- #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?

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: