MHB Clarification for Alternating Group

[Enzipino]

In class we had to show that ${A}_{5}$ is cyclic. So what we did was,

${A}_{5}$ is cyclic iff there is an $\alpha\in{A}_{5}$ with $<\alpha> = {A}_{5}$. So, the $ord(\alpha) = |<\alpha>| = |{A}_{5}| = \frac{5!}{2} = 60$. So, $60 = {2}^{2}*3*5$.

After this, we said that we could do a 4-cycle, 3-cycle, and 5-cycle, which would be in ${S}_{12}$ but not ${A}_{5}$. We concluded by saying ${A}_{5}$ is not cyclic.

The main thing I'm confused about is why the 4-cycle, 3-cycle, and 5-cycle is not in ${A}_{5}$. Could someone just clarify this?

 

[Fallen Angel]

Hi Enzipino,

I'm not sure what you are asking for, but I think it is why the product of a 3-cycle, a 4-cycle and a 5-cycle can't be in the alternating group.

A $n-$cycle can be decomposed as the product of $n-1$ 2-cycles, hence you got the product of $2+3+4=9$ 2-cycles, which is odd, so it is not in the alternating group.