I tried to provide an example of group elements $a,b,c$ that shows that in general $|abc|\neq |cba|$.
Please could you tell me if my example is correct?
Let $G$ be the dihedral group $D_6$ (the symmetries of a hexagon). Let us label the axes of symmetry with numbers $1$ to $6$ and call the corresponding reflections $F_1$ to $F_6$.
Now consider two "adjacent" reflections $F_1$ and $F_2$. Then
$$ F_1 F_2$$
is rotation clockwise by $120$ degrees and
$$ F_2 F_1$$
is roation by $120$ degrees counterclockwise.
Let $R$ be rotation by $120$ degrees counterclockwise. Then
$$ R F_1 F_2 = e$$
has order $1$ but
$$ F_2 F_1 R$$
is rotation by $240$ counterclockwise which has order not equal to $1$.
Take G to be any group, let x,y be elements of G and observe that the commutator [x,y] = x'y'xy (here x' denotes the inverse of x in G) can be written as a product abc such that cba=1. Indeed, just take a=x', b=y'x and c=y.
Now if x,y don't commute you get that abc is nontrivial, while cba is trivial and presto: you have infinitely many easy examples.
N.B. This trick also shows that in any group G, if a finite product of elements is trivial, then the product of the same elements in reverse order belongs to the derived subgroup G'.
Also: the elements of G' are those finite products of elements in G whose corresponding product in reverse order is the identity.
Hope this helps.