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Show that $ninmathbbN$ is square-free


Prove that a square matrix commutes with its inverseAre there homomorphisms of group algebras that don't come from a group homomorphism?Free Group generated by S is actually generated by S.Proving that the free group on two generators is the coproduct $mathbbZ*mathbbZ$ in $textbfGrp$Commuting elements in $rm PSL(2,mathbbZ)$ are powers of some elementI'm looking to construct an isomorphism to show: $mathbbZ_rs^timescong mathbbZ_r^times times mathbbZ_s^times$Abelian group between two free abelian groups of the same (finite) rank.Show that ($R^times$, $cdot$) is a group.proof verification of an elementary result about a subgroup of a free group using a technical methodRelation between abelian and free abelian group













0












$begingroup$



Show that $ninmathbbN$ is square-free if and only if there is a subset different from the null set, $LsubseteqmathbbZ_n$, with the property that summation and multiplication from $mathbb Z_n$ induces on $L$ a group structure.




I think that a proof ad-absurdum would provide a result, because the reciprocal seems easy to handle as I know that for a group $Z_k$, the invertible elements must be coprime with $k$ and I find so a ,,square-free" group. Any help to continue it?










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    The group structure induced on $L $ is additive or multiplicative?
    $endgroup$
    – Fabio Lucchini
    Mar 15 at 19:03















0












$begingroup$



Show that $ninmathbbN$ is square-free if and only if there is a subset different from the null set, $LsubseteqmathbbZ_n$, with the property that summation and multiplication from $mathbb Z_n$ induces on $L$ a group structure.




I think that a proof ad-absurdum would provide a result, because the reciprocal seems easy to handle as I know that for a group $Z_k$, the invertible elements must be coprime with $k$ and I find so a ,,square-free" group. Any help to continue it?










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    The group structure induced on $L $ is additive or multiplicative?
    $endgroup$
    – Fabio Lucchini
    Mar 15 at 19:03













0












0








0


1



$begingroup$



Show that $ninmathbbN$ is square-free if and only if there is a subset different from the null set, $LsubseteqmathbbZ_n$, with the property that summation and multiplication from $mathbb Z_n$ induces on $L$ a group structure.




I think that a proof ad-absurdum would provide a result, because the reciprocal seems easy to handle as I know that for a group $Z_k$, the invertible elements must be coprime with $k$ and I find so a ,,square-free" group. Any help to continue it?










share|cite|improve this question











$endgroup$





Show that $ninmathbbN$ is square-free if and only if there is a subset different from the null set, $LsubseteqmathbbZ_n$, with the property that summation and multiplication from $mathbb Z_n$ induces on $L$ a group structure.




I think that a proof ad-absurdum would provide a result, because the reciprocal seems easy to handle as I know that for a group $Z_k$, the invertible elements must be coprime with $k$ and I find so a ,,square-free" group. Any help to continue it?







abstract-algebra group-theory field-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 15 at 21:34









user26857

39.4k124183




39.4k124183










asked Mar 15 at 17:20







user651692














  • 1




    $begingroup$
    The group structure induced on $L $ is additive or multiplicative?
    $endgroup$
    – Fabio Lucchini
    Mar 15 at 19:03












  • 1




    $begingroup$
    The group structure induced on $L $ is additive or multiplicative?
    $endgroup$
    – Fabio Lucchini
    Mar 15 at 19:03







1




1




$begingroup$
The group structure induced on $L $ is additive or multiplicative?
$endgroup$
– Fabio Lucchini
Mar 15 at 19:03




$begingroup$
The group structure induced on $L $ is additive or multiplicative?
$endgroup$
– Fabio Lucchini
Mar 15 at 19:03










1 Answer
1






active

oldest

votes


















2












$begingroup$

This question is completely incoherent. I have no idea what the intended question is supposed to be. Here are all the possible interpretations I can form of it, and why they don't work.




$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a group under addition.




Fails because $BbbZ/(9)$ contains $(3)$.




$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a group under multiplication.




Fails because $BbbZ/(9)$ contains $langle 4rangle$, which has order 3.




$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a group under addition and multiplication.




We can always take $L=0$ to satisfy this, and if we require $Lne 0$, then
this still makes no sense, because in order for $L$ to be a group under addition, it must contain $0$, but then if $xin L$, $xne 0$, we have $xcdot 0 = 0$, so $L$ cannot be a group under multiplication.




$n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a field under addition and multiplication ($L$ is a group under addition, and $Lsetminus 0$ is a group under multiplication).




Let $xin L$, $xne 0$. Then $x$ must be invertible in $BbbZ/(n)$, which means that it is relatively prime to $BbbZ/(n)$. Hence $x$ generates the additive group of $BbbZ/(n)$, which implies that $L=BbbZ/(n)$. Then $BbbZ/(n)$ is a field if and only if $n$ is prime. Thus this doesn't work either.






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    2












    $begingroup$

    This question is completely incoherent. I have no idea what the intended question is supposed to be. Here are all the possible interpretations I can form of it, and why they don't work.




