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.

$\displaystyle G = \mathbb{C}^{x}$, the group of all nonzero complex numbers with operation given by multiplication of complex numbers.

$\displaystyle H = \left\{z \in \mathbb{C} \hspace{2mm} |\hspace{2mm} |z| = 1\right\}$

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

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 $\displaystyle a \in H$, there exists an $\displaystyle a^{-1} \in H$.

$\displaystyle |z|$ is defined as $\displaystyle |z|=\sqrt{a^2+b^2}$ where *a* and *b *come from the general complex number form $\displaystyle a+bi$.

Since $\displaystyle |z| = 1$, 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:

$\displaystyle 1+0i$

$\displaystyle 0+1i$

$\displaystyle -1+0i$

$\displaystyle 0-1i$

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 $\displaystyle G$ will satisfy

$\displaystyle a*e = e*a = a$

In this case, $\displaystyle e=1+0i$ intuitively satisfies this, and is the unique identity element of group G. $\displaystyle |e|=|1+0i|=1$, so the set H has the identity of G.

3) The inverse of a complex number is the conjugate. So a complex number $\displaystyle x = a+bi$ has the inverse $\displaystyle x^{-1}=a-bi$. Looking at the list of four elements from part (1), it can be seen that the first two elements are $\displaystyle x$ and $\displaystyle y$, and the following two are the conjugates (inverses) of $\displaystyle x$ and $\displaystyle y$, 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?

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

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.

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

Quote:

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.

$\displaystyle G = \mathbb{C}^{x}$, the group of all nonzero complex numbers with operation given by multiplication of complex numbers.

$\displaystyle H = \left\{z \in \mathbb{C} \hspace{2mm} |\hspace{2mm} |z| = 1\right\}$

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

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 $\displaystyle a \in H$, there exists an $\displaystyle a^{-1} \in H$.

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)).

Quote:

$\displaystyle |z|$ is defined as $\displaystyle |z|=\sqrt{a^2+b^2}$ where *a* and *b *come from the general complex number form $\displaystyle a+bi$.

Since $\displaystyle |z| = 1$, 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.

Quote:

Therefore, this set H only has four elements:

$\displaystyle 1+0i$

$\displaystyle 0+1i$

$\displaystyle -1+0i$

$\displaystyle 0-1i$

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.

Quote:

2) An identity of group $\displaystyle G$ will satisfy

$\displaystyle a*e = e*a = a$

In this case, $\displaystyle e=1+0i$ intuitively satisfies this, and is the unique identity element of group G. $\displaystyle |e|=|1+0i|=1$, so the set H has the identity of G.

3) The inverse of a complex number is the conjugate. So a complex number $\displaystyle x = a+bi$ has the inverse $\displaystyle x^{-1}=a-bi$. Looking at the list of four elements from part (1), it can be seen that the first two elements are $\displaystyle x$ and $\displaystyle y$, and the following two are the conjugates (inverses) of $\displaystyle x$ and $\displaystyle y$, 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.

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

Quote:

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?

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

suppose z = a+ib.

then $\displaystyle |z| = \sqrt{a^2+b^2}$ so

$\displaystyle |z|^2 = a^2+b^2$

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

$\displaystyle a^2+b^2 = 1$. for which pairs (a,b) is this true? this equation should ring a bell....

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

Quote:

Originally Posted by

**Deveno** suppose z = a+ib.

then $\displaystyle |z| = \sqrt{a^2+b^2}$ so

$\displaystyle |z|^2 = a^2+b^2$

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

$\displaystyle a^2+b^2 = 1$. 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.

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

Quote:

Originally Posted by

**tangibleLime** Problem: [/B]Determine whether the given subset H of the group G is a subgroup.

$\displaystyle G = \mathbb{C}^{x}$, the group of all nonzero complex numbers with operation given by multiplication of complex numbers.

$\displaystyle H = \left\{z \in \mathbb{C} :\hspace{2mm} |z| = 1\right\}$

The set $\displaystyle H$ is the whole unit circle in the complex plane.

Here is a neat theorem: If $\displaystyle H$ is a non-empty subset of a group $\displaystyle G$ then $\displaystyle H$ is subgroup if and only if for $\displaystyle \{x,y\}\subset H$ it is true that $\displaystyle xy^{-1}\in H$.

For this $\displaystyle H$ if $\displaystyle y\in H$ then $\displaystyle y^{-1}=\overline{y}$, the conjugate, and $\displaystyle |\overline{y}|=1$ .

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

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 :D

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

$\displaystyle e^{iy} = cos(y) + i sin(y)$ by definition.

so what is $\displaystyle |e^{iy}|$ ?

by direct calculation:

$\displaystyle |e^{iy}| = \sqrt{cos^2(y) + sin^2(y)} = \sqrt{1} = 1$

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)) = $\displaystyle \frac{a}{\sqrt{a^2+b^2}}$

and that sin t = sin(arctan(b/a)) = $\displaystyle \frac{b}{\sqrt{a^2+b^2}}$

since $\displaystyle \sqrt{a^2+b^2} = 1$,

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 = $\displaystyle \pi/2$, if z is in the top half-plane, or

t = $\displaystyle -\pi/2$, if z is in the lower half-plane.