# Algebra: abelian problem

1. Oct 16, 2015

### nuuskur

1. The problem statement, all variables and given/known data
Let $x,y,z$ be arbitrary integers and the GCD: greatest common divider between any two is $1$. Let $(G,\cdot )$ be abelian. Prove the implication:
$$a,b\in G\left (a^x = b^y = (ab)^z = e\Longrightarrow a=b=e\right )$$
where $e$ is the unit element of $G$.
2. Relevant equations
Corollary: For every integers $x,y$ there exist integers $u,v$ such that $xu+yv = GCD(x,y)$

3. The attempt at a solution
If we look at what $b$ is equal to:
$$b^y = e \Leftrightarrow b^yb^{1-y} = eb^{1-y} \Leftrightarrow b^1 = b^{1-y}$$
As $x,y,z$ have no common dividers, they have to be nonzero otherwise, for instance $GCD(x,0) = x$, which is not necessarely $1$.
At this point I'm stuck. If $b^1 = b^{1-y}$ and provided also $y\neq 0$, then $b\in G$ is equal to one of its powers. Cyclicity?

2. Oct 16, 2015

### RUber

Have you looked at $e^{xy}=e^{yz} = e^{xz} = a^{xyz} = b^{xyz} = (ab)^{xyz}$?