    $n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a group under addition.




    Fails because $BbbZ/(9)$ contains $(3)$.




    $n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a group under multiplication.




    Fails because $BbbZ/(9)$ contains $langle 4rangle$, which has order 3.




    $n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a group under addition and multiplication.




    We can always take $L=0$ to satisfy this, and if we require $Lne 0$, then
    this still makes no sense, because in order for $L$ to be a group under addition, it must contain $0$, but then if $xin L$, $xne 0$, we have $xcdot 0 = 0$, so $L$ cannot be a group under multiplication.




    $n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a field under addition and multiplication ($L$ is a group under addition, and $Lsetminus 0$ is a group under multiplication).




    Let $xin L$, $xne 0$. Then $x$ must be invertible in $BbbZ/(n)$, which means that it is relatively prime to $BbbZ/(n)$. Hence $x$ generates the additive group of $BbbZ/(n)$, which implies that $L=BbbZ/(n)$. Then $BbbZ/(n)$ is a field if and only if $n$ is prime. Thus this doesn't work either.






    share|cite|improve this answer









    $endgroup$

















      2












      $begingroup$

      This question is completely incoherent. I have no idea what the intended question is supposed to be. Here are all the possible interpretations I can form of it, and why they don't work.




      $n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a group under addition.




      Fails because $BbbZ/(9)$ contains $(3)$.




      $n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a group under multiplication.




      Fails because $BbbZ/(9)$ contains $langle 4rangle$, which has order 3.




      $n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a group under addition and multiplication.




      We can always take $L=0$ to satisfy this, and if we require $Lne 0$, then
      this still makes no sense, because in order for $L$ to be a group under addition, it must contain $0$, but then if $xin L$, $xne 0$, we have $xcdot 0 = 0$, so $L$ cannot be a group under multiplication.




      $n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a field under addition and multiplication ($L$ is a group under addition, and $Lsetminus 0$ is a group under multiplication).




      Let $xin L$, $xne 0$. Then $x$ must be invertible in $BbbZ/(n)$, which means that it is relatively prime to $BbbZ/(n)$. Hence $x$ generates the additive group of $BbbZ/(n)$, which implies that $L=BbbZ/(n)$. Then $BbbZ/(n)$ is a field if and only if $n$ is prime. Thus this doesn't work either.






      share|cite|improve this answer









      $endgroup$















        2












        2








        2





        $begingroup$

        This question is completely incoherent. I have no idea what the intended question is supposed to be. Here are all the possible interpretations I can form of it, and why they don't work.




        $n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a group under addition.




        Fails because $BbbZ/(9)$ contains $(3)$.




        $n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a group under multiplication.




        Fails because $BbbZ/(9)$ contains $langle 4rangle$, which has order 3.




        $n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a group under addition and multiplication.




        We can always take $L=0$ to satisfy this, and if we require $Lne 0$, then
        this still makes no sense, because in order for $L$ to be a group under addition, it must contain $0$, but then if $xin L$, $xne 0$, we have $xcdot 0 = 0$, so $L$ cannot be a group under multiplication.




        $n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a field under addition and multiplication ($L$ is a group under addition, and $Lsetminus 0$ is a group under multiplication).




        Let $xin L$, $xne 0$. Then $x$ must be invertible in $BbbZ/(n)$, which means that it is relatively prime to $BbbZ/(n)$. Hence $x$ generates the additive group of $BbbZ/(n)$, which implies that $L=BbbZ/(n)$. Then $BbbZ/(n)$ is a field if and only if $n$ is prime. Thus this doesn't work either.






        share|cite|improve this answer









        $endgroup$



        This question is completely incoherent. I have no idea what the intended question is supposed to be. Here are all the possible interpretations I can form of it, and why they don't work.




        $n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a group under addition.




        Fails because $BbbZ/(9)$ contains $(3)$.




        $n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a group under multiplication.




        Fails because $BbbZ/(9)$ contains $langle 4rangle$, which has order 3.




        $n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a group under addition and multiplication.




        We can always take $L=0$ to satisfy this, and if we require $Lne 0$, then
        this still makes no sense, because in order for $L$ to be a group under addition, it must contain $0$, but then if $xin L$, $xne 0$, we have $xcdot 0 = 0$, so $L$ cannot be a group under multiplication.




        $n$ is square free if and only if there exists $varnothing subsetneq Lsubseteq BbbZ/(n)$ such that $L$ is a field under addition and multiplication ($L$ is a group under addition, and $Lsetminus 0$ is a group under multiplication).




        Let $xin L$, $xne 0$. Then $x$ must be invertible in $BbbZ/(n)$, which means that it is relatively prime to $BbbZ/(n)$. Hence $x$ generates the additive group of $BbbZ/(n)$, which implies that $L=BbbZ/(n)$. Then $BbbZ/(n)$ is a field if and only if $n$ is prime. Thus this doesn't work either.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 15 at 20:06









        jgonjgon

        15.7k32143




        15.7k32143



























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