Let $G$ be a group and suppose that $a*b*c=e$ for all $a,b,c \in G$, show that $c*b*a=e$ I'm really in the dark here:
$a*b*c=e=identity$
$a*e=e*a, b*e=e*b, c*e=e*c$
$a*b*c=e$
$e=e$
$c*e*b=e*c*b$
$c*a*b*c*b=a*b*c*b*c*e=c*b$
$c*a*b*c*b=a*b*c*b*c*e=c*b*a*b*c$
$c*a*b*c*b=c*b*a*b*c$
$c*a*b*c*b*a*b*c=c*b*a*b*c$
$c*b*a*b=c*b*a*b$
$(a*b*c)^{-1}=c^{-1}*b^{-1}*a^{-1}$
$a*b*c=c^{-1}*b^{-1}*a^{-1}$
$a=c^{-1}*b^{-1}$
$b=c^{-1}*a^{-1}$
$c=b^{-1}*a^{-1}$
$c*b*a=b^{-1}*a^{-1}*c^{-1}*a^{-1}*c^{-1}*b^{-1}$
No matter what I try I can't seem to mirror the $a*b*c$  on one side and keep the identity on the other. I just get $c*b*a=c*b*a$
Edit: Did misread the assignment so changed title from
Let $G$ be a group and suppose that $a*b*c=e$, show that $c*b*a=e$ for all $a,b,c \in G$
to
Let $G$ be a group and suppose that $a*b*c=e$ for all $a,b,c \in G$, show that $c*b*a=e$
 A: Let 
$$I:=\begin{pmatrix}i&0\\0&-i\end{pmatrix}\text{, } J:=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\text{, } K:=\begin{pmatrix}0&i\\i&0\end{pmatrix}$$
Then :
$$IJK=\begin{pmatrix}0&-i\\-i&0\end{pmatrix}K=I_2 $$
Whereas :
$$KJI=K\begin{pmatrix}0&i\\i&0\end{pmatrix}=-I_2 $$
So the property is false in $G=GL_2(\mathbb{C})$ (in $\mathbb{Q}_8$ actually).
A: If the assumption is that $\forall a,b,c\in G$, $a*b*c=e$, then letting $a'=c,b'=b,c'=a$, we have: $c'*b'*a'=a*b*c=e$, and therefore, relabeling yields $c*b*a=e$.
Otherwise, if this only holds for some $a,b,c\in G$, then it is not true in general, as it would require $a*b=b*a$ by the uniqueness of inverses.
A: This is not true in general.  Let $G$ be $S_3$, and take  $a = (1,2)$, $b= (2, 3)$ and $c =(a*b)^{-1}= (1, 2, 3)^{-1}= (1, 3, 2)$.  Then $a*b*c = e$, but $c$ is not the inverse of $b*a = (1, 2,3 )$, so $c*b*a \neq e$.
For a general example, let $a$ and $b$ be any two non-commuting elements in a group $G$, and let $c = (a*b)^{-1}$.  Then $a*b *c = e$, but $c$ cannot also be  the inverse of $b*a$, so $c*b*a\neq e$.
