Thread: Determine if this set of complex numbers forms a subgroup

I think I answered this correctly, but I wanted to be sure.

Problem: Determine whether the given subset H of the group G is a subgroup.

, the group of all nonzero complex numbers with operation given by multiplication of complex numbers.



******************

For H to be a subgroup of G, the following three things must be satisfied:
1) Closed under the binary operation of G (in this case, multiplication of complex numbers)
2) H has the identity element of G
3) For all , there exists an .

is defined as where a and b come from the general complex number form .

Since , it must be true that only one of the pair a and b is 1, and the other needs to be zero.

Therefore, this set H only has four elements:






1) Multiplying two elements of this set will always return a result that is still within the set H, and therefore the set H is closed under the binary operation. Since there are always going to be two zero terms when foiling two of these terms, only the terms where +/-1 are multiplied together, always returning one of the four elements in the group.

2) An identity of group will satisfy



In this case, intuitively satisfies this, and is the unique identity element of group G. , so the set H has the identity of G.

3) The inverse of a complex number is the conjugate. So a complex number has the inverse . Looking at the list of four elements from part (1), it can be seen that the first two elements are and , and the following two are the conjugates (inverses) of and , and vice versa. Therefore, every element in H has an inverse in H.

Since (1), (2), and (3) are satisfied, H is a subgroup of G.

Am I correct? Is there anything here that I did wrong?

No, it's wrong. The elements are the circunference of points at the distance 1 from the point (0,0). They are of the form e^ix. try it again. Apart from that mistake, you had everything right.

Originally Posted by tangibleLime

I think I answered this correctly, but I wanted to be sure.

Problem: Determine whether the given subset H of the group G is a subgroup.

, the group of all nonzero complex numbers with operation given by multiplication of complex numbers.



******************

For H to be a subgroup of G, the following three things must be satisfied:
1) Closed under the binary operation of G (in this case, multiplication of complex numbers)
2) H has the identity element of G
3) For all , there exists an .

just a remark, here, that 1) & 3) together imply 2), so you don't need to show this. (if ab is in H when a,b are, and a^-1 is in H, then taking b = a^-1 proves 2)).

is defined as where a and b come from the general complex number form .

Since , it must be true that only one of the pair a and b is 1, and the other needs to be zero.

this is false. √2/2 + i√2/2 is also in H, as is easily verified.

Therefore, this set H only has four elements:






1) Multiplying two elements of this set will always return a result that is still within the set H, and therefore the set H is closed under the binary operation. Since there are always going to be two zero terms when foiling two of these terms, only the terms where +/-1 are multiplied together, always returning one of the four elements in the group.

all useless.

2) An identity of group will satisfy



In this case, intuitively satisfies this, and is the unique identity element of group G. , so the set H has the identity of G.

3) The inverse of a complex number is the conjugate. So a complex number has the inverse . Looking at the list of four elements from part (1), it can be seen that the first two elements are and , and the following two are the conjugates (inverses) of and , and vice versa. Therefore, every element in H has an inverse in H.

Since (1), (2), and (3) are satisfied, H is a subgroup of G.

Am I correct? Is there anything here that I did wrong?

the proper approach here, is to first show that |zw| = |z||w|, for any 2 complex numbers z = a+ib, w = c+id.

Originally Posted by ModusPonens

No, it's wrong. The elements are the circunference of points at the distance 1 from the point (0,0). They are of the form e^ix. try it again. Apart from that mistake, you had everything right.

I don't quite understand this. I thought that |z|, where z is a complex number, was the modulus of z and defined as sqrt(x^2+y^2). Why is it instead e^ix? That forms the roots of unity, correct?

suppose z = a+ib.

then so



if |z| = 1, we obtain the equation:

. for which pairs (a,b) is this true? this equation should ring a bell....

Originally Posted by Deveno

suppose z = a+ib.

then so



if |z| = 1, we obtain the equation:

. for which pairs (a,b) is this true? this equation should ring a bell....

Isn't that true for (1,0), (-1,0), (0,1), (0,-1) like I said? Why am I supposed to bring in e^ix? (from what ModusPonens said)

Edit: I see that those are just integer solutions... but I still don't see where e^ix is coming from.

Originally Posted by tangibleLime

Problem: [/B]Determine whether the given subset H of the group G is a subgroup.
, the group of all nonzero complex numbers with operation given by multiplication of complex numbers.

The set is the whole unit circle in the complex plane.
Here is a neat theorem: If is a non-empty subset of a group then is subgroup if and only if for it is true that .

For this if then , the conjugate, and .

As soon as you posted, I realized how I was thinking about it the wrong way. H is the set of complex numbers on the unit circle SUCH THAT |z|=1. Okay. Thanks

by definition.

so what is ?

by direct calculation:



now, let's show that every complex number z with |z| = 1, can be written in this form:

let z = a+ib, and define t = arctan(b/a),

note that cos t = cos(arctan(b/a)) =

and that sin t = sin(arctan(b/a)) =

since ,

cos t = a, sin t = b, and z = a+bi = cos t + i sin t.

one caveat: a might be 0, in which case arctan(b/a) is undefined.

in that case, use t = , if z is in the top half-plane, or

t = , if z is in the lower half-plane.