True or False: If the product of n elements of a group is the identity element, it remains so no matter in what order the terms are multiplied. I am working my way through Charles Pinter's book: A Book of Abstract Algebra. From recommendations on this site, I found a page/web address on Wisconsin University's Math Department that provides solutions to many (perhaps all) of the abundant exercises that are present in Pinter's book. 
One of the proposed solutions to Pinter's exercises is the following generalization:
If the product of n elements of a group is the identity element, it remains so no matter in what order the terms are multiplied.
I take issue with this claim and think it is only valid if the group is abelian. Using a simple 3 element example, consider:
$a\circ b \circ c = e$
Using the definition of inverses (and knowing that inverses commute...by defintion) and the associative law, I generated the following cases that must be true:
$a \circ (b \circ c) =e$ and therefore $(b \circ c) \circ a =e$
$(a \circ b) \circ c=e$ and therefore $c \circ (a \circ b)=e$
However, there are still several permutations of this list of elements that require consideration...for example:
$a \circ (c \circ b) =e$
It seems to me this can only be true if the group is abelian. Therefore, should the solution manual be amended to say: 
If the product of n elements of an abelian group is the identity element, it remains so no matter in what order the terms are multiplied.
?

What I was expecting the statement to say was:

If the product of $n$ elements in the group is the identity, it remains so under any cyclic permutation.

Perhaps this is what “order” is meant in the solution? In any case, this might be a reasonable statement to prove for yourself. 
