# Determining the number of subgroups

1. Mar 5, 2009

### duki

1. The problem statement, all variables and given/known data

How many subgroups does //567,000 have? (don't know the smart-text for a group)

2. Relevant equations

3. The attempt at a solution

From the notes I'm looking at, he did something like
$$2^3 * 3^4 * 5^3 * 7 = 567,000$$
and somehow this equates to 128 subgroups.

I have no clue how he came to this though... could someone give me a hand on finding the number of subroups in a group?

2. Mar 5, 2009

### Dick

You mean the cyclic group Z_567000, yes? It's just the number of divisors of 567000. I don't think it's 128. How many are there?

3. Mar 5, 2009

### duki

I'm not exactly sure how to find the number of divisors... I think there's a way to do something with the primes, but I never really understood it.

4. Mar 5, 2009

### Dick

The number of divisors is the number of integers of the form 2^i*3^j*5^k*7^l. i is in {0,1,2,3}, j is in {0,1,2,3,4} etc. Do you get my drift? Each choice is a different number because of unique prime factorization. How many are there?

5. Mar 5, 2009

### duki

I still don't understand =(
Maybe if we used a smaller number for me to get the hang of it? sorry

6. Mar 5, 2009

### Dick

Ok. How many divisors of 12? 12=2^2*3^1. So the divisors are 2^i*3^j where i is in {0,1,2} and j is {0,1}. That's 3 choices (for i) times 2 choices (for j)=6. And that is the number of divisors of 12, right? Same idea for 567000.

7. Mar 5, 2009

### duki

So for 567,000 I'm getting:

$$2^3 * 3^4 * 5^3 * 7 = 210 subgroups?$$

8. Mar 6, 2009

### Dick

No. How did you get 210?

9. Mar 6, 2009

### duki

2 * 3 * 5 * 7 ? =/

10. Mar 6, 2009