Showing $phi: mathbb Z to mathbb Z_{12}$ is a homomorphism and determining it's $Ker(phi)$ and $Im(phi)$












2












$begingroup$


Question: Group Theory: Let $phi: mathbb Z to mathbb Z_{12}$ be defined by $$phi(x) = x;(text{mod}; 12)$$



a). Show that $phi$ is a homomorphism.



My work: Let $phi: mathbb Z to mathbb Z_{12}$ be defined by $phi(x)$ = $x$ $mod$ $12$



$phi(x+y)$ $=$ $[x+y]_{12}$ $=$ $[x]_{12}$ $+$ $[y]_{12}$ $=$ $phi(x)$ $+$ $phi(y)$



$therefore$ $phi$ is a homomorphism.




b). Find $Ker(phi)$ and $Im(phi)$



My work: $Ker(phi)$ $=$ {$x in mathbb Z$ $| phi(x) = [0]_{12}$}
$=$ {$x in mathbb Z$ $|$$[x]_{12} = [0]_{12}$}
$=$ {$x in mathbb Z$ $|$ $12$ divides $x$}
$=$ $12mathbb Z$



$Im(phi)$ $=$ {$phi(x) in mathbb Z_{12} / x in mathbb Z$} $=$ $mathbb Z_{12}$



Is my reasoning correct? I am unsure if my intuition is correct. Any help, comments, concerns, appreciated!






abstract-algebra group-theory group-homomorphism







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    For part $(a)$ I would say that you did not really show anything. Why is $[x+y]_{12}=[x]+[y]$? As for finding the kernel, I think that you are all good. Maybe a word: $[x]=[0]$ if there exists an integer $n$ so that $12 cdot n=x$. For finding the image, I think that's fine.
    $endgroup$
    – Andres Mejia
    Dec 6 '18 at 3:15










  • $begingroup$
    You might want to mention this is in the context of group theory in the question, not just the tags. It took me a little while to spot that, and the same question would have different answers for, say, a ring homomorphism.
    $endgroup$
    – aschepler
    Dec 6 '18 at 3:20










  • $begingroup$
    I hate to be that guy but, you haven't proved that $phi$ is well-defined.
    $endgroup$
    – Yunus Syed
    Dec 6 '18 at 3:36
















2












$begingroup$


Question: Group Theory: Let $phi: mathbb Z to mathbb Z_{12}$ be defined by $$phi(x) = x;(text{mod}; 12)$$



a). Show that $phi$ is a homomorphism.



My work: Let $phi: mathbb Z to mathbb Z_{12}$ be defined by $phi(x)$ = $x$ $mod$ $12$



$phi(x+y)$ $=$ $[x+y]_{12}$ $=$ $[x]_{12}$ $+$ $[y]_{12}$ $=$ $phi(x)$ $+$ $phi(y)$



$therefore$ $phi$ is a homomorphism.




b). Find $Ker(phi)$ and $Im(phi)$



My work: $Ker(phi)$ $=$ {$x in mathbb Z$ $| phi(x) = [0]_{12}$}
$=$ {$x in mathbb Z$ $|$$[x]_{12} = [0]_{12}$}
$=$ {$x in mathbb Z$ $|$ $12$ divides $x$}
$=$ $12mathbb Z$



$Im(phi)$ $=$ {$phi(x) in mathbb Z_{12} / x in mathbb Z$} $=$ $mathbb Z_{12}$



Is my reasoning correct? I am unsure if my intuition is correct. Any help, comments, concerns, appreciated!






abstract-algebra group-theory group-homomorphism







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    For part $(a)$ I would say that you did not really show anything. Why is $[x+y]_{12}=[x]+[y]$? As for finding the kernel, I think that you are all good. Maybe a word: $[x]=[0]$ if there exists an integer $n$ so that $12 cdot n=x$. For finding the image, I think that's fine.
    $endgroup$
    – Andres Mejia
    Dec 6 '18 at 3:15










