Let $G$ be a group. Show that $\forall a, b, c \in G$, the elements $abc, bca, cab$ have the same order. 
Let $G$ be a group. Show that $\forall a, b, c \in G$, the elements
  $abc, bca, cab$ have the same order.

I thought that my solution ($?$) was enough to show that $abc, bca, cab$ have the same order, but my teacher told that it isn't, so I don't know what else to do here.
Attempt:
Let $o(abc) = n$. Then $(abc)^n = e$
Therefore
\begin{align}
abc(abc)^{n-2}abc &= e\\
(bc)\left[abc(abc)^{n-2}abc\right](cb)^{-1} &= e\\
(bca)^n &=e\\
bca(bca)^{n-2}bca &= e\\
(ca)\left[abca(bca)^{n-2}bca\right](ac)^{-1} &=e\\
(cab)^n &= e
\end{align}
Therefore $(abc)^n = (bca)^n = (cab)^n = e$
What else am I lacking after this?

In general, conjugate elements have the same order:

$o(ghg^{-1}) = o(h)$

because $ x \mapsto gxg^{-1}$ is an automorphism.
Now note that 
$bca = a^{-1} (abc) a$
and
$cab = c (abc) c^{-1}$
are conjugates of $abc$.
