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.
Best Answer
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.