Searching for an alternative definition of group homomorphism












1












$begingroup$


I am searching for an alternative definition of a group homomorphism using only their impact on subgroups. I thought of something like




Let $G$, $G'$ be groups. A map from $G$ to $G'$ is called homomorphism of groups if the image of every subgroup of $G$ is a subgroup in $G'$.




I guess this definition wouldn't work. It's maybe too weak to be a homomorphism.



Does anyone have an idea?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    It definitely is too weak. According to your definition any bijection $mathbb{Z}_ptomathbb{Z}_p$ that fixes $0$ would be a homomorphism. Why are you looking for an alternative if there is a simple classical definition?
    $endgroup$
    – freakish
    Dec 3 '18 at 9:33








  • 1




    $begingroup$
    Are you trying to do something that mimics the definition of a continuous map?
    $endgroup$
    – Tobias Kildetoft
    Dec 3 '18 at 9:44






  • 1




    $begingroup$
    @freakish When if first learned about normal subgroups I had absolutely no idea what this definition using conjugation really wants to tell me. It just seemed to be not very intuitive, but when I began to think of normal groups as the kernels of group homomorphism it really helped me a lot to understand what normal subgroups are all about.I hoped to find a definition which makes even clearer which structures of a group "survive" a homomorphism.
    $endgroup$
    – Dominik
    Dec 3 '18 at 9:52
















1












$begingroup$


I am searching for an alternative definition of a group homomorphism using only their impact on subgroups. I thought of something like




Let $G$, $G'$ be groups. A map from $G$ to $G'$ is called homomorphism of groups if the image of every subgroup of $G$ is a subgroup in $G'$.




I guess this definition wouldn't work. It's maybe too weak to be a homomorphism.



Does anyone have an idea?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    It definitely is too weak. According to your definition any bijection $mathbb{Z}_ptomathbb{Z}_p$ that fixes $0$ would be a homomorphism. Why are you looking for an alternative if there is a simple classical definition?
    $endgroup$
    – freakish
    Dec 3 '18 at 9:33








  • 1




    $begingroup$
    Are you trying to do something that mimics the definition of a continuous map?
    $endgroup$
    – Tobias Kildetoft
    Dec 3 '18 at 9:44






  • 1




    $begingroup$
    @freakish When if first learned about normal subgroups I had absolutely no idea what this definition using conjugation really wants to tell me. It just seemed to be not very intuitive, but when I began to think of normal groups as the kernels of group homomorphism it really helped me a lot to understand what normal subgroups are all about.I hoped to find a definition which makes even clearer which structures of a group "survive" a homomorphism.
    $endgroup$
    – Dominik
    Dec 3 '18 at 9:52














1












1








1





$begingroup$


I am searching for an alternative definition of a group homomorphism using only their impact on subgroups. I thought of something like




Let $G$, $G'$ be groups. A map from $G$ to $G'$ is called homomorphism of groups if the image of every subgroup of $G$ is a subgroup in $G'$.




I guess this definition wouldn't work. It's maybe too weak to be a homomorphism.



Does anyone have an idea?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$




I am searching for an alternative definition of a group homomorphism using only their impact on subgroups. I thought of something like




Let $G$, $G'$ be groups. A map from $G$ to $G'$ is called homomorphism of groups if the image of every subgroup of $G$ is a subgroup in $G'$.




I guess this definition wouldn't work. It's maybe too weak to be a homomorphism.



Does anyone have an idea?






abstract-algebra group-theory




abstract-algebra group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 9:29









José Carlos Santos

163k22130233




163k22130233










asked Dec 3 '18 at 9:26









DominikDominik

965




965








  • 2




    $begingroup$
    It definitely is too weak. According to your definition any bijection $mathbb{Z}_ptomathbb{Z}_p$ that fixes $0$ would be a homomorphism. Why are you looking for an alternative if there is a simple classical definition?
    $endgroup$
    – freakish
    Dec 3 '18 at 9:33








  • 1




    $begingroup$
    Are you trying to do something that mimics the definition of a continuous map?
    $endgroup$
    – Tobias Kildetoft
    Dec 3 '18 at 9:44






  • 1




    $begingroup$
    @freakish When if first learned about normal subgroups I had absolutely no idea what this definition using conjugation really wants to tell me. It just seemed to be not very intuitive, but when I began to think of normal groups as the kernels of group homomorphism it really helped me a lot to understand what normal subgroups are all about.I hoped to find a definition which makes even clearer which structures of a group "survive" a homomorphism.
    $endgroup$
    – Dominik
    Dec 3 '18 at 9:52














