Exact sequence with finite groups: $0 to A xrightarrow[]{alpha} mathbb{Z}^d xrightarrow[]{beta} mathbb{Z}...












4














I have this exact sequence of abelian groups:
$$0 to A xrightarrow{alpha} mathbb{Z}^d xrightarrow{beta} mathbb{Z} xrightarrow{gamma} B to0$$
with $A$ and $B$ finite abelian groups and $d in mathbb{N}$.



I would like to conclude $d=1$, but I have no idea how could use the information $A,B$ are finite. The only thing I noticed that $gamma$ cannot be injective, but I don't know how to proceed. I'm a bit rusty in Algebra...



Could someone give some help?



Thanks!






abstract-algebra group-theory modules







share|cite|improve this question




















  • 1




    You can completely remove $A$ from the sequence because $alpha$ is zero.
    – Matt Samuel
    Nov 25 '18 at 16:54
















4














I have this exact sequence of abelian groups:
$$0 to A xrightarrow{alpha} mathbb{Z}^d xrightarrow{beta} mathbb{Z} xrightarrow{gamma} B to0$$
with $A$ and $B$ finite abelian groups and $d in mathbb{N}$.



I would like to conclude $d=1$, but I have no idea how could use the information $A,B$ are finite. The only thing I noticed that $gamma$ cannot be injective, but I don't know how to proceed. I'm a bit rusty in Algebra...



Could someone give some help?



Thanks!






abstract-algebra group-theory modules







share|cite|improve this question




















  • 1




    You can completely remove $A$ from the sequence because $alpha$ is zero.
    – Matt Samuel
    Nov 25 '18 at 16:54














4












4








4


1





I have this exact sequence of abelian groups:
$$0 to A xrightarrow{alpha} mathbb{Z}^d xrightarrow{beta} mathbb{Z} xrightarrow{gamma} B to0$$
with $A$ and $B$ finite abelian groups and $d in mathbb{N}$.



I would like to conclude $d=1$, but I have no idea how could use the information $A,B$ are finite. The only thing I noticed that $gamma$ cannot be injective, but I don't know how to proceed. I'm a bit rusty in Algebra...



Could someone give some help?



Thanks!






abstract-algebra group-theory modules







share|cite|improve this question















I have this exact sequence of abelian groups:
$$0 to A xrightarrow{alpha} mathbb{Z}^d xrightarrow{beta} mathbb{Z} xrightarrow{gamma} B to0$$
with $A$ and $B$ finite abelian groups and $d in mathbb{N}$.



I would like to conclude $d=1$, but I have no idea how could use the information $A,B$ are finite. The only thing I noticed that $gamma$ cannot be injective, but I don't know how to proceed. I'm a bit rusty in Algebra...



Could someone give some help?



Thanks!






abstract-algebra group-theory modules




abstract-algebra group-theory modules






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 16:55









Shaun

8,750113680




8,750113680










asked Nov 25 '18 at 16:52









EmarJ

386114




386114








  • 1




    You can completely remove $A$ from the sequence because $alpha$ is zero.
    – Matt Samuel
    Nov 25 '18 at 16:54














  • 1




    You can completely remove $A$ from the sequence because $alpha$ is zero.
    – Matt Samuel
    Nov 25 '18 at 16:54








1




1




You can completely remove $A$ from the sequence because $alpha$ is zero.
– Matt Samuel
Nov 25 '18 at 16:54




You can completely remove $A$ from the sequence because $alpha$ is zero.
– Matt Samuel
Nov 25 '18 at 16:54










3 Answers
3






active

oldest

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1














You can first conclude that $alpha = 0$: if not, $A neq 0$, and any non-zero element of $A$ would be sent by $alpha$ to a torsion element of $mathbb{Z}^d$, but no such exist. So $alpha = 0$, and since the sequence is exact, $A = 0$. Thus, we require an injective homomorphism $beta: mathbb{Z}^d to mathbb{Z}$.



But if $dgeq 2$, taking $e_i$ to be the standard generators (all zeros with a one in the $i$th position), $beta(beta(e_1)e_2) = beta(e_1)beta(e_2) = beta(beta(e_2)e_1)$, so $beta$ cannot be an injection, so indeed, $d leq 1$ (and $d neq 0$ else $gamma$ would be an isomorphism).






share|cite|improve this answer





























    4














    Tensoring with the flat $mathbb{Z}$-module $mathbb{Q}$ gives an exact sequence $0 to mathbb{Q}^d to mathbb{Q} to 0$.






    share|cite|improve this answer



















    • 1




      This should be $0toBbb Q^dtoBbb Qto 0.$
      – Stahl
      Nov 25 '18 at 17:04










    • Yes, of course it should. Fixed.
      – anomaly
      Nov 25 '18 at 17:19



















    4














    Since $A$ is finite then there is no nontrivial homomorphism $Atomathbb{Z}^d$. Meaning $alpha=0$. This implies that $kerbeta=0$ and so $beta$ is injective. The case when $beta$ is injective is only possible when $dleq 1$.



