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First Isomorphism Theorem

  1. Jul 18, 2006 #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

    [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: Jul 18, 2006
  2. jcsd
  3. Jul 18, 2006 #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,*).
     
  4. Jul 19, 2006 #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.
     
  5. Jul 19, 2006 #4
    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.
     
  6. Jul 19, 2006 #5

    StatusX

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    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.
     
  7. Jul 19, 2006 #6

    matt grime

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    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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