A subgroup in Group theory












2












$begingroup$


I am currently studying group theory, and its a quite new concept for me.
When learning about subgroups, I bumped into something.



In a problem where one wants to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B?



i.e.



If we know for sure that B is a group,
and we show that some set A satisfies all axioms of being a subgroup of a group B,
can we conclude that A is a group?






group-theory







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Suppose set B is a group and A is a subset of B. If A is a subgroup, then A is also a group. So, yes.
    $endgroup$
    – Axion004
    Jan 5 at 14:29


















2












$begingroup$


I am currently studying group theory, and its a quite new concept for me.
When learning about subgroups, I bumped into something.



In a problem where one wants to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B?



i.e.



If we know for sure that B is a group,
and we show that some set A satisfies all axioms of being a subgroup of a group B,
can we conclude that A is a group?






group-theory







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Suppose set B is a group and A is a subset of B. If A is a subgroup, then A is also a group. So, yes.
    $endgroup$
    – Axion004
    Jan 5 at 14:29
















2












2








2


1



$begingroup$


I am currently studying group theory, and its a quite new concept for me.
When learning about subgroups, I bumped into something.



In a problem where one wants to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B?



i.e.



If we know for sure that B is a group,
and we show that some set A satisfies all axioms of being a subgroup of a group B,
can we conclude that A is a group?






group-theory







share|cite|improve this question









$endgroup$




I am currently studying group theory, and its a quite new concept for me.
When learning about subgroups, I bumped into something.



In a problem where one wants to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B?



i.e.



If we know for sure that B is a group,
and we show that some set A satisfies all axioms of being a subgroup of a group B,
can we conclude that A is a group?






group-theory




group-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 5 at 14:26









bladieblabladiebla

233




233








  • 2




    $begingroup$
    Suppose set B is a group and A is a subset of B. If A is a subgroup, then A is also a group. So, yes.
    $endgroup$
    – Axion004
    Jan 5 at 14:29
















  • 2




    $begingroup$
    Suppose set B is a group and A is a subset of B. If A is a subgroup, then A is also a group. So, yes.
    $endgroup$
    – Axion004
    Jan 5 at 14:29










2




2




$begingroup$
Suppose set B is a group and A is a subset of B. If A is a subgroup, then A is also a group. So, yes.
$endgroup$
– Axion004
Jan 5 at 14:29






$begingroup$
Suppose set B is a group and A is a subset of B. If A is a subgroup, then A is also a group. So, yes.
$endgroup$
– Axion004
Jan 5 at 14:29












3 Answers
3






active

oldest

votes


















4












$begingroup$

It can be done easier. Let $G$ be a group and $U$ be a nonempty subset of $G$.
Then $U$ forms a subgroup of $G$ iff (1) for all $g,hin U$, $ghin U$, and (2) for each $gin U$, $g^{-1}in U$.



Then $U$ forms a subgroup of $G$ without checking the axioms point by point.



For instance, the unit element $ein G$ lies in $U$, since $U$ is nonempty and so $U$ has an element $g$. Then $g^{-1}in U$ by (2) and so by (1) $e= gg^{-1}in U$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I'm sure you know this but for the OP: We can also collapse it down to just one condition: $U$ is a subgroup of $G$ $iff$ $forall (a in U) forall (b in U) ab^{-1} in U$ (this is often said verbally as "the subset is closed under subtraction", because in additive notation $ab^{-1}$ is denoted as $a-b$. And when you will study rings, you will switch to the additive notation for groups)
    $endgroup$
    – Ovi
    Jan 5 at 14:41












  • $begingroup$
    Surely, the ''subgroup criterion''.
    $endgroup$
    – Wuestenfux
    Jan 5 at 14:42










  • $begingroup$
    Yes ${}{}{}{}{}$
    $endgroup$
    – Ovi
    Jan 5 at 14:43



















1












$begingroup$

No, as associativity may not be satisfied within the set. Consider, for instance, the structure $(S,*)$, with $S={1,a,b}$, and the following definition table for $*$:
$$
begin{matrix}
* & bf1 & bf{a} & bf{b} \
bf1 & 1 & a & b \
bf{a} & a & 1 & b \
bf{b} & b & a & 1
end{matrix}
$$
This satisfies the conditions for being a subgroup (it is closed under * and inverses, and it is nonempty), except for the fact that it is not in a group. But $(S,*)$ is not a group, because associativity fails: $1=a(ba) neq (ab)a=a$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I don't see what this has to do with the question.
    $endgroup$
    – Matt Samuel
    Jan 5 at 19:16










