S= \{x^{2}, xy, xy^{-1}\}$. I could use a little help getting started on this problem. I am new to the idea of free groups.

share|cite|improve this question
Sorry... what is the even subgroup of a free group? – Arturo Magidin Nov 19 '11 at 21:50
From what I can understand, an even subfroup of a free group contains as its elements, words of even length. – Olga Nov 20 '11 at 2:14
  1. Check that the elements of $S$ are all in the group you want to generate.

  2. Think about an obvious set of generators of your group. (If this poses a problem you might want to write down the definition of your group.)

  3. Show that all elements of the obvious set are generated by $S$.

share|cite|improve this answer

You clarify that the "even subgroup" is the set of all reduced words of even length.

A reduced word of even length will be a (possibly empty) product of reduced words of length 2 (because every subword of a reduced word is reduced); the reduced words of length 2 are: $xx$, $xy$, $x{y^{-1}}$, $x^{-1}x^{-1}$, $x^{-1}y$, $x^{-1}y^{-1}$, $yx$, $yx^{-1}$, $yy$, $y^{-1}x$, $y^{-1}x^{-1}$, $y^{-1}y^{-1}$. So the group of even words is the subgroup generated by these twelve words. In fact, you can cut it down to six by removing elements that are inverses of other generators: just keep $x^2$, $xy$, $x^{-1}y$, $xy^{-1}$, $x^{-1}y^{-1}$, and $y^2$.

In order to show that your set generates the even subgroup, we need to show that each of those generators lies in the subgroup generated by $x^2$, $xy$, and $xy^{-1}$. Of the six generators, three are directly given by $x^2$, $xy$, $xy^{-1}$. Two more, $x^{-1}y$ and $x^{-1}y^{-1}$, are given by multiplying $xy^{-1}$ on the left by the inverse of $x^2$. That's five of the six generators.

The generator we are still missing is $y^2$. An easy way of getting some $y^2$ is to multiply $xy$ by the inverse of $xy^{-1}$; we get $xy^2x^{-1}$. Getting $y^2$ by multiplying this by suitable elements from $\{x^2,xy,xy^{-1}\}$ or their inverses should be simple enough now.

Now, letting $\mathbf{E}$ be the even subgroup of $\mathbf{F}_2$, we have: $$\mathbf{E} =\langle x^2, xy, xy^{-1}, x^{-1}y,x^{-1}y^{-1},y^2\rangle \subseteq \langle x^2,xy,xy^{-1}\rangle \subseteq \mathbf{E}.$$

share|cite|improve this answer
Thank you so much Arturo for the start. – Olga Nov 20 '11 at 3:31

A slightly different approach:

Let $G=\langle x, y; x^2, xy^{-1}, xy\rangle$. Clearly $G\cong C_2$. Thus, the normal subgroup of $F_2$ generated by $S=\{x^2, xy^{-1}, xy\}$ has index $2$ and thus clearly must contain the even subgroup. It is therefore sufficient to prove that $\langle S\rangle$ is normal in $F_2$.

Proving that $\langle S\rangle$ is normal is $F_2$ isn't too hard - just prove that elements of $S$ conjugated by $x^{\pm 1}$ and $y^{\pm 1}$ are in the subgroup generated by $S$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.