For a group $G$ and $a_1,a_2,a_3\in G$, if $a_1a_2a_3=e$, then does it follow that $a_2a_3a_1=e$ and that $a_3a_2a_1=e$? Question: Let $(G,\ast)$ be a group. If $a_1,a_2,a_3 \in G$ satisfies
$$a_1a_2a_3=e,$$
prove that
$$a_2a_3a_1=e.$$
Is it true that $a_3a_2a_1=e$?
My first thought:
$$a_1a_2a_3=e$$
$$a'_1a_1a_2a_3=a'_1e$$
$$ea_2a_3=a'_1e$$
$$a_2a_3=a'_1$$
$$a_2a_3a_1=a'_1a_1$$
$$a_2a_3a_1=e$$
For $a_3a_2a_1=e$, I think it is not true.
From the result,
$$a'_2a_2a_3a_1=a'_2e$$
$$ea_3a_1=a'_2e$$
$$a_3a_1=a'_2$$
$$a_3a_1a_2=a'_2a_2$$
$$a_3a_1a_2=e$$
But, $a_1a_2$ may not equal to $a_2a_1$.
However, a friend ask me that why the above proof "assume" $a'_1a_1=e$. The binary operation $\ast$ is not defined in the question. If I do the above proof, it seems I have assumed the binary operation $\ast$ is multiplication. Then I am in trouble now. How can I correct my mistakes?
 A: If $a_1a_2a_3 = e$ then $a_1a_2a_3a_1=ea_1 = a_1$ and so $a_2a_3a_1 = a_1^{-1}a_1a_2a_3a_1 = a_1^{-1}a_1 = e$
So that is true.
If $a_3a_2a_1 = e = a_2a_3a_1$ then $a_3a_2 = a_3a_2a_1*a_1^{-1} = a_2a_3a_1a_1a_1^{-1} = a_2a_3$ and there is utterly no reason that should be true.
It's easy to come up with a counter example.  Let $G$ be a non-abelian group and let $a_2a_3 \ne a_3a_2$.  Let $a_1 = (a_2a_3)^{-1}$.  Then $a_1a_2a_3 = e$ and $a_2a_3a_1 = e$.  But $a_2a_3 \ne a_3a_2$ so $a_2a_3a_1 \ne a_3a_2a_1$.
(Note: this basically is paraphrasing and using the same reasoning of the "cancelation law".  $a = b \iff ac = bc$ for all $c$ in the group.)
A: The assumption that $a_1a_2 = a_1 \ast a_2$ is fair. Writing without the operation is a standard way of more compactly notating these long products.
As for your proof, it looks good. Essentially, $a_1 a_2 a_3 = e$ implies that $a_2 a_3$ is a right-inverse of $a_1$, which must always be a (two-sided) inverse.
You also say "By the similar way, I can show that $a_3 a_2 a_1 = e$ is not true". This statement concerns me, because you can't really prove this in a similar way. It's something that may be true in specific cases, but will not be true in general. You'd need to formulate a counterexample in order to prove that $a_3 a_2 a_1 = e$ is not always true.
A: If $(G, *)$ is defined as a group, then for all $g \in G$, there should be a unique $g^{-1} \in G$ such that $g*g^{-1} = e$.
But we sometimes write $gg^{-1}$ by omitting the binary operation, which has the same meaning. Because of the same reason, in some books, for example it is said that "$G$ should be closed under multiplication" although the defined binary operation is not always multiplication. 
For $a_3a_2a_1 = e$ case, I can say that  is not always true. For example, take 
$$Q_8 = \{i,j,k,e'\ |\ (e')^2 = e, i^2 = j^2 = k^2 = ijk = e'\}$$
Then $kji = e$ but $ijk = e' \ne e$.