  • $begingroup$
    You might want to mention this is in the context of group theory in the question, not just the tags. It took me a little while to spot that, and the same question would have different answers for, say, a ring homomorphism.
    $endgroup$
    – aschepler
    Dec 6 '18 at 3:20










  • $begingroup$
    I hate to be that guy but, you haven't proved that $phi$ is well-defined.
    $endgroup$
    – Yunus Syed
    Dec 6 '18 at 3:36














2












2








2





$begingroup$


Question: Group Theory: Let $phi: mathbb Z to mathbb Z_{12}$ be defined by $$phi(x) = x;(text{mod}; 12)$$



a). Show that $phi$ is a homomorphism.



My work: Let $phi: mathbb Z to mathbb Z_{12}$ be defined by $phi(x)$ = $x$ $mod$ $12$



$phi(x+y)$ $=$ $[x+y]_{12}$ $=$ $[x]_{12}$ $+$ $[y]_{12}$ $=$ $phi(x)$ $+$ $phi(y)$



$therefore$ $phi$ is a homomorphism.




b). Find $Ker(phi)$ and $Im(phi)$



My work: $Ker(phi)$ $=$ {$x in mathbb Z$ $| phi(x) = [0]_{12}$}
$=$ {$x in mathbb Z$ $|$$[x]_{12} = [0]_{12}$}
$=$ {$x in mathbb Z$ $|$ $12$ divides $x$}
$=$ $12mathbb Z$



$Im(phi)$ $=$ {$phi(x) in mathbb Z_{12} / x in mathbb Z$} $=$ $mathbb Z_{12}$



Is my reasoning correct? I am unsure if my intuition is correct. Any help, comments, concerns, appreciated!






abstract-algebra group-theory group-homomorphism







share|cite|improve this question











$endgroup$




Question: Group Theory: Let $phi: mathbb Z to mathbb Z_{12}$ be defined by $$phi(x) = x;(text{mod}; 12)$$



a). Show that $phi$ is a homomorphism.



My work: Let $phi: mathbb Z to mathbb Z_{12}$ be defined by $phi(x)$ = $x$ $mod$ $12$



$phi(x+y)$ $=$ $[x+y]_{12}$ $=$ $[x]_{12}$ $+$ $[y]_{12}$ $=$ $phi(x)$ $+$ $phi(y)$



$therefore$ $phi$ is a homomorphism.




b). Find $Ker(phi)$ and $Im(phi)$



My work: $Ker(phi)$ $=$ {$x in mathbb Z$ $| phi(x) = [0]_{12}$}
$=$ {$x in mathbb Z$ $|$$[x]_{12} = [0]_{12}$}
$=$ {$x in mathbb Z$ $|$ $12$ divides $x$}
$=$ $12mathbb Z$



$Im(phi)$ $=$ {$phi(x) in mathbb Z_{12} / x in mathbb Z$} $=$ $mathbb Z_{12}$



Is my reasoning correct? I am unsure if my intuition is correct. Any help, comments, concerns, appreciated!






abstract-algebra group-theory group-homomorphism




abstract-algebra group-theory group-homomorphism






share|cite|improve this question















share|cite|improve this question













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edited Dec 6 '18 at 3:25







Trever

















asked Dec 6 '18 at 3:11









TreverTrever

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566








  • 2




    $begingroup$
    For part $(a)$ I would say that you did not really show anything. Why is $[x+y]_{12}=[x]+[y]$? As for finding the kernel, I think that you are all good. Maybe a word: $[x]=[0]$ if there exists an integer $n$ so that $12 cdot n=x$. For finding the image, I think that's fine.
    $endgroup$
    – Andres Mejia
    Dec 6 '18 at 3:15










  • $begingroup$
    You might want to mention this is in the context of group theory in the question, not just the tags. It took me a little while to spot that, and the same question would have different answers for, say, a ring homomorphism.
    $endgroup$
    – aschepler
    Dec 6 '18 at 3:20










