Clarification for Alternating Group

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SUMMARY

The discussion clarifies that the alternating group ${A}_{5}$ is not cyclic, as it cannot be generated by a single element. Specifically, the order of any element $\alpha \in {A}_{5}$ must equal 60, which is derived from the formula $|{A}_{5}| = \frac{5!}{2}$. The confusion arises regarding the inclusion of 4-cycles, 3-cycles, and 5-cycles in ${A}_{5}$; these cycles cannot be represented as products of an even number of transpositions, thus disqualifying them from membership in the group.

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  • Basic grasp of the symmetric group ${S}_{n}$ and its relationship to ${A}_{n}$.
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  • Study the structure and properties of alternating groups, focusing on ${A}_{n}$ for various values of n.
  • Learn about the decomposition of cycles into transpositions and how this affects group membership.
  • Explore the concept of group generators and cyclic groups in depth.
  • Investigate the relationship between symmetric groups ${S}_{n}$ and alternating groups ${A}_{n}$.
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Enzipino
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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?
 
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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.
 

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