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I have to prove $(a^m)^n=a^{mn}$ for all $a\in G$ and $m,n\in\mathbb{Z}$ where $G$ is a group. Is it enough to just expand $(a^m)^n=(a^m***a^m)$- $n$ times. And then from here we can expand it a bit more to there there are $mn$ amount of $a's$? Or do I need to break it up into cases. I felt if I did I'll have atleast 3 cases and a few subcases. As of right now I have two cases one where $m=n=0$ and the other where $a,b\neq 0$.

Here is what I have so far.

New proof #2

Case 1: Let $m>0$ and $n>0$ We will proceed by induction. We fix m and induct on n. Base case: Let n=1. We see that $a^m=a^m$. Inductive case: Suppose that $(a^m)^k=a^{mk}$ We shall prove $(a^m)^{k+1}=a^{m(k+1)}$. It follows immediately from assumption that $(a^m)^{k+1}=a^{m(k+1)}$.

Case 2: $m=n=0$. It is immediately obvious that $(a^m)^n=a^{mn}$

Case 3: $m<0$ and $n<0$. Let $m=-t$ and $n=-r$ where $t,r>0$. Then $(a^m)^n=(a^{-t})^{-r})=(a^{-1})^t)^r)^{-1}=(a^{-1})^{rt})^{-1}=(a^{-nt})^{-1}=(a^{nt})^{-1}=a^{n*(-1)t}=a^{mn}$

user60887
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    We will need to deal with $m$ and/or $n$ that may be negative. Separate argument will be needed, since the definition of negative powers is different from the definition of positive powers. – André Nicolas Sep 21 '13 at 05:30
  • well I was thinking even with negative powers we are just making sure the total number of $a's$ one side is the same as the other side despite $m,n$ being negative or postitive – user60887 Sep 21 '13 at 05:33
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    Yes, but $a^{-1}$ is defined as the inverse of $a$, so there are a couple of lemmas to prove before we can use a counting argument. – André Nicolas Sep 21 '13 at 05:38
  • Ah ok that is true. So basically I have extend the fact that $(ab)^{-1}=b^{-1}a^{-1}$ and show that it is true by induction. And then I can use a counting argument? – user60887 Sep 21 '13 at 05:43
  • One can use the reults for positives to prove for negatives. Several cases. For non-negatives, can use induction on $n$ as a formal replacement for counting. – André Nicolas Sep 21 '13 at 05:46
  • oh ok thanks. I'll do that and post what I got afterwards – user60887 Sep 21 '13 at 05:50
  • OK. Am not planning to write out a solution, lots of typing. – André Nicolas Sep 21 '13 at 05:52
  • Oh no I want to do this myself. – user60887 Sep 21 '13 at 06:23
  • Oh yeah I posted my proof. I'm not sure if its entirely correct. – user60887 Sep 22 '13 at 00:55

2 Answers2

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Case 1: Let $m>0$ and $n>0$ We will proceed by induction. We fix m and induct on n. Base case: Let n=1. We see that $a^m=a^m$. Inductive case: Suppose that $(a^m)^k=a^{mk}$ We shall prove $(a^m)^{k+1}=a^{m(k+1)}$. It follows immediately from assumption that $(a^m)^{k+1}=a^{m(k+1)}$.

Case 2: $m=n=0$. It is immediately obvious that $(a^m)^n=a^{mn}$

Case 3: $m<0$ and $n<0$. Let $m=-t$ and $n=-r$ where $t,r>0$. Then $(a^m)^n=(a^{-t})^{-r})=(a^{-1})^t)^r)^{-1}=(a^{-1})^{rt})^{-1}=(a^{-nt})^{-1}=(a^{nt})^{-1}=a^{n*(-1)t}=a^{mn}$

user60887
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This is more a comment than an answer, but you nowhere say how you define $a^n$. You also do not say if the group is Abelian.

My guess is that the definition is something like this: $a^0 = e$ (the identity element). If $n$ is an non-negative integer such that $a^n$ is defined, then $a^{n+1} =a(a^n) $ (or maybe $(a^n)a$). If $n < 0$, then $a^n =(a^{-1})^{-n} $.

To me, this looks like the problem of defining multiplication in terms of addition.

I recommend Landau's "Foundations of Analysis" for a bracingly rigorous discussion of these and similar matters.

marty cohen
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  • (+1) for reference, can you prove the main question in a set which satisfies all field axioms? – NewBornMATH Mar 25 '19 at 19:41
  • The thing is i have proved for real numbers $(ab)^{-1}=a^{-1}b^{-1}$ and want it to generalise to $(ab)^{-n}=a^{-n}b^{-n}$. So i want to prove this(the op's question) entirely using field axioms, any help will be greatly appreciated ! : ) – NewBornMATH Mar 25 '19 at 19:44
  • Are you sure Landau works with arbitrary groups? There are some shortcuts you can make in those basic proofs if you assume commutativity. – darij grinberg Jul 31 '19 at 18:34