2
$\begingroup$

I'm trying to learn abstract algebra, and I'm stuck on these two problem, both about free groups - specifically, both about the free group of 2 generators.

So let the group be $\mathbb{F}_2=\langle x,y \rangle$. The questions are:

First problem

The first question is to show is that the subgroup $G=\langle x^2, xy\rangle$ is also free. I have no idea how to approach it.

Second problem

The second problem is to find a subgroup of $\mathbb{F}_2$ which is isomorph to $\mathbb{F}_\omega = \langle u_i \rangle_{i\in\mathbb{N}}$

My approach has been this - I want to find a formula for the generator $u_i$ in terms of $x,y$ - so basically to "encode" the group with $x,y$. However, everytime I try, either exponentiating a generator doesn't do what it should, or multiplying two distinct generators doesn't do what it should, or inversion doesn't work, etc.

I've asked this in another math-help forum, and despite getting some pointers, nothing really helped me.

Thanks in advance.

$\endgroup$
  • $\begingroup$ 1. Subgroups of free groups are free, but if you don't know that, any relatipn between $x^2$ and $xy$ would also be be trivial in $F(x,y)$. That means, at the least, the exponents of $x$ and $y$ would be zero. What does that tell you about your relation? $\endgroup$ – Steve D Jan 19 '17 at 20:38
  • $\begingroup$ 2. Consider the group generated y $x^iyx^{-i}$ for all positive integers $i$. Can you show this is free? $\endgroup$ – Steve D Jan 19 '17 at 20:40
0
$\begingroup$

Hints:

First problem: Let $a=x^2$ and $b=xy$. Then $G\cong F(a,b)/\langle\langle R\rangle\rangle$ (where $F(a,b)\cong\mathbb{F}_2$ is the free group on $a$ and $b$). So an element in $G$ looks like $x=x_0^{\pm 1}x_1^{\pm 1}\cdots x_n^{\pm 1}$ where $x_i\in\{a,b\}$ for each $i$. Show by induction on $n$ that $x=1$ in $G$ if and only if $x=1$ in $F(a,b)$.

Second problem: Consider conjugation of $x$ by $y$ (that is $y^{-1}xy$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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