Generating $A_4$ with two distinct $3$-cycles $x$ and $y$ in $S_4$

Question

Given two distinct $3$-cycles $x$ and $y$ in $S_4$ such that $x\neq y^{-1}$, I would like to show that $A_4$ is generated by any such pair of $x$ and $y$.

The only times where $3$-cycles in $S_4$ can be the inverse of each other is when they can be written in the form $(a\;b\;c),(a\;c\;b)$. Excluding this case, the possible pairs of cycles should only be of the form $$(a\;b\;c),(a\;b\;d)\qquad\text{or}\qquad(a\;b\;c),(a\;d\;b).$$ Suppose the two $3$-cycles are of the form $(a\;b\;c),(a\;b\;d)$, then I can say the element $$(a\;b\;c)(a\;b\;d)=(c\;a)(d\;b)$$ is in the subgroup generated by $(a\;b\;c),(a\;b\;d)$.

From here I am not sure how I should go about showing that the generated set must be $a_4$. Should I keep discovering every possible element in the generated set display that the sets are equivalent? Or is there a more clever way than brute forcing it like this for every valid pair of 3-cycles?

Answer 1

Given a pair $x,y$ with property $x \neq y^{-1}$ you have distinct elements $e,x,y,x^2,y^2$. It is sufficient to show that $xy$ and $x^2y^2$ are two new elements distinct from $e,x,y,x^2,y^2$. And we obtain 7 elements in the respective subgroup of $A_4$. So, this subgroup is $A_4$ (by Lagrange's theorem).

Answer 2

First of all, note that the two cases you distinguish are really the same, because in the second one, the subgroup $H$ generated by the two $3$-cycles contains $(a\ d\ b)^{2} = (a\ b\ d)$.

One way to show this is to note that under these hypotheses $H$ is a subgroup of $A_{4}$ of order greater than $3$, and this forces it to have order $12$, as $A_{4}$ has no subgroup of order $4$.

Alternatively, once you have shown as you did that $(a\ c) (b \ d) \in H$, note that also the elements $$ (b\ a) (c\ d) = ((a\ c) (b \ d))^{(a\ b\ c)}, \qquad (c\ b) (a\ d) = ((a\ c) (b \ d))^{(a\ c\ b)} $$ are in $H$, so $H$ contains the subgroup $$ \{ \mathrm{Id}, (a\ b)(c\ d), (a\ c)(b\ d), (a\ d)(b\ c) \}, $$ so $H$ has order divisible by $4$ and $3$, so $H$ has order $12$.