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I was reading this answer. I understand almost all of it. However, there is still one thing that continues to puzzle me.

How should I prove for sure that, in this example, if $m_1\neq m_2$ and $a^{m_1}=a^{m_2}$, then $a^{m_1-m_2}=e$.

This proof expects me to know exactly why this holds true. Am I missing something?

I don't quite get why this is the case. What is the proof behind this assertion?

I guess I can see why $a^0=e$ but how can I be certain that this holds true for cases such as the one presented above?

Thank you!

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One way to look at $a^{-m_2}$ is $$a^{-m_2}=\underbrace{a^{-1}\times \dots \times a^{-1}}_{m_2\text{ times}},$$ so if we multiply $a^{m_1}=a^{m_2}$ on one side, say, the right, then we have

$$\begin{align} a^{m_1}a^{-m_2}&=\underbrace{a\times \dots \times a}_{m_1\text{ times}}\times \underbrace{a^{-1}\times \dots\times a^{-1}}_{m_2\text{ times}} \\ &=a^{m_1-m_2} \\ &=\underbrace{a\times \dots \times a}_{m_2\text{ times}}\times\underbrace{a^{-1}\times \dots\times a^{-1}}_{m_2\text{ times}} \\ &=a^{m_2}a^{-m_2}\\ &=e. \end{align}$$

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If $$a^{m_1} = a^{m_2}$$ then by multiplying by $a^{-m_2}$ both sides we get $$a^{m_1}a^{-m_2} = a^{m_2}a^{-m_2}$$ The left hand side turns out to be $a^{m_1-m_2}$ while the right hand side is $e$.

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    $\begingroup$ For the "turns out to be" part, it might help to expand out the powers. By definition, $a^{m_1} = \underbrace{a a a \ldots a}_{m_1 \text{ times}}$, etc. $\endgroup$ – 6005 Mar 20 at 17:18
  • $\begingroup$ @6005 Looks like Shaun already done that. Good idea though. $\endgroup$ – Yanko Mar 21 at 9:41

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