So like I want to know the elements of order 6 and the Euler function of 6 is 2, I conclude that there are only 2 elements of order 6 in H, am I right?

@Maialen Yes you are right. Notice that this only works since the subgroup is cyclic. Try to think for example to the number of transposition in $S_{3}$

Oh thanks! But how can I know which are those 2 elements of order 6?

@Maialen In general you can't. In this case you are seeking for elements of order six and since there are two, once noticed that the square of $\sigma$ works, the other one is free since $\sigma$ is not the inverse of itself. Of course consideration about the cycle structure can be done in order to semplify the calculus.

And the inverse is (1 6 4)(2 5 3)(7 10)(8 9) right?

@Maialen Yes exactly, you just have to invert the $3$-cycles

And I've got other question, How can I know if the (7 8)H and (8 10 9)H are equal?

I do not understood you notation, could you please explain it ?

Oh the quotient

the left class of (7 8) respect H and the left class of (8 10 9) respect H

I have to know if those two are equal

But for it, I have to multiply all the elements of H with (7 8) and then with (8 10 9)?

Okay they can't be, you know that in a finite group the index of the subgroup is equal to the cardinality of the quotient group, since $H$ quotient by $\langle (12) \rangle$ has cardinality 6 and $H$ quotient by $\langle (8 10 9) \rangle$ has cardinality 4

How do you know those cardinalities? have you calculated all the elements and count them?

H in question has order 12, the subgroup generated by (8,9,10) has 3 instead (9,10) has 2

If this doesn't answer your question it's possible that i didn't understood the question

I think I've understood it, thank you very much! :)

Hi again, I've got a doubt.. Can I say that the quotient groups (7 8)H and (8 10 9)H are not equal owing to the fact that the element (7 8)*identity (cause identity is an element from H) is into (7 8)H but is not into (8 9 10)H ???