# Why does $a^{m_1}=a^{m_2}$ imply $a^{m_1-m_2}=e$?

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!

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}
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 $$e$$.
• 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. – 6005 Mar 20 at 17:18