Left coset of a subgroup of Complex numbers.

In summary: To summarize, for (ii), the left cosets of H in G are half-lines coming out of the origin, with the angle between 0 and pi, and the radius ranging over all positive real numbers.
  • #1
Silversonic
130
1

Homework Statement



For H [itex]\leq[/itex] G as specified, determine the left cosets of H in G.

(ii) G = [itex]\mathbb{C}[/itex]* H = [itex]\mathbb{R}[/itex]*

(iii) G = [itex]\mathbb{C}[/itex]* H = [itex]\mathbb{R}[/itex][itex]_{+}[/itex]

The Attempt at a Solution



I have the answers, it's just a little inconsistency I don't understand.

For (ii) left cosets are

{r(cos∅ + isin∅); r [itex]\in[/itex] (0,∞)} ∅ [itex]\in[/itex] [0, 2[itex]\pi[/itex])

For (iii)

{r(cos∅ + isin∅); r [itex]\in[/itex] [itex]\mathbb{R}[/itex] \ {0} } ∅ [itex]\in[/itex] [0, [itex]\pi[/itex])I'm told that the answers are different because the range of r and ∅ are different. It says in (ii) they are "half lines" coming out of the origin and in (iii) they are lines through the origin but excluding the origin itself. What I don't get, though, is that surely the answer for (ii) should be the answer for (iii)? And vice versa? Basically in (ii) we have H is the set of all the real numbers, while G is the set of all the complex numbers. So when we multiply an element of H by an element of G (and the constant multiplying the euler's forumla is positive), surely r would then range over all the real numbers (excluding zero).

Yet in (iii) we have H is the set of all the positive real numbers, while G is still the set of all the complex numbers (and the constant multiplying the euler's forumla is positive). So when we multiply an element of H by an element of G, surely r would only range over the positive real numbers, as opposed to all the real numbers exluding zero like the answer says?

Does anyone understand my problem?

Thanks.
 
Last edited:
Physics news on Phys.org
  • #2
The question does not even make sense. R+ is NOT a subgroup of C*. Are you sure you have copied the problem correctly? R+ is a subgroup of C+.
 
  • #3
HallsofIvy said:
The question does not even make sense. R+ is NOT a subgroup of C*. Are you sure you have copied the problem correctly? R+ is a subgroup of C+.

Yeah, I've double checked and I've copied it correctly.

When saying R+, I assumed it was talking about the multiplication of positive real numbers (as opposed to addition of R+, which cannot be a group let alone a subgroup).

Why would R+ under multiplication not be a subgroup of C*? Surely every possible value of R+ on the positive real line is some form a complex number. All R+ is in C*, as well as gh (where g and h are elements of R+) are elements of R+, and lastly the inverse of an element g, is 1/g which is in R+).
 
  • #4
Yes, you are right. R+ is a group under multiplication.
And I agree, solutions (ii) and (iii) should be swapped around.
Here's my interpretation of (ii).

Let's consider one specific element of C*, say z=a cis(phi).
Multiply it with R* to get the left coset.
This is a line excluding zero.

Now if we consider z=a cis(phi + pi) we get the same coset.
Indeed if pick any z in C* on the line, we get the same coset.
So the coset is uniquely defined by an angle between 0 and pi, but is independent of a.

So a specific coset is: {r cis(phi) | r in R*} 0 ≤ phi < pi
 

1. What is a left coset of a subgroup of complex numbers?

A left coset of a subgroup of complex numbers is a set of complex numbers that can be obtained by multiplying every element in the subgroup by a single complex number from the left. This results in a new set of numbers that is not necessarily a subgroup, but has the same number of elements as the original subgroup.

2. How is a left coset different from a subgroup?

A subgroup of complex numbers is a subset of the complex numbers that forms a group under multiplication. A left coset, on the other hand, is a set of complex numbers obtained by multiplying a subgroup by a single complex number from the left. While a subgroup is a group itself, a left coset is not necessarily a group.

3. Can a left coset be a subgroup?

Yes, a left coset can be a subgroup if the complex number used to multiply the subgroup by from the left is also in the subgroup. In this case, the resulting set will have the same number of elements as the original subgroup and will also be closed under multiplication, making it a subgroup.

4. What is the purpose of studying left cosets of subgroups of complex numbers?

Studying left cosets of subgroups of complex numbers is important in group theory and abstract algebra. It helps us understand the structure of groups and how they relate to each other. In particular, left cosets are useful in proving theorems and understanding the properties of groups.

5. How do left cosets relate to normal subgroups?

A normal subgroup is a subgroup that is invariant under conjugation by any element in the larger group. This means that the left cosets of a normal subgroup are also right cosets, and can be used to form a quotient group. Additionally, normal subgroups have a special property where the left cosets form equivalence classes, which is useful in proving important theorems in group theory.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
522
  • Calculus and Beyond Homework Help
Replies
1
Views
433
  • Calculus and Beyond Homework Help
Replies
3
Views
743
  • Calculus and Beyond Homework Help
Replies
3
Views
451
  • Calculus and Beyond Homework Help
Replies
4
Views
3K
  • Calculus and Beyond Homework Help
Replies
21
Views
2K
  • Topology and Analysis
Replies
11
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
648
  • Calculus and Beyond Homework Help
Replies
3
Views
482
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
Back
Top