Do the laws of exponents apply to a Group as for real numbers? I would like to know if this:
$$ a^k \ast a^{-m} = a^{k-m}$$
is true for any given group G, $\ast$ 
Thanks.
 A: $a^k*a^{-m} = a^{k-m}$ is true, simple to prove and is almost (but not quite) entirely notation and has little to do with group theory and has nothing to do with any particular group.
$a^k; k\in \mathbb N; k> 1$ simply means $a*a*a*.... *a$ times and as the group is associative that is meaningful notation.  And $a^1 = a$ and $a^0 = e$ is certainly definable.  That $a^k*a^n = (a*a*a....*a)*(a*a*a*....*a) = a*a*a....*a*a*=a^{k +n}$ follows directly by associativity and has nothing to do with the group but only to do with the notation.
$a^{-1}$ is defined to be that inverse of $a$ so that $a*a^{-1} = a^{-1}*a = e$.  As $G$ is a group, then it is a basic definition that such an element will exist and be unique.
$a^{-m}$ can either be defined to be $(a^{-1})^m = a^{-1} * a^{-1}*......*a^{-1}$ or as $(a^m)^{-1}$ i.e. the unique element, $d$ , so that $a^m*d = d*a^m = e$.  I is easy to prove that those two different expressions are equal.  via $a^m*(a^{-1})^m = a(a(a(......(aa^{-1})a^{-1}) ..... ) a^{-1}=a(a((....a(e)a^{-1})a^{-1}).....) = e$ so $(a^m)^{-1} = (a^{-1})^m$ so writing this number as $a^{-m}$ is unambiguous and "makes sense".
If finally:
Case 1:  If $n = m$ then $a^na^{-m} = a^n(a^n)^{-1} = e = a^0 = a^{n-m}$.
Case 2:  If $n > m$ then $a^n = a*a*a*a*....*a = a^{n-m}a^m$ so $a^n *a^{-m} = a^{n-m}a^ma^{-m} = a^{n-m}e = a^{n-m}$ by case 1:
Case 3: If $n < m$ and $a^{-m} = (a^{-1})^m= (a^{-1})^n(a^{-1})^{m-n}$
So $a^n*a^{-m} = a^n*(a^{-1})^n(a^{-1})^{m-n} = e(a^{-1})^{m-n} = a^{n-m}$ by case 1.
A: The specific law you mention does hold for all groups, but in general no: the laws of exponents do not apply to a group as for real numbers. 
To be specific the following does hold in any group:
$$ x^p x^q = x^{p+q} $$
$$ (x^p)^q = x^{pq} $$
The following only holds in general for abelian groups:
$$ (xy)^p = x^py^p $$
Regarding the laws involving division there's a question about what you mean by division. If you define them as being multiplication by the inverse of the divisor it's straight forward for abelian groups, but for non-abelian you would need to define which side you multiply from. Regardless the following would hold for any group:
$$ x^p / x^q = x^{p-q}$$
While the following requires abelian group:
$$ (x/y)^p = x^p/y^p$$
$$ (x/y)^{-p} = (y/x)^p$$
The reason the laws that require abelian group doesn't hold in general is of course that $(xy)^2 = xyxy$ and without commutativity you're not guaranteed this to be equal to $xxyy$.
A: Yes it is true , but be careful. Not all the laws hold. For example, $$(ab)^k=a^kb^k$$ is not always true unless the group is abelian.
