Is a function that preserves normal subgroups a group morphism?

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Proposition 2.10.4 in Michael Artin's Algebra says that the inverse image of a normal subgroup under a group morphism is a normal subgroup (of the morphism's domain, of course)



Motivated by analogy with topology, my question is then if the converse implication holds: given a function between groups such that inverse images of normal subgroups are always normal, is such a function in fact a group morphism?










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    Proposition 2.10.4 in Michael Artin's Algebra says that the inverse image of a normal subgroup under a group morphism is a normal subgroup (of the morphism's domain, of course)



    Motivated by analogy with topology, my question is then if the converse implication holds: given a function between groups such that inverse images of normal subgroups are always normal, is such a function in fact a group morphism?










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      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Proposition 2.10.4 in Michael Artin's Algebra says that the inverse image of a normal subgroup under a group morphism is a normal subgroup (of the morphism's domain, of course)



      Motivated by analogy with topology, my question is then if the converse implication holds: given a function between groups such that inverse images of normal subgroups are always normal, is such a function in fact a group morphism?










      share|cite|improve this question













      Proposition 2.10.4 in Michael Artin's Algebra says that the inverse image of a normal subgroup under a group morphism is a normal subgroup (of the morphism's domain, of course)



      Motivated by analogy with topology, my question is then if the converse implication holds: given a function between groups such that inverse images of normal subgroups are always normal, is such a function in fact a group morphism?







      abstract-algebra group-theory






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      asked 1 hour ago









      alkchf

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          No, and here is a silly reason: there are many groups whose only normal subgroups are the trivial group $1$ and the entire group! Such groups are called simple. Basic examples include the cyclic groups $C_p$ for primes $p$, and the alternating groups $A_n$ for $n neq 4$. If you take any two big simple groups $G, G'$, and any mapping $f: G to G'$ such that $f(a) = 1 iff a = 1$, then $f$ will satisfy your requirements. Obviously $f$ can still be a very weird function then; certainly it will not need to be a group homomorphism.






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          • I think you also need $f$ to be bijective for your example to work.
            – C Monsour
            37 mins ago










          • @CMonsour, no, this is not necessary. Note that we are talking about inverse images under $f$, not images. So e.g. the mapping $f: A_5 to A_5$ defined by $f(mathrmid) = mathrmid$ and $f(alpha) = (1, 2)$ when $alpha neq mathrmid$ is an example.
            – Mees de Vries
            35 mins ago










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          up vote
          5
          down vote



          accepted










          No, and here is a silly reason: there are many groups whose only normal subgroups are the trivial group $1$ and the entire group! Such groups are called simple. Basic examples include the cyclic groups $C_p$ for primes $p$, and the alternating groups $A_n$ for $n neq 4$. If you take any two big simple groups $G, G'$, and any mapping $f: G to G'$ such that $f(a) = 1 iff a = 1$, then $f$ will satisfy your requirements. Obviously $f$ can still be a very weird function then; certainly it will not need to be a group homomorphism.






          share|cite|improve this answer




















          • I think you also need $f$ to be bijective for your example to work.
            – C Monsour
            37 mins ago










          • @CMonsour, no, this is not necessary. Note that we are talking about inverse images under $f$, not images. So e.g. the mapping $f: A_5 to A_5$ defined by $f(mathrmid) = mathrmid$ and $f(alpha) = (1, 2)$ when $alpha neq mathrmid$ is an example.
            – Mees de Vries
            35 mins ago














          up vote
          5
          down vote



          accepted










          No, and here is a silly reason: there are many groups whose only normal subgroups are the trivial group $1$ and the entire group! Such groups are called simple. Basic examples include the cyclic groups $C_p$ for primes $p$, and the alternating groups $A_n$ for $n neq 4$. If you take any two big simple groups $G, G'$, and any mapping $f: G to G'$ such that $f(a) = 1 iff a = 1$, then $f$ will satisfy your requirements. Obviously $f$ can still be a very weird function then; certainly it will not need to be a group homomorphism.






          share|cite|improve this answer




















          • I think you also need $f$ to be bijective for your example to work.
            – C Monsour
            37 mins ago










          • @CMonsour, no, this is not necessary. Note that we are talking about inverse images under $f$, not images. So e.g. the mapping $f: A_5 to A_5$ defined by $f(mathrmid) = mathrmid$ and $f(alpha) = (1, 2)$ when $alpha neq mathrmid$ is an example.
            – Mees de Vries
            35 mins ago












          up vote
          5
          down vote



          accepted







          up vote
          5
          down vote



          accepted






          No, and here is a silly reason: there are many groups whose only normal subgroups are the trivial group $1$ and the entire group! Such groups are called simple. Basic examples include the cyclic groups $C_p$ for primes $p$, and the alternating groups $A_n$ for $n neq 4$. If you take any two big simple groups $G, G'$, and any mapping $f: G to G'$ such that $f(a) = 1 iff a = 1$, then $f$ will satisfy your requirements. Obviously $f$ can still be a very weird function then; certainly it will not need to be a group homomorphism.






          share|cite|improve this answer












          No, and here is a silly reason: there are many groups whose only normal subgroups are the trivial group $1$ and the entire group! Such groups are called simple. Basic examples include the cyclic groups $C_p$ for primes $p$, and the alternating groups $A_n$ for $n neq 4$. If you take any two big simple groups $G, G'$, and any mapping $f: G to G'$ such that $f(a) = 1 iff a = 1$, then $f$ will satisfy your requirements. Obviously $f$ can still be a very weird function then; certainly it will not need to be a group homomorphism.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          Mees de Vries

          14.9k12450




          14.9k12450











          • I think you also need $f$ to be bijective for your example to work.
            – C Monsour
            37 mins ago










          • @CMonsour, no, this is not necessary. Note that we are talking about inverse images under $f$, not images. So e.g. the mapping $f: A_5 to A_5$ defined by $f(mathrmid) = mathrmid$ and $f(alpha) = (1, 2)$ when $alpha neq mathrmid$ is an example.
            – Mees de Vries
            35 mins ago
















          • I think you also need $f$ to be bijective for your example to work.
            – C Monsour
            37 mins ago










          • @CMonsour, no, this is not necessary. Note that we are talking about inverse images under $f$, not images. So e.g. the mapping $f: A_5 to A_5$ defined by $f(mathrmid) = mathrmid$ and $f(alpha) = (1, 2)$ when $alpha neq mathrmid$ is an example.
            – Mees de Vries
            35 mins ago















          I think you also need $f$ to be bijective for your example to work.
          – C Monsour
          37 mins ago




          I think you also need $f$ to be bijective for your example to work.
          – C Monsour
          37 mins ago












          @CMonsour, no, this is not necessary. Note that we are talking about inverse images under $f$, not images. So e.g. the mapping $f: A_5 to A_5$ defined by $f(mathrmid) = mathrmid$ and $f(alpha) = (1, 2)$ when $alpha neq mathrmid$ is an example.
          – Mees de Vries
          35 mins ago




          @CMonsour, no, this is not necessary. Note that we are talking about inverse images under $f$, not images. So e.g. the mapping $f: A_5 to A_5$ defined by $f(mathrmid) = mathrmid$ and $f(alpha) = (1, 2)$ when $alpha neq mathrmid$ is an example.
          – Mees de Vries
          35 mins ago

















           

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