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Proposition:Let $a$ be an element of a group so, for any $m,n\in\mathbb{Z}$,

$a^ma^n=a^{m+n}$ and $(a^m)^n=a^{mn}$

I want to prove the proposition by induction.

Is the following statement enough:

$a^{m+2}=a^{m+1}a=a^{m}a=a^ma^2$, so if we take $n\in\mathbb{N}$ we have $a^{m}a^n=a^{m+n}$.

I was thinking of an analogous way to prove the other identity.

Questions:

Am I doing the proof right?

If not. What should I do?

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Pedro Gomes
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    How do you define $a^m$ in the first place? A proof of those identities by induction should work just fine, if the definition of $a^m$ is an inductive definition. – Lee Mosher Oct 29 '17 at 16:47
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    Here is almost the same question, although since it has no answer one cannot say the current question is a duplicate: https://math.stackexchange.com/questions/767423/proof-verification-that-group-elements-follow-law-of-exponents?rq=1 – Lee Mosher Oct 29 '17 at 16:48
  • Everything comes down to associativity of group operation. Basically, writing $a \cdot b \cdot c \cdot d$ is fine as long as group operation goes. – user340297 Oct 29 '17 at 16:52
  • @LeeMosher In the material I am studying $a^m$ is defined as usual. $a^m=aaa...a$ m times. – Pedro Gomes Oct 29 '17 at 16:54
  • That's not really a true definition, more of an intuition. A proper definition by induction would be: Base case $a^1=a$; Inductive case $a^{n+1}=a^n \cdot a$. – Lee Mosher Oct 29 '17 at 16:55
  • I think one subtle thing is that this really has nothing to do with group theory but just to to with natural number indexing. $a^ma^n = a^{m+n}$ true for the exact same reason as having n apples next to m apples means you have $m + n$ apples. Induction should work fine but ... I don't see how you know that $a^{m+2}=a^{m+1}a$. Was that part of you definition. – fleablood Oct 29 '17 at 17:02
  • @LeeMosher I think it can be considered a proper definition if it is assumed basic arithmetic on integers has been established and "exists" independently and practical. But if so there is nothing to prove. $a^ma^n = a^{m+n}$ because $m$ $a$ and $n$ more $a$s is $m + n$ $a$s. – fleablood Oct 29 '17 at 17:07
  • What about negative powers? – Randall Oct 29 '17 at 17:07
  • $a^{-1}$ is the $k$ so that $ka = e$. There's an easy to prove proposition that $(a^n)^{-1} = (a^{-1})^k$. So $a^{-n} = (a^{-1})^n$ and so $a^{m-n}=a^{m-n}e=a^{m-n}(a^na^{-n}) = (a^{m-n}a^n)a^{-n} = a^{m-n+n}a^{-n} = a^m*a^{-n}$ for $m, n > 0$. – fleablood Oct 29 '17 at 17:16
  • @fleablood: although I do agree, one could of course get a rigorous proof without assuming basic arithmetic and instead mimicking the proof of the exact same identities in Peano arithmetic. – Lee Mosher Oct 29 '17 at 17:32
  • @LeeMosher exactly! But in all practicality.... that's kind of overkill. Or maybe pertinent to another area of study. – fleablood Oct 29 '17 at 17:48

2 Answers2

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Here is a proof as I allude to in my comments, although this proof depends on having a more rigorous inductive definition of exponentiation as follows:

  • Base case: $a^0 = \text{Id}$, the identity element of the group
  • Inductive case: assuming $a^n$ is defined for some integer $n \ge 0$, define $a^{n+1} = a^n \cdot a$.

Then you can prove each identity by induction on one of the exponents. For instance, to prove $a^m a^n = a^{m+n}$ one can induct on $n$:

  • Base case: $$a^m \cdot a^0 = a^m \cdot \text{Id} = a^m = a^{m+0} $$
  • Inductive case: $$a^m \cdot a^{n+1} = a^m \cdot (a^n \cdot a) = (a^m \cdot a^n) \cdot a = a^{m+n} \cdot a = a^{(m+n)+1} $$
Lee Mosher
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using induction you have have to show that if $a^ma^n=a^{m+n}$ then $a^ma^{n+1}=a^{m+n+1}$ and then because $a^ma^0=a^m=a^{m+0}$ it is true

so if $a^ma^n=a^{m+n}$ is true then $a^ma^{n+1}=a^m\left(a^na\right)=\left(a^ma^n\right)a=a^{m+n}a=a^{m+n+1}$

and done. this is showing it is true for specific $m$ and all $n$, you can do the same thing with $a^{m+1}a^n$ and you are done for both $m$ and $n$. ofc if you can say that $kp=pk$ then you dont need to do the last part. but this is depends on what you can assume to be true

ℋolo
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