Simple question on associativity and commutativity in a group This seems to me to be a rather silly question, but I want to be sure that I'm getting it right.
Say that I have a group $G$, and I consider the element $a \in G$ and specifically the element $a^5$. It is certainly the case that
$$a^5 = a^3 a^2.$$
Is the justification for this step associativitiy in the group? Or am I using the fact that any element of a group commutes with powers of itself? 
 A: To clarify things, you should first think about this question: what do $a^3$ and $a^5$ mean?
You may want to define them recursively: $a^n$ is by definition $a^{n - 1}\cdot a$. This would lead to the following definition: $$a^3 = (a\cdot a)\cdot a.$$
But why not $a^3 = a \cdot (a \cdot a)$, which looks equally good as a definition?
Things get worse for $a^5$: it could be $((((a \cdot a)\cdot a)\cdot a)\cdot a)$, or $(a\cdot(a\cdot(a\cdot(a\cdot a))))$, or $(a \cdot a) \cdot ((a \cdot a) \cdot a)$, or many many other possibilities.

The reason that we don't have to deal with this complication is exactly associativity.
By the associativity axiom, all the above definitions should give the same element, hence justifying the notations $a^3$ and $a^5$, i.e. we can put parentheses anywhere in the defining formula.
And after that, it becomes trivial to see that $a^5 = a^3 \cdot a^2$.
A: Associativity is what is required here. Without associativity, this doesn't work. Let's say the operation is subtraction on the integers, which is not associative. We compute
$$1^5 = -3$$
$$1^3 = -1$$
$$1^2 = 0$$
So
$$1^31^2 = -1\neq 1^5$$
If you want to single out associativity we can use $xy = 2|x-y|$, since this operation is commutative. Then we get
$$1^5 = 2$$
$$1^3 = 2$$
$$1^2 = 0$$
$$1^31^2 = 4\neq 1^5$$
A: If we define $a^{n+1}=a^n\cdot a$, then the claim is
$$(((a\cdot a)\cdot a)\cdot a)\cdot a = ((a\cdot a)\cdot 
a)\cdot(a\cdot a)$$
and this is associativity applied to $(a\cdot a)\cdot 
a, a, a$.
A: You may consider the following simple example of a magma to see that being accotiativity is vital for your specific conclusion:
+|a b 
-----
a|b b  
b|b a 

