# Lattice of Subgroups of A4

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 $$A_4$$ 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 $$A_4$$ 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 $$A_4$$.

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 $$A_4$$.

• For the subgroups lattice, use gap. See also here. Mar 1, 2017 at 13:55

Using the answers of this MSE-question, we see that the subgroup lattice of the alternating group $$A_4$$ is as follows:
On the left-hand side, we also see that the group generated by $$(12)(34)$$ and $$(13)(24)$$ is not the full group.
• $1$ goes to $3$ by $(134)$ and $3$ then goes to $1$ by $(123)$, so $1$ is fixed. This way we see that $(123)(134)=(234)$. Mar 1, 2017 at 14:58