  • $begingroup$
    I hate to be that guy but, you haven't proved that $phi$ is well-defined.
    $endgroup$
    – Yunus Syed
    Dec 6 '18 at 3:36














  • 2




    $begingroup$
    For part $(a)$ I would say that you did not really show anything. Why is $[x+y]_{12}=[x]+[y]$? As for finding the kernel, I think that you are all good. Maybe a word: $[x]=[0]$ if there exists an integer $n$ so that $12 cdot n=x$. For finding the image, I think that's fine.
    $endgroup$
    – Andres Mejia
    Dec 6 '18 at 3:15










  • $begingroup$
    You might want to mention this is in the context of group theory in the question, not just the tags. It took me a little while to spot that, and the same question would have different answers for, say, a ring homomorphism.
    $endgroup$
    – aschepler
    Dec 6 '18 at 3:20










  • $begingroup$
    I hate to be that guy but, you haven't proved that $phi$ is well-defined.
    $endgroup$
    – Yunus Syed
    Dec 6 '18 at 3:36








2




2




$begingroup$
For part $(a)$ I would say that you did not really show anything. Why is $[x+y]_{12}=[x]+[y]$? As for finding the kernel, I think that you are all good. Maybe a word: $[x]=[0]$ if there exists an integer $n$ so that $12 cdot n=x$. For finding the image, I think that's fine.
$endgroup$
– Andres Mejia
Dec 6 '18 at 3:15




$begingroup$
For part $(a)$ I would say that you did not really show anything. Why is $[x+y]_{12}=[x]+[y]$? As for finding the kernel, I think that you are all good. Maybe a word: $[x]=[0]$ if there exists an integer $n$ so that $12 cdot n=x$. For finding the image, I think that's fine.
$endgroup$
– Andres Mejia
Dec 6 '18 at 3:15












$begingroup$
You might want to mention this is in the context of group theory in the question, not just the tags. It took me a little while to spot that, and the same question would have different answers for, say, a ring homomorphism.
$endgroup$
– aschepler
Dec 6 '18 at 3:20




$begingroup$
You might want to mention this is in the context of group theory in the question, not just the tags. It took me a little while to spot that, and the same question would have different answers for, say, a ring homomorphism.
$endgroup$
– aschepler
Dec 6 '18 at 3:20












$begingroup$
I hate to be that guy but, you haven't proved that $phi$ is well-defined.
$endgroup$
– Yunus Syed
Dec 6 '18 at 3:36




$begingroup$
I hate to be that guy but, you haven't proved that $phi$ is well-defined.
$endgroup$
– Yunus Syed
Dec 6 '18 at 3:36










1 Answer
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oldest

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0












$begingroup$

It is unclear what you asking since you've not explained 'where we are at'.
So the answer here will be general, forgetting all group theory and only assuming an understanding of equivalence relations.



Fact 1: The relation $x equiv y text{ (mod 12) }$ is an equivalence relation, which by definition means that $x - y$ is divisible by $12$.



The block containing $0$ is equal to the set ${dots,-36,-24,-12,0,12,24,36,dots }$; we denote this set by $12mathbb Z$ and the quotient set by



$tag 1 mathbb Z text{/} 12mathbb Z$



General theory tells us that we have a projection (or quotient) mapping



$tag 2 phi: mathbb Z to mathbb Z text{/} 12mathbb Z$



The projection is by definition a surjective mapping.



Notice that $0$ gets mapped to $12mathbb Z in mathbb Z text{/} 12mathbb Z$, and of course $phi(x) = 12mathbb Z$ just means that $x in 12mathbb Z $.



So this takes care of the questions on the image and the kernel without having to know any group theory.






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    1 Answer
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    1 Answer
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    0












    $begingroup$

    It is unclear what you asking since you've not explained 'where we are at'.
    So the answer here will be general, forgetting all group theory and only assuming an understanding of equivalence relations.