  • $begingroup$
    The question reads: "...to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B...". I provide a counterexample to show that it is not enough. The structure (S,*) satisfies the conditions for a subgroup, but is not a group.
    $endgroup$
    – pendermath
    Jan 5 at 19:37












  • $begingroup$
    But if it's already a subset of a group with the same operation, it's automatically associative. We're not talking about an arbitrary operation.
    $endgroup$
    – Matt Samuel
    Jan 5 at 19:38










  • $begingroup$
    To quote the question, "If we know for sure that B is a group, and we show that some set A satisfies all axioms of being a subgroup of a group B, can we conclude that A is a group?"
    $endgroup$
    – Matt Samuel
    Jan 5 at 19:39










  • $begingroup$
    I am not assuming that is a subset of a group. Maybe I missed the question.
    $endgroup$
    – pendermath
    Jan 5 at 19:39



















0












$begingroup$

If $G$ is a group and $Hsubseteq G$, then $H$ is a subgroup of $G$ iff it is nonvoid and closed under multiplication and inversion. You do not need to verify all of the axioms in this case because much is inherited from $G$.






share|cite|improve this answer









$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    It can be done easier. Let $G$ be a group and $U$ be a nonempty subset of $G$.
    Then $U$ forms a subgroup of $G$ iff (1) for all $g,hin U$, $ghin U$, and (2) for each $gin U$, $g^{-1}in U$.



    Then $U$ forms a subgroup of $G$ without checking the axioms point by point.



    For instance, the unit element $ein G$ lies in $U$, since $U$ is nonempty and so $U$ has an element $g$. Then $g^{-1}in U$ by (2) and so by (1) $e= gg^{-1}in U$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I'm sure you know this but for the OP: We can also collapse it down to just one condition: $U$ is a subgroup of $G$ $iff$ $forall (a in U) forall (b in U) ab^{-1} in U$ (this is often said verbally as "the subset is closed under subtraction", because in additive notation $ab^{-1}$ is denoted as $a-b$. And when you will study rings, you will switch to the additive notation for groups)
      $endgroup$
      – Ovi
      Jan 5 at 14:41












    • $begingroup$
      Surely, the ''subgroup criterion''.
      $endgroup$
      – Wuestenfux
      Jan 5 at 14:42










    • $begingroup$
      Yes ${}{}{}{}{}$
      $endgroup$
      – Ovi
      Jan 5 at 14:43
















    4












    $begingroup$

    It can be done easier. Let $G$ be a group and $U$ be a nonempty subset of $G$.
    Then $U$ forms a subgroup of $G$ iff (1) for all $g,hin U$, $ghin U$, and (2) for each $gin U$, $g^{-1}in U$.



    Then $U$ forms a subgroup of $G$ without checking the axioms point by point.



    For instance, the unit element $ein G$ lies in $U$, since $U$ is nonempty and so $U$ has an element $g$. Then $g^{-1}in U$ by (2) and so by (1) $e= gg^{-1}in U$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I'm sure you know this but for the OP: We can also collapse it down to just one condition: $U$ is a subgroup of $G$ $iff$ $forall (a in U) forall (b in U) ab^{-1} in U$ (this is often said verbally as "the subset is closed under subtraction", because in additive notation $ab^{-1}$ is denoted as $a-b$. And when you will study rings, you will switch to the additive notation for groups)
      $endgroup$
      – Ovi
      Jan 5 at 14:41












    • $begingroup$
      Surely, the ''subgroup criterion''.
      $endgroup$
      – Wuestenfux
      Jan 5 at 14:42










    • $begingroup$
      Yes ${}{}{}{}{}$
      $endgroup$
      – Ovi
      Jan 5 at 14:43














    4












    4








    4





    $begingroup$

    It can be done easier. Let $G$ be a group and $U$ be a nonempty subset of $G$.
    Then $U$ forms a subgroup of $G$ iff (1) for all $g,hin U$, $ghin U$, and (2) for each $gin U$, $g^{-1}in U$.



