First Isomorphism Theorem

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  • #1
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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

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

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

Could somebody give me some pointers please?
 
Last edited:

Answers and Replies

  • #2
StatusX
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That map is not a homomorphism for (C,+) (eg, |1+(-1)]=0 [itex]\neq[/itex]|1|+|-1|=2), but it is for (C,*).
 
  • #3
matt grime
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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
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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
StatusX
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matt grime said:
StatusX, (C,*) is not a group, either.
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
matt grime
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StatusX said:
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.
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.
 

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