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I am having trouble with this problem. We just started learning it and I am unsure of how to obtain all the elements and how to form them into cyclic subgroups.

The alternating group A4 consists of the identity together with eight 3-cycles and three pairs of 2-cycles. Each 3-cycle generates a cyclic subgroup H of order 3. Each pair of 2-cycles also generates a cyclic subgroup K of order 2.

Write down all 12 elements and arrange them into their cyclic subgroups. Pick any two elements from two separate 3-cycles (e.g. α= (123) and β = (13)(24) and show you can generate all of A4 from these two elements. Express all 12 elements as products of ↵ and .

Show that the 3-cycle α = (123) together with the 2-cycle β = (13)(24) also generates all of A4.

What do you generate from the elements (13)(24) and (12)(34)? Is it the entire group or is it a subgroup?

Carefully draw a subgroup lattice diagram for A4.

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  • $\begingroup$ For the subgroups lattice, use gap. See also here. $\endgroup$ – Dietrich Burde Mar 1 '17 at 13:55
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Using the answers of this MSE-question, we see that the subgroup lattice of the alternating group $A_4$ is as follows: https://crazyproject.files.wordpress.com/2010/04/subgroups-of-a4.png

We also see, that the group generated by $(12(34)$ and $(13)(24)$ is not the full group.

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  • $\begingroup$ Thank you! Would you mind showing me how (123)(134) would look like? Would it become (13)(2)(4)? $\endgroup$ – Maria Arias Mar 1 '17 at 14:15
  • $\begingroup$ $1$ goes to $3$ by $(134)$ and $3$ then goes to $1$ by $(123)$, so $1$ is fixed. This way we see that $(123)(134)=(234)$. $\endgroup$ – Dietrich Burde Mar 1 '17 at 14:58

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