What is the implication of this algebraic problem?

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SUMMARY

The discussion focuses on proving the implication that if \( a^x = b^y = (ab)^z = e \) in an abelian group \( (G, \cdot) \) where the GCD of \( x, y, z \) is 1, then \( a = b = e \). The participants analyze the relationships between \( a \) and \( b \) using properties of group theory and the definition of the unit element \( e \). Key insights include the necessity of \( x, y, z \) being non-zero integers without common divisors, which is critical for establishing the uniqueness of the elements involved.

PREREQUISITES
  • Understanding of group theory, specifically abelian groups
  • Knowledge of greatest common divisor (GCD) and its properties
  • Familiarity with integer powers in the context of group elements
  • Basic algebraic manipulation and proof techniques
NEXT STEPS
  • Study the properties of abelian groups and their implications on element orders
  • Learn about the GCD and its role in number theory
  • Explore cyclic groups and their characteristics
  • Investigate the implications of the unit element in group theory
USEFUL FOR

This discussion is beneficial for students and researchers in abstract algebra, particularly those studying group theory and its applications in mathematics. It is also useful for anyone looking to deepen their understanding of the relationships between group elements and their powers.

nuuskur
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Homework Statement


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:
<br /> a,b\in G\left (a^x = b^y = (ab)^z = e\Longrightarrow a=b=e\right )<br />
where ##e## is the unit element of ##G##.

Homework Equations


Corollary: For every integers ##x,y## there exist integers ##u,v## such that ##xu+yv = GCD(x,y)##

The Attempt at a Solution


If we look at what ##b## is equal to:
<br /> b^y = e \Leftrightarrow b^yb^{1-y} = eb^{1-y} \Leftrightarrow b^1 = b^{1-y}<br />
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?
 
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Have you looked at ## e^{xy}=e^{yz} = e^{xz} = a^{xyz} = b^{xyz} = (ab)^{xyz} ##?
 

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