Show that there exists no element $\alpha$ in $S_n$ where $S_n$ is the symmetric group on $n$ symbols such that $(123)=\alpha^3$.
Suppose that $\alpha^3=(123)$ for some $\alpha\in S_n$.
Then $o(123)=o(\alpha^3)\implies 3=\dfrac{o(a)}{\gcd(3,o(a))}$
CASE I :$\gcd(3,o(a))=3\implies o(a)=9\implies $
$\text{either $a$ is a 9-cycle or it is product of 9 cycles}.$
Now both are false because a 9-cycle or a product of 9 cycles don't give a 3cycle when multiplied 3 times.
CASE II:
$\gcd(3,o(a))=1\implies o(a)=3\implies $ $\text{either $a$ is a 3-cycle or it is product of 3 cycles}.$
Now both the conditions are false as a 3-cycle has order $3$ and hence it's cube cant be $(123)$.
Am I right?Please correct me if I'm wrong.