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.
?
 A: Since this is a true or false question, it is not that the question is phrased incorrectly, but rather that the answer is that it is false. 
Your claim that it can only hold if the group is abelian is not true for all such $a, b, c$, which we can see in any group by $a=b=c=e$ and other less trivial examples. What you need to do to show it is false is to pick a specific nonabelian group and find three specific elements where it doesn't work.
Edit: I just noticed that this was not an exercise, but part of the solution manual.  In $S_3$, the elements $(12),(23),(321)$ show the original claim is invalid. In fact, if $abc=e$ and $bc\neq cb$, then it is never true that $acb=e$ because $bc$ is the unique inverse of $a$. By similar reasoning, if the claim holds for $a, b, c$, then all three elements commute pairwise. You can show that the group is abelian if this holds true for all triples such that $abc=e$. 
A: 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. 
A: Counter examples could be found in $2\times 2$ invertible matrices.
For example $A= \begin{bmatrix}1&2\\2&1\end{bmatrix}$,$B= \begin{bmatrix}3&2\\1&5\end{bmatrix}$
$C=(AB)^{-1}$ then obviously $$ABC=I$$ while $$CBA\ne I$$ 
