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.


Let $o(abc) = n$. Then $(abc)^n = e$


\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?

  • 4
    $\begingroup$ In fact, you may try to show that $|ab|=|ba|$ for $a,b\in G$. And your problem is the consequence of this property. $\endgroup$ Apr 17, 2017 at 15:36

3 Answers 3


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 $abc$.

  • 1
    $\begingroup$ So, as I understand, since an isomorphism preserves order of elements it follows (immediately?) that the elements $abc, bca, cab$ have the same order. Is the argument as simple as it sounds? $\endgroup$
    – asd
    Apr 17, 2017 at 18:47
  • $\begingroup$ This argument is so simple yet powerful! +1 $\endgroup$
    – Paramanand Singh
    Apr 18, 2017 at 9:57

Notice that you just need to show that $ o(ab) = o(ba) $. Let $ n=o(ab) $. You have

$$ (ba)^{n+1} = b(ab)^{n}a = ba $$

Thus, by multiplying by $ (ba)^{-1} $, you get $ o(ba) \leq o(ab) $. Symmetry then gives the desired equality.

In your proof, you only showed that $ o(ba) \leq o(ab)$.


Two things you should consider:

  1. Your first step only works if $n \ge 2$. You should handle the case $n = 1$ separately.

  2. You've shown that if $(abc)^n = 1$ then $(bca)^n = e$ and $(cab)^n = e$. What this shows is that $|abc| \ge |bca|$ and $|abc| \ge |cab|$ because, at least in principle, $(bca)^m$ might equal $e$ for some $m < n$.


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