# Math Help - Determine if this set of complex numbers forms a subgroup

1. ## 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.

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

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

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

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

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

Since $|z| = 1$, 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:

$1+0i$
$0+1i$
$-1+0i$
$0-1i$

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 $G$ will satisfy

$a*e = e*a = a$

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

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

2. ## 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.

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

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.

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

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

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

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

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

Since $|z| = 1$, 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.

Therefore, this set H only has four elements:

$1+0i$
$0+1i$
$-1+0i$
$0-1i$

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.

2) An identity of group $G$ will satisfy

$a*e = e*a = a$

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

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

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

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?

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

suppose z = a+ib.

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

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

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

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

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

Originally Posted by Deveno
suppose z = a+ib.

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

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

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

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

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

Originally Posted by tangibleLime
Problem: [/B]Determine whether the given subset H of the group G is a subgroup.
$G = \mathbb{C}^{x}$, the group of all nonzero complex numbers with operation given by multiplication of complex numbers.

$H = \left\{z \in \mathbb{C} :\hspace{2mm} |z| = 1\right\}$
The set $H$ is the whole unit circle in the complex plane.
Here is a neat theorem: If $H$ is a non-empty subset of a group $G$ then $H$ is subgroup if and only if for $\{x,y\}\subset H$ it is true that $xy^{-1}\in H$.

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

8. ## 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

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

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

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

by direct calculation:

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

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

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

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

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

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