    Then $U$ forms a subgroup of $G$ without checking the axioms point by point.



    For instance, the unit element $ein G$ lies in $U$, since $U$ is nonempty and so $U$ has an element $g$. Then $g^{-1}in U$ by (2) and so by (1) $e= gg^{-1}in U$.






    share|cite|improve this answer









    $endgroup$



    It can be done easier. Let $G$ be a group and $U$ be a nonempty subset of $G$.
    Then $U$ forms a subgroup of $G$ iff (1) for all $g,hin U$, $ghin U$, and (2) for each $gin U$, $g^{-1}in U$.



    Then $U$ forms a subgroup of $G$ without checking the axioms point by point.



    For instance, the unit element $ein G$ lies in $U$, since $U$ is nonempty and so $U$ has an element $g$. Then $g^{-1}in U$ by (2) and so by (1) $e= gg^{-1}in U$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 5 at 14:31









    WuestenfuxWuestenfux

    4,2281413




    4,2281413












    • $begingroup$
      I'm sure you know this but for the OP: We can also collapse it down to just one condition: $U$ is a subgroup of $G$ $iff$ $forall (a in U) forall (b in U) ab^{-1} in U$ (this is often said verbally as "the subset is closed under subtraction", because in additive notation $ab^{-1}$ is denoted as $a-b$. And when you will study rings, you will switch to the additive notation for groups)
      $endgroup$
      – Ovi
      Jan 5 at 14:41












    • $begingroup$
      Surely, the ''subgroup criterion''.
      $endgroup$
      – Wuestenfux
      Jan 5 at 14:42










    • $begingroup$
      Yes ${}{}{}{}{}$
      $endgroup$
      – Ovi
      Jan 5 at 14:43


















    • $begingroup$
      I'm sure you know this but for the OP: We can also collapse it down to just one condition: $U$ is a subgroup of $G$ $iff$ $forall (a in U) forall (b in U) ab^{-1} in U$ (this is often said verbally as "the subset is closed under subtraction", because in additive notation $ab^{-1}$ is denoted as $a-b$. And when you will study rings, you will switch to the additive notation for groups)
      $endgroup$
      – Ovi
      Jan 5 at 14:41












    • $begingroup$
      Surely, the ''subgroup criterion''.
      $endgroup$
      – Wuestenfux
      Jan 5 at 14:42










    • $begingroup$
      Yes ${}{}{}{}{}$
      $endgroup$
      – Ovi
      Jan 5 at 14:43
















    $begingroup$
    I'm sure you know this but for the OP: We can also collapse it down to just one condition: $U$ is a subgroup of $G$ $iff$ $forall (a in U) forall (b in U) ab^{-1} in U$ (this is often said verbally as "the subset is closed under subtraction", because in additive notation $ab^{-1}$ is denoted as $a-b$. And when you will study rings, you will switch to the additive notation for groups)
    $endgroup$
    – Ovi
    Jan 5 at 14:41






    $begingroup$
    I'm sure you know this but for the OP: We can also collapse it down to just one condition: $U$ is a subgroup of $G$ $iff$ $forall (a in U) forall (b in U) ab^{-1} in U$ (this is often said verbally as "the subset is closed under subtraction", because in additive notation $ab^{-1}$ is denoted as $a-b$. And when you will study rings, you will switch to the additive notation for groups)
    $endgroup$
    – Ovi
    Jan 5 at 14:41














    $begingroup$
    Surely, the ''subgroup criterion''.
    $endgroup$
    – Wuestenfux
    Jan 5 at 14:42




    $begingroup$
    Surely, the ''subgroup criterion''.
    $endgroup$
    – Wuestenfux
    Jan 5 at 14:42












