1
$\begingroup$

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.

  • 0
    You appear to be right. Only, an element of order $9$ is a product of a $9$-cycle, and then possibly $9$-cycles and $2$-cycles, all the cycles being disjoint - this is something you should also spell out. – 2017-01-07
  • 0
    I don't get that;how is an element of order $9$ a product of a 2-cycle and a 9-cycle – 2017-01-07
  • 0
    Sorry, I meant $3$, not $2$, and it's to late to edit the comment. – 2017-01-07

0 Answers 0