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

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.

• One another remark I have to make $:$ Disjoint cycles in $S_n$ commute. May be that's the reason why it is the case. – Anacardium Sep 18 '20 at 13:46
• 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. – halrankard2 Sep 18 '20 at 13:56
• Related – 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.
• @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". – Angina Seng Sep 18 '20 at 14:12