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Proof

(related to Proposition: Unique Solvability of $a\ast x=b$ in Groups)

• By hypothesis, $(G,\ast)$ is a group.
• We first show that $a^{-1}\ast b$ is the solution of the equation $a\ast x=b.$
  • Replacing \(x\) by \((a^{-1}\ast b)\) in the equation leads to $a\ast (a^{-1}\ast b)=b.$
  • Since "$\ast$" is associative, we get $(a\ast a^{-1})\ast b=b.$
  • Since the inverse element $a^{-1}$ is unique, it follows $e\ast b=b,$ where $e$ is a neutral element in $G$ (even "the" neutral element in $G,$ since it is also unique in $G.$)
  • It follows $b=b$ which shows that $a^{-1}\ast b$ is a solution.
• Now, we show, the solution $a^{-1}\ast b$ is unique.
  • Assume, $c\in G$ is a solution of the equation $a\ast x=b,$ i.e. $a\ast c=b.$
  • Then $a^{-1}\ast (a\ast c)=a^{-1}\ast b.$
  • Since "$\ast$" is associative, we get $(a\ast a^{-1})\ast c=a^{-1}\ast b.$
  • Thus, $e\ast c=a^{-1}\ast b.$
  • Thus, $c=a^{-1}\ast b.$
    ∎

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References

Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983