If $n>m$ and $a^n=a^m$ for any $n,m \in \mathbb{N}$, then $a^{n-(m-1)}\cdot a^{m-1} = a\cdot a^{m-1}$.

Question

Please bear with me. Undergraduate student. Regarding the question and answer here.

If $S$ is a semi-group and for each $a \in S$, in the first part of the proof by contradiction: If $n>m$ and $a^n=a^m$ for any positive integers $n$ and $m$, how does this imply that $a^{n-(m-1)}\cdot a^{m-1} = a\cdot a^{m-1}$. I'm struggling to understand how one gets to this equation from $a^n=a^m$? Do you first multiply on the left and right with something?

Thank you for the help.

Answer 1

$a^k=\underbrace{a·a·\ldots·a}_\text{$k$ times}$ so

$a^{n-(m-1)}a^{m-1}=\big(\underbrace{a·\ldots·a}_\text{$n-(m-1)$ times}\big)\big(\underbrace{a·\ldots·a}_\text{$m-1$ times}\big)=\underbrace{a·\ldots·a}_\text{$n-(m-1)$ times}·\underbrace{a·\ldots·a}_\text{$m-1$ times}=\underbrace{a·\ldots·a}_\text{$n-(m-1)+(m-1)$ times}=\underbrace{a·\ldots·a}_\text{$n-(m-1)+(m-1)$ times}=\underbrace{a·\ldots·a}_\text{$n$ times}=a^n$

by associativity.

On the other hand, $a·a^{m-1}=a·\big(\underbrace{a·\ldots·a}_\text{$m-1$ times}\big)=a·\underbrace{a·\ldots·a}_\text{$m-1$ times}=\underbrace{a·\ldots·a}_\text{$m$ times}=a^m$

by associativity too.

Since $a^n=a^m$, we have the result.

Answer 2

You're just regrouping the $a$'s that you already have. Multiply the left-hand side by adding the exponents and you'll see that you have $a^n$ because $n-(m-1)+(m-1)=n$. Similarly, multiply the right-hand side by adding exponents and you'll see that you have $a^m$ because $m=1+(m-1)$. You need $n \gt m$ to be sure that all exponents in sight are positive.

Then using the cancellation law you have $1 \cdot a = a = a^{n-(m-1)}=a^{(n-m)+1}=a^{n-m}a$ and using the cancellation law again you see $a^{n-m}=1$.