- #1

- 1,086

- 2

## Homework Statement

How many subgroups of index 10 are in [itex]\mathbb{Z}_{20} \times \mathbb{Z}_{9} \times \mathbb{Z}_{35}[/itex]?

## The Attempt at a Solution

So I figure the order of this group would be 20 x 9 x 35 = 6300, and so any subgroup of index 10 would have order 630. By the fundamental theorem of finite commutative groups, you get that such a subgroup is isomorphic to one of these two:

- [itex]\mathbb{Z}_{2} \times \mathbb{Z}_{9} \times \mathbb{Z}_{5} \times \mathbb{Z}_{7}[/itex]

- [itex]\mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3} \times \mathbb{Z}_{5} \times \mathbb{Z}_{7}[/itex]

On the other hand, the elementary divisor decomposition of the main group is [itex]\mathbb{Z}_{4} \times \mathbb{Z}_{5} \times \mathbb{Z}_{9} \times \mathbb{Z}_{5} \times \mathbb{Z}_{7}.[/itex]

I can see some of the elements in the decompositions match, but I don't know where to go from here. First I thought of just counting the number of elements of order 630, but I'm not sure this is the right approach, and I also don't really know how to account for all of them with so many different groups in the decomposition.

Any thoughts on this?