Proof that $|x||y|=|xy|$ where $|a|$ is the order of $a$

Question

Let $G$ be a finite group, $|a|$ the order of $a$ defined as the minimum positive integer such that $a^{p}=1$

Given $x,y\in G$, meeting

$(1):xy=yx$,

$(2):gcd(|x|,|y|)=1$

Prove that $|x||y|=|xy|$

.

My attempt was:

Let $|x|=n, |y|=m$

Using $(1)$ I get that $(xy)^{nm}=x^{nm}y^{nm}=1$

Now I need to prove that $\nexists p\in \mathbb{Z} ^+$ such that $(xy)^p=1, p<mn$

In other similar exercises I assumed that $p$ existed and found a contradiction

I thought of doing:

$1=(xy)^p=x^p y^p \implies x^p=1$ and $y^p=1$

Then the rest would be easy, but I don´t know if that last implication is true and haven´t found a way to prove it.

Could you tell me whether that is true or not, or a hint for another way to solve the problem? Thanks

.

Edit:

I know that the last implication is not true in general, but I am assuming the condition $(2)$

Answer 1

If $\gcd(|x|,|y|)=1$, and $g\in\langle x\rangle\cap\langle y\rangle$, then $g=e$.

To see that, note that $|g|$ divides $|x|$ (because $g\in\langle x\rangle$); and $|g|$ also divides $|y|$ (because $g\in\langle y\rangle$), hence $|g|$ divides $\gcd(|x|,|y|)=1$. Thus, $|g|=1$ so $g=e$.

So you already know that $|xy|$ divides $|x||y|$. Let’s do a direct proof instead of a proof by contradiction. We show that if $(xy)^k=e$, then $|x||y|$ divides $k$.

Assume that $k\gt0$ is such that $(xy)^k=e$. Then $e=(xy)^k=x^ky^k$; hence $x^k=y^{-k}$, so $x^k\in\langle x\rangle\cap\langle y\rangle$. Thus, $x^k=e$, so $|x|$ divides $k$. Symmetrically, $y^{-k}\in\langle x\rangle \cap\langle y\rangle$, hence $y^{-k}=e$, so $|y|$ divides $k$.

Therefore, $|x|$ divides $k$, and $|y|$ divides $k$, so $\mathrm{lcm}(|x|,|y|)$ divides $k$. Since $\gcd(|x|,|y|)=1$, you have $\mathrm{lcm}(|x|,|y|) = |x||y|$, so $|x||y|$ divides $k$.

Thus, $(xy)^{|x||y|}=e$, and if $(xy)^k=e$ with $k\gt 0$, then $|x||y|$ divides $k$. Hence $|x||y|$ is the smallest positive integer $r$ such that $(xy)^r=e$, so $|xy|=|x||y|$.

Answer 2

Consider $a$ and $a^{-1}$. Then $\operatorname{ord}(a) = \operatorname{ord}(a^{-1})$, but $\operatorname{ord}(a \cdot a^{-1}) = \operatorname{ord}(e) = 1$

Answer 3

Just because the product of two elements is the identity, doesn't make them both the identity. But you do get $x^p=y^{-p}\implies x^p\in \langle x\rangle\cap\langle y\rangle$. This means $x^p=e=y^{-p}$. (Here I have used that the cyclic groups generated by $x$ and $y$ have relatively prime orders.)