    The case when $d=0$ is impossible because that would imply that $gamma$ is an isomorphism (and it cannot be because $B$ is finite). Hence $d=1$.



    EDIT: Actually unless $A=0$ then no such sequence can exist to begin with since $alpha=0$ but $0 to A xrightarrow{alpha} mathbb{Z}^d$ implies it is injective.






    share|cite|improve this answer























    • You have a typo: you mean $d leq 1$ at the end of your first line.
      – user3482749
      Nov 25 '18 at 17:02










    • @user3482749 Yes, thank you.
      – freakish
      Nov 25 '18 at 17:02











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    3 Answers
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    3 Answers
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    1














    You can first conclude that $alpha = 0$: if not, $A neq 0$, and any non-zero element of $A$ would be sent by $alpha$ to a torsion element of $mathbb{Z}^d$, but no such exist. So $alpha = 0$, and since the sequence is exact, $A = 0$. Thus, we require an injective homomorphism $beta: mathbb{Z}^d to mathbb{Z}$.



    But if $dgeq 2$, taking $e_i$ to be the standard generators (all zeros with a one in the $i$th position), $beta(beta(e_1)e_2) = beta(e_1)beta(e_2) = beta(beta(e_2)e_1)$, so $beta$ cannot be an injection, so indeed, $d leq 1$ (and $d neq 0$ else $gamma$ would be an isomorphism).






    share|cite|improve this answer


























      1














      You can first conclude that $alpha = 0$: if not, $A neq 0$, and any non-zero element of $A$ would be sent by $alpha$ to a torsion element of $mathbb{Z}^d$, but no such exist. So $alpha = 0$, and since the sequence is exact, $A = 0$. Thus, we require an injective homomorphism $beta: mathbb{Z}^d to mathbb{Z}$.



      But if $dgeq 2$, taking $e_i$ to be the standard generators (all zeros with a one in the $i$th position), $beta(beta(e_1)e_2) = beta(e_1)beta(e_2) = beta(beta(e_2)e_1)$, so $beta$ cannot be an injection, so indeed, $d leq 1$ (and $d neq 0$ else $gamma$ would be an isomorphism).






      share|cite|improve this answer
























        1












        1








        1






        You can first conclude that $alpha = 0$: if not, $A neq 0$, and any non-zero element of $A$ would be sent by $alpha$ to a torsion element of $mathbb{Z}^d$, but no such exist. So $alpha = 0$, and since the sequence is exact, $A = 0$. Thus, we require an injective homomorphism $beta: mathbb{Z}^d to mathbb{Z}$.



        But if $dgeq 2$, taking $e_i$ to be the standard generators (all zeros with a one in the $i$th position), $beta(beta(e_1)e_2) = beta(e_1)beta(e_2) = beta(beta(e_2)e_1)$, so $beta$ cannot be an injection, so indeed, $d leq 1$ (and $d neq 0$ else $gamma$ would be an isomorphism).






        share|cite|improve this answer












        You can first conclude that $alpha = 0$: if not, $A neq 0$, and any non-zero element of $A$ would be sent by $alpha$ to a torsion element of $mathbb{Z}^d$, but no such exist. So $alpha = 0$, and since the sequence is exact, $A = 0$. Thus, we require an injective homomorphism $beta: mathbb{Z}^d to mathbb{Z}$.



        But if $dgeq 2$, taking $e_i$ to be the standard generators (all zeros with a one in the $i$th position), $beta(beta(e_1)e_2) = beta(e_1)beta(e_2) = beta(beta(e_2)e_1)$, so $beta$ cannot be an injection, so indeed, $d leq 1$ (and $d neq 0$ else $gamma$ would be an isomorphism).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 '18 at 17:01









        user3482749

        2,553414




        2,553414























            4














            Tensoring with the flat $mathbb{Z}$-module $mathbb{Q}$ gives an exact sequence $0 to mathbb{Q}^d to mathbb{Q} to 0$.






            share|cite|improve this answer



















            • 1




              This should be $0toBbb Q^dtoBbb Qto 0.$
              – Stahl
              Nov 25 '18 at 17:04










            • Yes, of course it should. Fixed.
              – anomaly
              Nov 25 '18 at 17:19
















            4














            Tensoring with the flat $mathbb{Z}$-module $mathbb{Q}$ gives an exact sequence $0 to mathbb{Q}^d to mathbb{Q} to 0$.






            share|cite|improve this answer



















            • 1




              This should be $0toBbb Q^dtoBbb Qto 0.$
              – Stahl
              Nov 25 '18 at 17:04










            • Yes, of course it should. Fixed.
              – anomaly
              Nov 25 '18 at 17:19














            4












            4








            4






            Tensoring with the flat $mathbb{Z}$-module $mathbb{Q}$ gives an exact sequence $0 to mathbb{Q}^d to mathbb{Q} to 0$.






