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Left coset of a subgroup of Complex numbers.

  • #1
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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:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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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
130
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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
I like Serena
Homework Helper
6,577
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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
 

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