I want to prove that if $G$ is a group and $a\in G$, $n,m\in \Bbb Z$, then $a^na^m=a^{n+m}$. I think, that it's easier to prove the case when $n,m\in \Bbb N$. I found this question: Induction (over 2 variables) and I've been trying to use what the answer says.
Lets consider m fixed, then we do the induction on n.
case $n=1$) $a^1a^m=a*(a*...*a)=a*a*...*a=a^{1+m}$. Is this correct? I feel like I'm using what we want to prove, maybe I'm using the ellipsis wrong.
What I thought was a little bit more complicated, but I don't know if it's correct.
case $n=1$) $a^1a^m$, then we leave n=1 fixed and prove by induction on m that $a^1a^m=a^{1+m}$:
subcase $m=1$) $a^1a^1=a^2=a^{1+1}$
Lets assume this is valid for any $m\in \Bbb N$, ie, $a^1a^m=a^{m+1}$.
Lets prove that $a^1a^{m+1}=a^{m+2}$. $a^1a^{m+1}=a^1(a^1a^m)=(a^1a^1)a^m=a^2a^m=a^{m+2}$.
It's pretty obvious that this way is pretty long, I haven't finish it, and I want to know if this is valid.
If you want to proceed with two-variable induction:
Those two cases prove $\forall m\in\mathbb N, P(1,m)$ (i.e in your case $a\cdot a^m = a^{m+1}$). This is where you stopped, and just to be sure I'll tell you the last step:
In your first attempt (with a single induction), some people would say this is perfectly fine (but then, when don't you just say $a^n\cdot a^m = \underbrace{(a\cdot a \cdot \ldots \cdot a)}_n \cdot \underbrace{(a\cdot a \cdot \ldots\cdot a)}_m = a\cdot \ldots\cdot a = a^{m+n}$?). The purpose of these simple questions is (for the teacher) to be sure that you understood what's going on behind those dots. The most formal way to do this is to do the two-variables induction, because it doesn't appeal to any "$\dots$".