# Abstract Algebra: Abelian group order

1. May 12, 2014

### Airman

1. The problem statement, all variables and given/known data

Let G be an abelian group and let x, y be elements in G. Suppose that x and y are of finite order. Show that xy is of finite order and that, in fact, o(xy) divides o(x)o(y). Assume in addition that (o(x),(o(y)) = 1. Prove that o(xy) = o(x)o(y).

3. The attempt at a solution

I was able to prove the first part, that xy is of finite order and that o(xy) divides o(x)o(y). I'm having trouble with the second part, proving that o(xy) = o(x)o(y) if the greatest common factor of o(x) and o(y) is 1.

2. May 12, 2014

### jbunniii

If $(xy)^k = 1$, then $x^k$ and $y^k$ are inverses of each other. Can this happen if $x^k$ and $y^k$ are not the identity element, and $((o(x),o(y)) = 1$?

Last edited: May 12, 2014
3. May 12, 2014

### Airman

So since (xy)^(o(xy)) = e, x^(o(xy)) and y^(o(xy)) are inverses of each other. We also know x^(o(xy)) = e because it's the same as x^(o(x))^(o(y), and the same goes for y^(o(xy)) = e. I feel like I'm really close but I'm still not grasping why it has to be the same.

4. May 12, 2014

### jbunniii

The question is to find out whether $(xy)^k = e$ is possible where $k$ is a proper divisor of $o(x)o(y)$. We want to show that this is impossible if $(o(x),o(y)) = 1$.

So suppose $(xy)^k = e$. Since the group is abelian, this means $x^k y^k = e$, and so $x^k$ and $y^k$ are inverses of each other. I claim that this implies that $x^k = y^k = e$. Therefore, assuming this claim is true, $k$ must be a multiple of both $o(x)$ and $o(y)$ and therefore a multiple of $o(x)o(y)$ since $(o(x),o(y)) = 1$.

So the key is to prove the claim. This brings us back to the question I asked in my first post. Can $x^k$ and $y^k$ be inverses of each other if they are not the identity?

Hint: what can you say about the intersection of $\langle x\rangle$ and $\langle y \rangle$, the subgroups generated by $x$ and $y$ respectively?

Last edited: May 12, 2014
5. May 18, 2014

### Airman

Ok, I've spent a lot of time struggling with this one. I don't know why it's so hard for me! I understand completely how if $x^k = y^k = e$, then the desired result that $o(xy) = o(x)o(y)$ is true. So, to prove your claim. I think this works, please let me know if it does not:

$x^k$ and $y^k$ are inverses of each other. An element must have the same order as its inverse, so $o(x^k) = o(y^k)$. But $o(x^k) = o(x)/(k,o(x))$, and $o(y^k) = o(y)/(k,o(y))$, so $o(x)/(k,o(x)) = o(y)/(k,o(y)) = p$. If $p > 1$, both $o(x)$ and $o(y)$ are divisible by $p$. But since $(o(x),o(y)) = 1$, $p = 1$, so both sides of the equation must equal $1$, which means $x^k = y^k = e$.

Does this proof make sense? And even if it does, is there a more elegant way to prove this? Thank you so much for your help!

6. May 18, 2014

### jbunniii

I think your proof is OK, but the argument I had in mind may be simpler because it doesn't require the formula involving the gcd: $x^k y^k = e$, so $x^k$ and $y^k$ are inverses of each other. Now $\langle x\rangle$ must contain both $x^k$ and its inverse $y^k$, and similarly, $\langle y \rangle$ must contain both $x^k$ and $y^k$. This means that $x^k$ and $y^k$ are both elements of $\langle x \rangle \cap \langle y \rangle$. Since $\langle x \rangle \cap \langle y \rangle$ is a subgroup of both $\langle x \rangle$ and $\langle y \rangle$, its order must divide $o(x)$ and $o(y)$. But $o(x)$ and $o(y)$ are relatively prime, so this forces $|\langle x \rangle \cap \langle y \rangle| = 1$ and therefore $\langle x \rangle \cap \langle y \rangle = \{e\}$. We conclude that $x^k = y^k = e$.

Last edited: May 18, 2014
7. May 19, 2014

### micromass

Follow-up question: Is the same true in non-abelian groups

Generalization of the question: Prove that if in a group $G$ the elements $x$ and $y$ commute, then
$$o(xy)~\text{divides}~\textrm{lcm}(o(x),o(y)) = \frac{o(xy)}{\textrm{gcd}(o(x),o(y))}$$
and
$$\frac{o(xy)}{\textrm{gcd}(o(x),o(y))^2}~\text{divides}~o(xy)$$

Find a counterexample where $o(xy) \neq \textrm{lcm}(o(x),o(y))$ even if they $x$ and $y$ commute.