            share|cite|improve this answer














            Tensoring with the flat $mathbb{Z}$-module $mathbb{Q}$ gives an exact sequence $0 to mathbb{Q}^d to mathbb{Q} to 0$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 25 '18 at 17:19

























            answered Nov 25 '18 at 17:03









            anomaly

            17.3k42663




            17.3k42663








            • 1




              This should be $0toBbb Q^dtoBbb Qto 0.$
              – Stahl
              Nov 25 '18 at 17:04










            • Yes, of course it should. Fixed.
              – anomaly
              Nov 25 '18 at 17:19














            • 1




              This should be $0toBbb Q^dtoBbb Qto 0.$
              – Stahl
              Nov 25 '18 at 17:04










            • Yes, of course it should. Fixed.
              – anomaly
              Nov 25 '18 at 17:19








            1




            1




            This should be $0toBbb Q^dtoBbb Qto 0.$
            – Stahl
            Nov 25 '18 at 17:04




            This should be $0toBbb Q^dtoBbb Qto 0.$
            – Stahl
            Nov 25 '18 at 17:04












            Yes, of course it should. Fixed.
            – anomaly
            Nov 25 '18 at 17:19




            Yes, of course it should. Fixed.
            – anomaly
            Nov 25 '18 at 17:19











            4














            Since $A$ is finite then there is no nontrivial homomorphism $Atomathbb{Z}^d$. Meaning $alpha=0$. This implies that $kerbeta=0$ and so $beta$ is injective. The case when $beta$ is injective is only possible when $dleq 1$.



            The case when $d=0$ is impossible because that would imply that $gamma$ is an isomorphism (and it cannot be because $B$ is finite). Hence $d=1$.



            EDIT: Actually unless $A=0$ then no such sequence can exist to begin with since $alpha=0$ but $0 to A xrightarrow{alpha} mathbb{Z}^d$ implies it is injective.






            share|cite|improve this answer























            • You have a typo: you mean $d leq 1$ at the end of your first line.
              – user3482749
              Nov 25 '18 at 17:02










            • @user3482749 Yes, thank you.
              – freakish
              Nov 25 '18 at 17:02
















            4














            Since $A$ is finite then there is no nontrivial homomorphism $Atomathbb{Z}^d$. Meaning $alpha=0$. This implies that $kerbeta=0$ and so $beta$ is injective. The case when $beta$ is injective is only possible when $dleq 1$.



            The case when $d=0$ is impossible because that would imply that $gamma$ is an isomorphism (and it cannot be because $B$ is finite). Hence $d=1$.



            EDIT: Actually unless $A=0$ then no such sequence can exist to begin with since $alpha=0$ but $0 to A xrightarrow{alpha} mathbb{Z}^d$ implies it is injective.






            share|cite|improve this answer























            • You have a typo: you mean $d leq 1$ at the end of your first line.
              – user3482749
              Nov 25 '18 at 17:02










            • @user3482749 Yes, thank you.
              – freakish
              Nov 25 '18 at 17:02














            4












            4








            4






            Since $A$ is finite then there is no nontrivial homomorphism $Atomathbb{Z}^d$. Meaning $alpha=0$. This implies that $kerbeta=0$ and so $beta$ is injective. The case when $beta$ is injective is only possible when $dleq 1$.



            The case when $d=0$ is impossible because that would imply that $gamma$ is an isomorphism (and it cannot be because $B$ is finite). Hence $d=1$.



            EDIT: Actually unless $A=0$ then no such sequence can exist to begin with since $alpha=0$ but $0 to A xrightarrow{alpha} mathbb{Z}^d$ implies it is injective.






            share|cite|improve this answer














            Since $A$ is finite then there is no nontrivial homomorphism $Atomathbb{Z}^d$. Meaning $alpha=0$. This implies that $kerbeta=0$ and so $beta$ is injective. The case when $beta$ is injective is only possible when $dleq 1$.



            The case when $d=0$ is impossible because that would imply that $gamma$ is an isomorphism (and it cannot be because $B$ is finite). Hence $d=1$.



            EDIT: Actually unless $A=0$ then no such sequence can exist to begin with since $alpha=0$ but $0 to A xrightarrow{alpha} mathbb{Z}^d$ implies it is injective.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 25 '18 at 17:51

























            answered Nov 25 '18 at 17:01









            freakish

            11.4k1629




            11.4k1629












            • You have a typo: you mean $d leq 1$ at the end of your first line.
              – user3482749
              Nov 25 '18 at 17:02










            • @user3482749 Yes, thank you.
              – freakish
              Nov 25 '18 at 17:02


















            • You have a typo: you mean $d leq 1$ at the end of your first line.
              – user3482749
              Nov 25 '18 at 17:02










            • @user3482749 Yes, thank you.
              – freakish
              Nov 25 '18 at 17:02
















            You have a typo: you mean $d leq 1$ at the end of your first line.
            – user3482749
            Nov 25 '18 at 17:02




            You have a typo: you mean $d leq 1$ at the end of your first line.
            – user3482749
            Nov 25 '18 at 17:02












            @user3482749 Yes, thank you.
            – freakish
            Nov 25 '18 at 17:02




            @user3482749 Yes, thank you.
            – freakish
            Nov 25 '18 at 17:02


















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