How to determine if a surjective homomorphism exists between two groups?

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My question sheet asks about whether a surjective homomorphism exists between various symmetric groups and various $Z_n$ groups, for example between $S_3$ and $Z_3$, or $A_4$ and $Z_3$. To be honest, I don't really know where to start at all - I've been racking my brain for some property of homomorphisms to do with cardinality, or something like that, but I basically have no idea what I'm doing. Any pointers?










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    up vote
    3
    down vote

    favorite












    My question sheet asks about whether a surjective homomorphism exists between various symmetric groups and various $Z_n$ groups, for example between $S_3$ and $Z_3$, or $A_4$ and $Z_3$. To be honest, I don't really know where to start at all - I've been racking my brain for some property of homomorphisms to do with cardinality, or something like that, but I basically have no idea what I'm doing. Any pointers?










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      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      My question sheet asks about whether a surjective homomorphism exists between various symmetric groups and various $Z_n$ groups, for example between $S_3$ and $Z_3$, or $A_4$ and $Z_3$. To be honest, I don't really know where to start at all - I've been racking my brain for some property of homomorphisms to do with cardinality, or something like that, but I basically have no idea what I'm doing. Any pointers?










      share|cite|improve this question













      My question sheet asks about whether a surjective homomorphism exists between various symmetric groups and various $Z_n$ groups, for example between $S_3$ and $Z_3$, or $A_4$ and $Z_3$. To be honest, I don't really know where to start at all - I've been racking my brain for some property of homomorphisms to do with cardinality, or something like that, but I basically have no idea what I'm doing. Any pointers?







      group-theory permutations group-homomorphism






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      asked 4 hours ago









      cal

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          2 Answers
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          Sounds like a fun problem. Basically, your strategy is going to be to write one down, or prove it doesn't exist.



          The biggest thing that's useful is knowing how many normal subgroups a group has. For example, $S_n$ has only itself, $1$, and $A_n$. By the first isomorphism theorem, the number of groups that can receive a surjective homomorphism from $S_n$ is severely limited.






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            A good place to start would be the first isomorphism theorem: if $ f: Grightarrow H$ is a homomorphism, then $G/ker fsimeq operatornameIm f$. So if $f : G rightarrow H$ is a surjective homomorphism, then $G / ker f simeq H$ since surjectivity implies $operatornameIm f = H$. So to determine whether such a homomorphism exists, you should determine whether or not there exists a normal subgroup $G’ leq G$ such that $G/ G’ simeq H$. If such a $G’$ exists, $varphi : G/G’ rightarrow H$ is an isomorphism, and $pi : G rightarrow G/G’$ is the quotient map, then $varphi circ pi$ is a surjective homomorphism from $G$ to $H$.






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            • And use this in tandem with Lagrange: divisibility issues will often show it is impossible to have certain surjections
              – Randall
              3 hours ago











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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

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            active

            oldest

            votes








            up vote
            4
            down vote













            Sounds like a fun problem. Basically, your strategy is going to be to write one down, or prove it doesn't exist.



            The biggest thing that's useful is knowing how many normal subgroups a group has. For example, $S_n$ has only itself, $1$, and $A_n$. By the first isomorphism theorem, the number of groups that can receive a surjective homomorphism from $S_n$ is severely limited.






            share|cite|improve this answer
























              up vote
              4
              down vote













              Sounds like a fun problem. Basically, your strategy is going to be to write one down, or prove it doesn't exist.



              The biggest thing that's useful is knowing how many normal subgroups a group has. For example, $S_n$ has only itself, $1$, and $A_n$. By the first isomorphism theorem, the number of groups that can receive a surjective homomorphism from $S_n$ is severely limited.






              share|cite|improve this answer






















                up vote
                4
                down vote










                up vote
                4
                down vote









                Sounds like a fun problem. Basically, your strategy is going to be to write one down, or prove it doesn't exist.



                The biggest thing that's useful is knowing how many normal subgroups a group has. For example, $S_n$ has only itself, $1$, and $A_n$. By the first isomorphism theorem, the number of groups that can receive a surjective homomorphism from $S_n$ is severely limited.






                share|cite|improve this answer












                Sounds like a fun problem. Basically, your strategy is going to be to write one down, or prove it doesn't exist.



