Prove that $(1\ 2\ 3)$ cannot be a cube of any element in the symmetric group $S_n.$

If such an element do exist say $a$ then $a^3 = (1\ 2\ 3).$ Let $\text {ord}\ (a) = m.$ So we have $$3 = \text {ord}\ ((1\ 2\ 3)) = \text {ord}\ \left (a^3 \right ) = \frac {m} {\text {gcd}\ (3,m)}.$$ Then it is clear from the above equality that $3\ \mid\ m.$ But this shows that $\text {gcd}\ (3,m) = 3.$ So we have $\text {ord}\ (a) = m = 9.$ This means if $a$ is written as a product of disjoint cycles in $S_n$ then one of the cycles has to be a $9$-cycle. Certainly $a$ is not a $9$-cycle for otherwise $a^3$ is the product of three disjoint $3$-cycles, a contradiction to the given hypothesis. How do I analyze all the other possibilities that may arise here?

Any help in this regard will be highly appreciated. Thanks in advance.

  • $\begingroup$ One another remark I have to make $:$ Disjoint cycles in $S_n$ commute. May be that's the reason why it is the case. $\endgroup$ – Anacardium Sep 18 '20 at 13:46
  • $\begingroup$ That's all you need to analyze the other possibilities. If $a$ is a product of disjoint cycles $c_1,\ldots,c_k$, one of which is a 9-cycle, then, since disjoint cycles commute, you can apply the same reasoning to the 9cycle. $\endgroup$ – halrankard2 Sep 18 '20 at 13:56
  • $\begingroup$ Related $\endgroup$ – Jyrki Lahtonen Sep 18 '20 at 14:01

If $\sigma^3=(1\,2\,3)$ then $\sigma^9$ is the identity. So the decomposition of $\sigma$ into disjoint cycles consists of $a_9$ $9$-cycles, $a_3$ $3$-cycles and $a_1$ $1$-cycles, where $n=9a_9+3a_3+a_1$. I reckon then that $\sigma^3$ would have $3a_9$ $3$-cycles and $3a_3+a_1$ $1$-cycles.

  • $\begingroup$ You are right. That's what I mean in my comment. $\endgroup$ – Anacardium Sep 18 '20 at 14:00
  • 1
    $\begingroup$ Yup. This is probably the simplest argument. $\endgroup$ – Jyrki Lahtonen Sep 18 '20 at 14:01
  • $\begingroup$ @Anacardium I should have pointed out that this method readily generalises to solve any problem of the form "is $\tau\in S_n$ a $k$-th power". $\endgroup$ – Angina Seng Sep 18 '20 at 14:12

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