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This is what I know: (G,.) is a group $$a^0=e \\ a^n=a^{n-1}a\\a^{-n}=(a^n)^{-1}$$

I need to prove for n and m integers $$i)\ a^{m+n}=a^ma^n\\ii)\ (a^m)^n=a^{mn}$$

For i), my attemp was trying to separate the problem into the following cases:

  • $m>0$ and $n>0$: proved this one by induction over n. It applies for $m>0$ and $n=0$ or $n=1$

Assuming $a^{k+m}=a^ka^m$, I want to prove $a^{m+(k+1)}=a^ma^{k+1}$ $$a^{m+(k+1)}=a^{(m+k)+1}=a^{m+k}a^{1}=a^ma^ka^1=a^ma^{k+1}$$ Is this part ok?

  • $m<0$ and $n<0$: this one I really don't know. I tried the following $$p=-m\ q=-n\\a^{m+n}=a^{-p-q}=a^{-(p+q)}=(a^{p+q})^{-1}=(a^pa^q)^{-1}$$ $$(a^pa^q)^{-1}=a^{-p}a^{-q}$$
  • If m and n have different signs then I don't know... maybe trying with the absolute value??

For ii) I did it by induction similar to the first case, would this be correct?

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both first and second part are ok, why not use a similar inverse technique on the last part? – gt6989b Feb 20 '14 at 23:24
up vote 0 down vote accepted

for the last part, here is a hint $$ a^{m + (-n)} a^n = a^{m + (-n) + n} = a^m, $$ so $a^{m + (-n)} = a^m \cdot (a^n)^{-1}$.

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How do I know $m+(-n)<0$ ? – Shomar Feb 21 '14 at 18:48
@Shomar why do you need it? Make 2 cases, inverses of the positives will yield the negative case. – gt6989b Feb 21 '14 at 19:09

The relevant definitions ensure that $x^{m+n}=x^m x^n$ whenever $m\geq 0$ and $n=0$ or $1$. Suppose that $x^{m+k}=x^m x^k$ for some positive integer $k$. Then, $$x^{m+k+1}=x^{m+k} x = (x^m x^k)x = x^m (x^k x)=x^m x^{k+1}. $$

It therefore follows by induction on $n$ that $x^{m+n}=x^m x^n$ for non-negative integers $m$ and $n$.

Let $p$ and $q$ be non-negative integers. Then, $$x^{-p-q}=(x^{p+q})^{-1}=(x^p x^q)^{-1}=(x^p)^{-1}(x^q)^{-1}=x^{-p} x^{-q} $$

We deduce from this that $x^{m+n}=x^m x^n$ when $m$ and $n$ are both negative.

Now let $p$ and $q$ be non-negative integers with $p\leq q$. Then $x^q=x^p x^{q-p}$ and $x^q=x^{q-p} x^p$. On multiplying these identities by $x^{-p}$ on the left and right respectively, we deduce that $x^{q-p}=x^q x^{-p}$ and $x^{q-p} =x^{-p} x^q$. On taking inverses, we see also that $x^{p-q}=x^p x^{-q}$ and $x^{p-q}=x^{-q}x^p$. We can apply these formulas either with $p=|m|$ and $q=|n|$ (in the case when $|m|<|n|$), or with $p=|n|$ and signs. Combining this result with the corresponding results when $m$ and $n$ have the same sign, we deduce that $x^{m+n}=x^m x^n,\quad \forall m,n \in \mathbb{Z}$.

The identity $x^{mn}=x^m x^n$ follows immediately from the definitions when $n=0,1$ or $-1$. If $x^{mk}=(x^m)^k$ for some integer $k$, then $$(x^{m})^{k}=x^{mk+m}=x^{mk} x^m=(x^m)^{k}x^m=(x^m)^{k+1}. $$ It follows by induction on $n$ that $x^{mn}=(x^m)^{n}$ for all positive integers $n$. The result when $n$ is a negative integer then follows on taking inverses. Thus, $x^{mn}=(x^m)^{n} \quad \forall m,n \in \mathbb{Z}$, as required.

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