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I need to find image, kernel and quotient group of these homomorphism

$$ \begin{array} .\phi_1: & \mathbb{C} & \to & \mathbb{C} \\ &z &\mapsto& z + 3iz \end{array}$$ and $$ \begin{array} .\phi_2: & \mathbb{C}^* & \to & \mathbb{C}^* \\ &z &\mapsto& z\bar{z} \end{array}$$

I've had a go at the first already. For the first I have the image as all complex numbers C, kernel as 0 due to that being the member of the domain that maps to 0 the identity of C. If all that I've said so far is correct, which i doubt it is, then I need to find the quotient group of G/N where G is all complex numbers C and N is 0. I have no idea how to do this I've only dealt with quotient groups of finite sets in the past.

Any help would be much appreciated.

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Put $\;z=x+iy\;$ , so

$$\phi_1(z)=z+3iz=(x+iy)-(3y-3ix)=(x-3y)+(y-3x)=0$$

Thus, for the kernel you need to solve the linear system

$$\begin{cases}&\;\;\;\;x-3y=0\\{}\\&-3x+y=0\end{cases}\;\implies x=y=0$$

For the second one

$$\phi_2(z)=z\overline z=|z|^2=1\iff |z|=1\implies $$

$$\implies \ker\phi_2=S^1:=\{\;z\in\Bbb Z\;:\;|z|=1\;\}=\{\;z\in\Bbb C\;:\;z=e^{it}\;,\;\;t\in\Bbb R\;\}$$

What's the image here?

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  • $\begingroup$ Is the image +1 and -1? $\endgroup$ – Brad Mar 6 '15 at 21:03
  • $\begingroup$ @Brad No, of course not. Just observe that $\;\phi_2(r)=r^2\;,\;\;\forall\,r\in\Bbb R\;$ , so we already have lots of other real positive (!) elements in the image. $\endgroup$ – Timbuc Mar 6 '15 at 21:05
  • $\begingroup$ All positive numbers R? $\endgroup$ – Brad Mar 6 '15 at 21:07
  • $\begingroup$ @Brad What do you think?! Is any positive real number the square of another real number? $\endgroup$ – Timbuc Mar 6 '15 at 21:07
  • $\begingroup$ Yes? Is it all real numbers then? Sorry if I'm being an idiot, i've been staring at maths literally all day. $\endgroup$ – Brad Mar 6 '15 at 21:11
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I'm going to walk you through the second one. Firstly, $z \bar{z} = |z|^2$. Timbuc has the kernel calculation correct, so that leaves image and quotient group. The image is the set of all values that $\phi_2(z)$ can take, and you worked out that the image is $\mathbb{R}^+$, the positive real numbers.

That leaves us to work out the quotient group. So, what are the members of the quotient group? They're sets of the form $z\ker(\phi_2) = \{ zw : w\in \ker(\phi_2) \}$ where $z\in \mathbb{C}^*$. So, if we let $z=1$ you get the set $\ker(\phi_2)$, and this is the identity in the quotient group. What about some other elements? Well, whatever $z$ is, it has the form $r e^{i\theta}$ with $r>0$, so $$z\ker(\phi_2) = \{ r e^{i\theta} e^{i\psi} : \psi\in[0, 2\pi)\} = \{ r e^{i\psi} : \psi\in[0, 2\pi)\} = \{ w\in \mathbb{C}^* : |w| = r\}= C_r.$$ Therefore, the elements of the quotient group are the subsets of the complex numbers which all have the same modulus,except for the set $\{0\}$ which isn't a subset of $\mathbb{C}^*$. I'm also labeling these sets $C_r$, just for convenience. What about the group operation? Well, the way that these sets are created guarantees that if we pick an element of from one set and and element from another, then multiply them in $\mathbb{C}^*$, the product of them is in the same element of the quotient group, no matter which elements we picked. This means that the group operation is such that $C_{r_1} C_{r_2} = C_{r_1 r_2}$ because $r_1 \in C_{r_1}$, $r_2\in C_{r_2}$ and $r_1 r_2 \in C_{r_1} C_{r_2}$.

Using this example, and the fact that you had the image and kernel of $\phi_1$ correct, what are the elements of the quotient group of $\phi_1$? What is their group operation? Its much easier than working out what's happening with $\phi_2$.

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