    Fact 1: The relation $x equiv y text{ (mod 12) }$ is an equivalence relation, which by definition means that $x - y$ is divisible by $12$.



    The block containing $0$ is equal to the set ${dots,-36,-24,-12,0,12,24,36,dots }$; we denote this set by $12mathbb Z$ and the quotient set by



    $tag 1 mathbb Z text{/} 12mathbb Z$



    General theory tells us that we have a projection (or quotient) mapping



    $tag 2 phi: mathbb Z to mathbb Z text{/} 12mathbb Z$



    The projection is by definition a surjective mapping.



    Notice that $0$ gets mapped to $12mathbb Z in mathbb Z text{/} 12mathbb Z$, and of course $phi(x) = 12mathbb Z$ just means that $x in 12mathbb Z $.



    So this takes care of the questions on the image and the kernel without having to know any group theory.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      It is unclear what you asking since you've not explained 'where we are at'.
      So the answer here will be general, forgetting all group theory and only assuming an understanding of equivalence relations.



      Fact 1: The relation $x equiv y text{ (mod 12) }$ is an equivalence relation, which by definition means that $x - y$ is divisible by $12$.



      The block containing $0$ is equal to the set ${dots,-36,-24,-12,0,12,24,36,dots }$; we denote this set by $12mathbb Z$ and the quotient set by



      $tag 1 mathbb Z text{/} 12mathbb Z$



      General theory tells us that we have a projection (or quotient) mapping



      $tag 2 phi: mathbb Z to mathbb Z text{/} 12mathbb Z$



      The projection is by definition a surjective mapping.



      Notice that $0$ gets mapped to $12mathbb Z in mathbb Z text{/} 12mathbb Z$, and of course $phi(x) = 12mathbb Z$ just means that $x in 12mathbb Z $.



      So this takes care of the questions on the image and the kernel without having to know any group theory.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        It is unclear what you asking since you've not explained 'where we are at'.
        So the answer here will be general, forgetting all group theory and only assuming an understanding of equivalence relations.



        Fact 1: The relation $x equiv y text{ (mod 12) }$ is an equivalence relation, which by definition means that $x - y$ is divisible by $12$.



        The block containing $0$ is equal to the set ${dots,-36,-24,-12,0,12,24,36,dots }$; we denote this set by $12mathbb Z$ and the quotient set by



        $tag 1 mathbb Z text{/} 12mathbb Z$



        General theory tells us that we have a projection (or quotient) mapping



        $tag 2 phi: mathbb Z to mathbb Z text{/} 12mathbb Z$



        The projection is by definition a surjective mapping.



        Notice that $0$ gets mapped to $12mathbb Z in mathbb Z text{/} 12mathbb Z$, and of course $phi(x) = 12mathbb Z$ just means that $x in 12mathbb Z $.



        So this takes care of the questions on the image and the kernel without having to know any group theory.






        share|cite|improve this answer









        $endgroup$



        It is unclear what you asking since you've not explained 'where we are at'.
        So the answer here will be general, forgetting all group theory and only assuming an understanding of equivalence relations.



        Fact 1: The relation $x equiv y text{ (mod 12) }$ is an equivalence relation, which by definition means that $x - y$ is divisible by $12$.



        The block containing $0$ is equal to the set ${dots,-36,-24,-12,0,12,24,36,dots }$; we denote this set by $12mathbb Z$ and the quotient set by



        $tag 1 mathbb Z text{/} 12mathbb Z$



        General theory tells us that we have a projection (or quotient) mapping



        $tag 2 phi: mathbb Z to mathbb Z text{/} 12mathbb Z$



        The projection is by definition a surjective mapping.



        Notice that $0$ gets mapped to $12mathbb Z in mathbb Z text{/} 12mathbb Z$, and of course $phi(x) = 12mathbb Z$ just means that $x in 12mathbb Z $.



        So this takes care of the questions on the image and the kernel without having to know any group theory.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 4:14









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