Finding all of the right cosets of H in G

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  • #1
hb1547
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Homework Statement


"Write out all the right cosets of H in G where G = (a) is a cyclic group of order 10 and H = (a^2) is the subgroup of G generated by a^2."


Homework Equations


- If G = (a), then G = {a^i | i=0,-1,1,-2,2...}.
- A right coset is the set Hb = {hb | h is in H}
- Order of G is 10, so G has ten elements. Order of H has to divide that.
- G is the disjoint union of all its right cosets.
- The right cosets of H in G are disjoint.
- The

The Attempt at a Solution


I guess I'm a bit confused, due to the order being ten but not being given any elements to work with.

I reasoned that H = {a^(2i) | i=0,-1,1...}, which led to me thinking that at least one right coset was:
Ha = {a^(3i) | i = 0, -1, 1,...}

But I'm not sure where I should stop with the i's, leading to me being confused as to how many elements would be in that right coset (if that even is one of the right cosets).

Am I even on the right track?
 

Answers and Replies

  • #2
Dick
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Let's be a little more concrete here. If G has order 10 and is generated by a then G={a^0,a^1,...a^9}. Let's call a^0=1. The group identity. So a^10=a^0=1. Now, what is H?
 
  • #3
hb1547
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Then that makes H = G, doesn't it?

H = {a^0, a^3, a^6, a^9, a^2, a^5, a^8, a^1, a^4, a^7}, if I keep adding 3 to the powers and cycling through once they get to ten. Then it's clear that H = G.
 
  • #4
Dick
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Then that makes H = G, doesn't it?

H = {a^0, a^3, a^6, a^9, a^2, a^5, a^8, a^1, a^4, a^7}, if I keep adding 3 to the powers and cycling through once they get to ten. Then it's clear that H = G.

H is generated by a^2. Not a^3.
 
  • #5
hb1547
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Ahh right. So would it be:

H = {a^0, a^2, a^4, a^6, a^8}?

Then that leads me to infer that the other right coset must be:
{a^1, a^3, a^5, a^7, a^9}

But I'm not sure how quite to show that -- is it because the first coset is He, where it's acting on the identity, and the other one is Ha, where it's acting on a?
 
  • #6
Dick
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Ahh right. So would it be:

H = {a^0, a^2, a^4, a^6, a^8}?

Then that leads me to infer that the other right coset must be:
{a^1, a^3, a^5, a^7, a^9}

But I'm not sure how quite to show that -- is it because the first coset is He, where it's acting on the identity, and the other one is Ha, where it's acting on a?

Yes, I think you got it. There are two cosets.
 
  • #7
hb1547
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Okay awesome, thank you very much.

I guess what I'm confused about is that the definition for a cyclic group said,
{a^i | i=0,-1,1,-2,2...}.

Here, we only used the positive values of i, and I guess I'm unclear as to why we did that.
 
  • #8
Dick
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Because if the group has order 10 then a^(-1)=a^9. Right? You don't need the negative powers and you don't need powers over 9. They are duplicates of the other powers.
 
  • #9
hb1547
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Ahh, so the negative powers cycle through as well. Okay thanks! That helps a lot!
 

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