  • 2




    $begingroup$
    It definitely is too weak. According to your definition any bijection $mathbb{Z}_ptomathbb{Z}_p$ that fixes $0$ would be a homomorphism. Why are you looking for an alternative if there is a simple classical definition?
    $endgroup$
    – freakish
    Dec 3 '18 at 9:33








  • 1




    $begingroup$
    Are you trying to do something that mimics the definition of a continuous map?
    $endgroup$
    – Tobias Kildetoft
    Dec 3 '18 at 9:44






  • 1




    $begingroup$
    @freakish When if first learned about normal subgroups I had absolutely no idea what this definition using conjugation really wants to tell me. It just seemed to be not very intuitive, but when I began to think of normal groups as the kernels of group homomorphism it really helped me a lot to understand what normal subgroups are all about.I hoped to find a definition which makes even clearer which structures of a group "survive" a homomorphism.
    $endgroup$
    – Dominik
    Dec 3 '18 at 9:52








2




2




$begingroup$
It definitely is too weak. According to your definition any bijection $mathbb{Z}_ptomathbb{Z}_p$ that fixes $0$ would be a homomorphism. Why are you looking for an alternative if there is a simple classical definition?
$endgroup$
– freakish
Dec 3 '18 at 9:33






$begingroup$
It definitely is too weak. According to your definition any bijection $mathbb{Z}_ptomathbb{Z}_p$ that fixes $0$ would be a homomorphism. Why are you looking for an alternative if there is a simple classical definition?
$endgroup$
– freakish
Dec 3 '18 at 9:33






1




1




$begingroup$
Are you trying to do something that mimics the definition of a continuous map?
$endgroup$
– Tobias Kildetoft
Dec 3 '18 at 9:44




$begingroup$
Are you trying to do something that mimics the definition of a continuous map?
$endgroup$
– Tobias Kildetoft
Dec 3 '18 at 9:44




1




1




$begingroup$
@freakish When if first learned about normal subgroups I had absolutely no idea what this definition using conjugation really wants to tell me. It just seemed to be not very intuitive, but when I began to think of normal groups as the kernels of group homomorphism it really helped me a lot to understand what normal subgroups are all about.I hoped to find a definition which makes even clearer which structures of a group "survive" a homomorphism.
$endgroup$
– Dominik
Dec 3 '18 at 9:52




$begingroup$
@freakish When if first learned about normal subgroups I had absolutely no idea what this definition using conjugation really wants to tell me. It just seemed to be not very intuitive, but when I began to think of normal groups as the kernels of group homomorphism it really helped me a lot to understand what normal subgroups are all about.I hoped to find a definition which makes even clearer which structures of a group "survive" a homomorphism.
$endgroup$
– Dominik
Dec 3 '18 at 9:52










2 Answers
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$begingroup$

Others have pointed out the trouble with this definition. This is the way one learns anyway. Try to see why this definition and not this gives a clear and critical understanding.



Let me point out another example where your suggested alternative definition fails: Take a group which is NOT abelian, say symmeric group on $n$ letters, . $n>3$. Consider the function $xmapsto x^{-1}$. As a function from a group to itself, it satisfies your condition, image of any subgroup is a subgroup (in fact it is the same subgroup). However it is not true in general that $(xy)^{-1}= x^{-1}y^{-1}$. So it is not a homomorphism.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    This definition is far too weak! you need compatibility with the group structure, which gets completely forgotten by your definition, for example, consider the following map: $$mathbb{Z}/pmathbb{Z} to mathbb{Z}/qmathbb{Z}\
    x mapsto [x]$$     where we interpret $mathbb{Z}/pmathbb{Z} $ as ${0,1,...,p-1}$.
    Now if $q le p$ this is would be by your "definition" a "groupmorphism", but all the properties of the elements, like order gets messed up, and you lose all control!






    share|cite|improve this answer









    $endgroup$













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      2 Answers
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      2 Answers
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      2












      $begingroup$

      Others have pointed out the trouble with this definition. This is the way one learns anyway. Try to see why this definition and not this gives a clear and critical understanding.



