Why should $b$ groups of $a$ apples be the same as $a$ groups of $b$ apples? Why should $b$ groups of $a$ apples be the same as $a$ groups of $b$ apples?

We where taught this so it seems rather trivial but the more I think about it the more I feel that it is not. 

I'm trying to avoid an argument that uses the fact that multiplication is commutative. Because I see that I am trying to PROVE that in $\mathbb{Z}^{+}-0$ multiplication is commutative if we define multiplication by repeated addition. 

I would accept arguments using the fact that:

$a+b=b+a$ because if we define $+$ to be the operation combining to quantities then it should be rather trivial that $a$ apples and $b$ apples is the same as $b$ apples and $a$ apples.

Is it enough to draw $a$ groups of $b$ (1 by 1) squares and rotate this to show that it is the same as $b$ groups of $a$ (1 by 1) squares. It does not seem good enough for me because it uses a picture, and I was taught before that pictures in math do not prove anything.
The usual way of proving it is by induction.

First, define multiplication recursively: $$a\cdot 1 = a\\a\cdot(b+1)=a\cdot b + a$$

Next, show, by induction on $b$ that $1\cdot b=b\cdot 1$. That's relatively easy to do.

Next, prove by induction on $b$ that that $(a+1)\cdot b = a\cdot b + b$ by induction on $b$.

Finally, prove by induction on $a$ that for all $b$, $a\cdot b = b\cdot a$.
Multiplication is commutative because a collection of objects that has $a$ groups of $b$ objects in each group can be partitioned also into $b$ groups of $a$ objects. This is the equivalent to repeated addition:
$a+a+...+a$  ($b$ times) is the same as
$b+b+...+b$  ($a$ times).