                The biggest thing that's useful is knowing how many normal subgroups a group has. For example, $S_n$ has only itself, $1$, and $A_n$. By the first isomorphism theorem, the number of groups that can receive a surjective homomorphism from $S_n$ is severely limited.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                hunter

                13.7k22337




                13.7k22337




















                    up vote
                    1
                    down vote













                    A good place to start would be the first isomorphism theorem: if $ f: Grightarrow H$ is a homomorphism, then $G/ker fsimeq operatornameIm f$. So if $f : G rightarrow H$ is a surjective homomorphism, then $G / ker f simeq H$ since surjectivity implies $operatornameIm f = H$. So to determine whether such a homomorphism exists, you should determine whether or not there exists a normal subgroup $G’ leq G$ such that $G/ G’ simeq H$. If such a $G’$ exists, $varphi : G/G’ rightarrow H$ is an isomorphism, and $pi : G rightarrow G/G’$ is the quotient map, then $varphi circ pi$ is a surjective homomorphism from $G$ to $H$.






                    share|cite|improve this answer




















                    • And use this in tandem with Lagrange: divisibility issues will often show it is impossible to have certain surjections
                      – Randall
                      3 hours ago















                    up vote
                    1
                    down vote













                    A good place to start would be the first isomorphism theorem: if $ f: Grightarrow H$ is a homomorphism, then $G/ker fsimeq operatornameIm f$. So if $f : G rightarrow H$ is a surjective homomorphism, then $G / ker f simeq H$ since surjectivity implies $operatornameIm f = H$. So to determine whether such a homomorphism exists, you should determine whether or not there exists a normal subgroup $G’ leq G$ such that $G/ G’ simeq H$. If such a $G’$ exists, $varphi : G/G’ rightarrow H$ is an isomorphism, and $pi : G rightarrow G/G’$ is the quotient map, then $varphi circ pi$ is a surjective homomorphism from $G$ to $H$.






                    share|cite|improve this answer




















                    • And use this in tandem with Lagrange: divisibility issues will often show it is impossible to have certain surjections
                      – Randall
                      3 hours ago













                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    A good place to start would be the first isomorphism theorem: if $ f: Grightarrow H$ is a homomorphism, then $G/ker fsimeq operatornameIm f$. So if $f : G rightarrow H$ is a surjective homomorphism, then $G / ker f simeq H$ since surjectivity implies $operatornameIm f = H$. So to determine whether such a homomorphism exists, you should determine whether or not there exists a normal subgroup $G’ leq G$ such that $G/ G’ simeq H$. If such a $G’$ exists, $varphi : G/G’ rightarrow H$ is an isomorphism, and $pi : G rightarrow G/G’$ is the quotient map, then $varphi circ pi$ is a surjective homomorphism from $G$ to $H$.






                    share|cite|improve this answer












                    A good place to start would be the first isomorphism theorem: if $ f: Grightarrow H$ is a homomorphism, then $G/ker fsimeq operatornameIm f$. So if $f : G rightarrow H$ is a surjective homomorphism, then $G / ker f simeq H$ since surjectivity implies $operatornameIm f = H$. So to determine whether such a homomorphism exists, you should determine whether or not there exists a normal subgroup $G’ leq G$ such that $G/ G’ simeq H$. If such a $G’$ exists, $varphi : G/G’ rightarrow H$ is an isomorphism, and $pi : G rightarrow G/G’$ is the quotient map, then $varphi circ pi$ is a surjective homomorphism from $G$ to $H$.







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                    share|cite|improve this answer










                    answered 3 hours ago









                    Oiler

                    1,9041820




                    1,9041820











                    • And use this in tandem with Lagrange: divisibility issues will often show it is impossible to have certain surjections
                      – Randall
                      3 hours ago

















                    • And use this in tandem with Lagrange: divisibility issues will often show it is impossible to have certain surjections
                      – Randall
                      3 hours ago
















                    And use this in tandem with Lagrange: divisibility issues will often show it is impossible to have certain surjections
                    – Randall
                    3 hours ago





                    And use this in tandem with Lagrange: divisibility issues will often show it is impossible to have certain surjections
                    – Randall
                    3 hours ago


















                     

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