Originally Posted by

**johng** Dan,

I believe the result, but I don't understand your post. You are told that x and y are both **3** cycles and you have written x and y as 5 cycles?

Here's some hints to get you started:

As you point out, you can assume x=(123). Let y be a 3 cycle:

1. Suppose y is disjoint from x. Then "easily" <x,y> is the direct product of 2 cyclic groups of order 3, viz. $Z_3\times Z_3$

2. Suppose y has exactly 2 letters in common with x, say 1 and 2; you can assume the other letter of y is 4. Say y=(124). Then <x,y> is a subgroup H of $A_4$ such that 3 divides the order of H and the order of H is greater than 3. So the order of H is 6 or 12. Can order H be 6? No. Any group of order 6 has a normal subgroup of order 3 and so if $A_4$ contained a subgroup of order 6, it would contain a normal subgroup of order 3, contradiction. So the order of H is 12 and hence $H=A_4$.

3. Suppose y has all letters in common with x. Easy. <x,y> is cyclic of order 3.

4. "Hardest" case. Suppose y has exactly 1 letter in common with x, say 1. As in 2, you can assume y's other two letters are 4 and 5. Then <x,y> is $A_5$. I'll leave you here. If you need help, just say so.