      Let me point out another example where your suggested alternative definition fails: Take a group which is NOT abelian, say symmeric group on $n$ letters, . $n>3$. Consider the function $xmapsto x^{-1}$. As a function from a group to itself, it satisfies your condition, image of any subgroup is a subgroup (in fact it is the same subgroup). However it is not true in general that $(xy)^{-1}= x^{-1}y^{-1}$. So it is not a homomorphism.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Others have pointed out the trouble with this definition. This is the way one learns anyway. Try to see why this definition and not this gives a clear and critical understanding.



        Let me point out another example where your suggested alternative definition fails: Take a group which is NOT abelian, say symmeric group on $n$ letters, . $n>3$. Consider the function $xmapsto x^{-1}$. As a function from a group to itself, it satisfies your condition, image of any subgroup is a subgroup (in fact it is the same subgroup). However it is not true in general that $(xy)^{-1}= x^{-1}y^{-1}$. So it is not a homomorphism.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Others have pointed out the trouble with this definition. This is the way one learns anyway. Try to see why this definition and not this gives a clear and critical understanding.



          Let me point out another example where your suggested alternative definition fails: Take a group which is NOT abelian, say symmeric group on $n$ letters, . $n>3$. Consider the function $xmapsto x^{-1}$. As a function from a group to itself, it satisfies your condition, image of any subgroup is a subgroup (in fact it is the same subgroup). However it is not true in general that $(xy)^{-1}= x^{-1}y^{-1}$. So it is not a homomorphism.






          share|cite|improve this answer











          $endgroup$



          Others have pointed out the trouble with this definition. This is the way one learns anyway. Try to see why this definition and not this gives a clear and critical understanding.



          Let me point out another example where your suggested alternative definition fails: Take a group which is NOT abelian, say symmeric group on $n$ letters, . $n>3$. Consider the function $xmapsto x^{-1}$. As a function from a group to itself, it satisfies your condition, image of any subgroup is a subgroup (in fact it is the same subgroup). However it is not true in general that $(xy)^{-1}= x^{-1}y^{-1}$. So it is not a homomorphism.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 3 '18 at 10:42

























          answered Dec 3 '18 at 10:35









          P VanchinathanP Vanchinathan

          15.2k12136




          15.2k12136























              0












              $begingroup$

              This definition is far too weak! you need compatibility with the group structure, which gets completely forgotten by your definition, for example, consider the following map: $$mathbb{Z}/pmathbb{Z} to mathbb{Z}/qmathbb{Z}\
              x mapsto [x]$$               where we interpret $mathbb{Z}/pmathbb{Z} $ as ${0,1,...,p-1}$.
              Now if $q le p$ this is would be by your "definition" a "groupmorphism", but all the properties of the elements, like order gets messed up, and you lose all control!






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                This definition is far too weak! you need compatibility with the group structure, which gets completely forgotten by your definition, for example, consider the following map: $$mathbb{Z}/pmathbb{Z} to mathbb{Z}/qmathbb{Z}\
                x mapsto [x]$$                 where we interpret $mathbb{Z}/pmathbb{Z} $ as ${0,1,...,p-1}$.
                Now if $q le p$ this is would be by your "definition" a "groupmorphism", but all the properties of the elements, like order gets messed up, and you lose all control!






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  This definition is far too weak! you need compatibility with the group structure, which gets completely forgotten by your definition, for example, consider the following map: $$mathbb{Z}/pmathbb{Z} to mathbb{Z}/qmathbb{Z}\
                  x mapsto [x]$$                   where we interpret $mathbb{Z}/pmathbb{Z} $ as ${0,1,...,p-1}$.
                  Now if $q le p$ this is would be by your "definition" a "groupmorphism", but all the properties of the elements, like order gets messed up, and you lose all control!






                  share|cite|improve this answer









                  $endgroup$



                  This definition is far too weak! you need compatibility with the group structure, which gets completely forgotten by your definition, for example, consider the following map: $$mathbb{Z}/pmathbb{Z} to mathbb{Z}/qmathbb{Z}\
                  x mapsto [x]$$                   where we interpret $mathbb{Z}/pmathbb{Z} $ as ${0,1,...,p-1}$.
                  Now if $q le p$ this is would be by your "definition" a "groupmorphism", but all the properties of the elements, like order gets messed up, and you lose all control!







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 3 '18 at 10:18









                  EnkiduEnkidu

                  1,36719




                  1,36719






























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