# First Isomorphism Theorem

1. Jul 18, 2006

### sqrt(-1)

I'm slightly confused with the following function so I was wondering if anybody could give me some hints as to the next step.

A function f is defined as

$$f:\mathbb{C} \longrightarrow \mathbb{C} \\$$
$$~~z \longmapsto |z|$$
where
$$\mathbb{C} = (\mathbb{C},+)$$
assuming the function is a homomorphism, I've gone on to find
$$\rm{Ker}(f) = \{0\}$$
$$\rm{Im}(f) = \mathbb{R}^+ \cup \{0\}$$
by the first isomorphism theorem I would have made the following conclusion(this part doesn't make sense to me):
$$\mathbb{C}/\rm{Ker}(f) \cong (\mathbb{R}^+ \cup \{0\}, + )$$

But firstly
$$(\mathbb{R}^+ \cup \{0\}, + )$$
is not a group since there is no inverse element, and secondly since Ker(f) = the trivial group
$$\mathbb{C}/\rm{Ker}(f) \cong (\mathbb{C}, + )$$

Could somebody give me some pointers please?

Last edited: Jul 18, 2006
2. Jul 18, 2006

### StatusX

That map is not a homomorphism for (C,+) (eg, |1+(-1)]=0 $\neq$|1|+|-1|=2), but it is for (C,*).

3. Jul 19, 2006

### matt grime

Firstly, when talking about groups it is best to specify whenever necessary what the group laws are before hand, and secondly. The image set is not a group in any obvious way, the map is not a homomorphism of groups, so applying the 1st isomorphism theorem won't get you anywhere.

StatusX, (C,*) is not a group, either.

The mod map is a homomorphism from the non-zero complex numbers to the reals, the kernel is the unit circle, and indeed the 1st isomorphism theorem holds.

4. Jul 19, 2006

### sqrt(-1)

Thanks for the reply - now that I've looked at it with the information you provided I realise I made a howler of a mistake in the first part of my working.

5. Jul 19, 2006

### StatusX

Right, sorry. Although I didn't really say it was, I just said that the map is a homomorphism, ie, structure preserving, not a group homomorphism. Or maybe by C I was referring to the extended complex numbers, I just couldn't write it in this font. I know, I'm reaching.

6. Jul 19, 2006

### matt grime

The extended complex numbers under multiplication still aren't a group: 0*infinity is not even defined. I don't know, from the context, if you genuinely think that they are, but I would hope you didn't think they were. You simply made a small oversight in your first post, that's all. We all do it.