Subgroups of Z30xZ5: Order 10

  • Thread starter lola1990
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In summary, there are six subgroups of order 10 in Z30xZ5 that contain the 24 elements of order 10. Every 4 elements generate the same group.
  • #1
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0

Homework Statement


3. How many subgroups of order 10 are there in Z30 x Z5?

Homework Equations


the order of an element must divides the order of the group

The Attempt at a Solution



3. I have no idea where to start with this, especially since Z30 x Z5 is not cyclic, since 30 and 5 are not relatively prime.
 
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  • #2
find at least one subgroup of order 10. or two. do you know what the order of an element is? (in this group?)

oh you do. you showed this. well how many elements of order 10 are there in a given subgroup of order 10?
 
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  • #3
There are 4 elements of order 10 in a given subgroup of order 10. I did this by brute force though; is there a more elegant way to show this? I concluded that there are 6 subgroups of order 10 that contain the 24 elements of order 10. Now I'm looking at elements of order 2 and 5 to see if I can generate any others.

I was wondering if there is a better way to find out how many rather than just listing them all.
 
  • #4
"I concluded that there are 6 subgroups of order 10 that contain the 24 elements of order 10. Now I'm looking at elements of order 2 and 5 to see if I can generate any others."

huhh?
 
  • #5
suppose we had a subgroup H of order 10 in Z30xZ5. 2 and 5 are prime divisors of 10.

so we have a subgroup A of H of order 2, and a subgroup B of order 5 (cauchy's theorem for abelian groups).

now H is obviously abelian, so H = AB (chinese remainder theorem, or even the more elementary: |AB| = |A||B|/|A∩B|).

so H is cyclic, since if a generates A, and b generates B, ab generates H

((ab)m = ambm, right? (H is abelian). so if (ab)m = e, 5|m and 2|m, since H∩K = {e}).

so every subgroup of order 10 of Z30xZ5 is generated by an element of order 10. you already know how many elements of order 10 you have, now just don't count elements that generate the same subgroup of Z30xZ5 as some earlier one on your list of 24.

for example, any element of order 10 of the form (x,0) gives the same subgroup of order 10, isomorphic to Z10 x {0}.

(this group is generated by (3,0)).

use mathwonk's suggestion to speed up the counting...
 
  • #6
ok, so as I said, I have 6 unique subgroups of order ten generated by the 24 elements of order 10 (ie, every 4 elements generate the same group). So, according to the logic from Deveno, this is it, right?
 
  • #7
there are no elements of order 2 in Z5 because 2 does not divide 5. elements in Z5 have order 5 or 1. If the element from Z5 has order 1, the element from Z30 must have order 10 so the lcm=10. If the element from Z5 has order 5, the element from Z30 could have order 2 or 10 for the lcm to be 10. I'm pretty sure that 10&1, 10&5, and 2&5 are the only possibilities.
 

1. What is the order of the subgroup Z30xZ5?

The order of the subgroup Z30xZ5 is 10, as stated in the title.

2. How is the subgroup Z30xZ5 defined?

The subgroup Z30xZ5 is defined as a subset of the direct product of two groups, Z30 and Z5, with elements of order 10.

3. Can you provide an example of an element in the subgroup Z30xZ5?

One example of an element in the subgroup Z30xZ5 is (3,2), where 3 is an element of Z30 and 2 is an element of Z5. This element has an order of 10 since (3,2)^10 = (0,1), the identity element of the subgroup.

4. How many elements are in the subgroup Z30xZ5?

The subgroup Z30xZ5 has 10 elements, as its order is 10.

5. What is the significance of the subgroup Z30xZ5 in mathematics?

The subgroup Z30xZ5 is significant in mathematics because it demonstrates the concept of direct products and subgroups, which are important in group theory and abstract algebra.

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