    $begingroup$
    Yes ${}{}{}{}{}$
    $endgroup$
    – Ovi
    Jan 5 at 14:43




    $begingroup$
    Yes ${}{}{}{}{}$
    $endgroup$
    – Ovi
    Jan 5 at 14:43











    1












    $begingroup$

    No, as associativity may not be satisfied within the set. Consider, for instance, the structure $(S,*)$, with $S={1,a,b}$, and the following definition table for $*$:
    $$
    begin{matrix}
    * & bf1 & bf{a} & bf{b} \
    bf1 & 1 & a & b \
    bf{a} & a & 1 & b \
    bf{b} & b & a & 1
    end{matrix}
    $$
    This satisfies the conditions for being a subgroup (it is closed under * and inverses, and it is nonempty), except for the fact that it is not in a group. But $(S,*)$ is not a group, because associativity fails: $1=a(ba) neq (ab)a=a$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      I don't see what this has to do with the question.
      $endgroup$
      – Matt Samuel
      Jan 5 at 19:16










    • $begingroup$
      The question reads: "...to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B...". I provide a counterexample to show that it is not enough. The structure (S,*) satisfies the conditions for a subgroup, but is not a group.
      $endgroup$
      – pendermath
      Jan 5 at 19:37












    • $begingroup$
      But if it's already a subset of a group with the same operation, it's automatically associative. We're not talking about an arbitrary operation.
      $endgroup$
      – Matt Samuel
      Jan 5 at 19:38










    • $begingroup$
      To quote the question, "If we know for sure that B is a group, and we show that some set A satisfies all axioms of being a subgroup of a group B, can we conclude that A is a group?"
      $endgroup$
      – Matt Samuel
      Jan 5 at 19:39










    • $begingroup$
      I am not assuming that is a subset of a group. Maybe I missed the question.
      $endgroup$
      – pendermath
      Jan 5 at 19:39
















    1












    $begingroup$

    No, as associativity may not be satisfied within the set. Consider, for instance, the structure $(S,*)$, with $S={1,a,b}$, and the following definition table for $*$:
    $$
    begin{matrix}
    * & bf1 & bf{a} & bf{b} \
    bf1 & 1 & a & b \
    bf{a} & a & 1 & b \
    bf{b} & b & a & 1
    end{matrix}
    $$
    This satisfies the conditions for being a subgroup (it is closed under * and inverses, and it is nonempty), except for the fact that it is not in a group. But $(S,*)$ is not a group, because associativity fails: $1=a(ba) neq (ab)a=a$






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      I don't see what this has to do with the question.
      $endgroup$
      – Matt Samuel
      Jan 5 at 19:16










    • $begingroup$
      The question reads: "...to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B...". I provide a counterexample to show that it is not enough. The structure (S,*) satisfies the conditions for a subgroup, but is not a group.
      $endgroup$
      – pendermath
      Jan 5 at 19:37












    • $begingroup$
      But if it's already a subset of a group with the same operation, it's automatically associative. We're not talking about an arbitrary operation.
      $endgroup$
      – Matt Samuel
      Jan 5 at 19:38










    • $begingroup$
      To quote the question, "If we know for sure that B is a group, and we show that some set A satisfies all axioms of being a subgroup of a group B, can we conclude that A is a group?"
      $endgroup$
      – Matt Samuel
      Jan 5 at 19:39










    • $begingroup$
      I am not assuming that is a subset of a group. Maybe I missed the question.
      $endgroup$
      – pendermath
      Jan 5 at 19:39














    1












    1








    1





    $begingroup$

    No, as associativity may not be satisfied within the set. Consider, for instance, the structure $(S,*)$, with $S={1,a,b}$, and the following definition table for $*$:
    $$
    begin{matrix}
    * & bf1 & bf{a} & bf{b} \
    bf1 & 1 & a & b \
    bf{a} & a & 1 & b \
    bf{b} & b & a & 1
    end{matrix}
    $$
    This satisfies the conditions for being a subgroup (it is closed under * and inverses, and it is nonempty), except for the fact that it is not in a group. But $(S,*)$ is not a group, because associativity fails: $1=a(ba) neq (ab)a=a$






    share|cite|improve this answer











    $endgroup$



    No, as associativity may not be satisfied within the set. Consider, for instance, the structure $(S,*)$, with $S={1,a,b}$, and the following definition table for $*$:
    $$
    begin{matrix}
    * & bf1 & bf{a} & bf{b} \
    bf1 & 1 & a & b \
    bf{a} & a & 1 & b \
    bf{b} & b & a & 1
    end{matrix}
    $$
    This satisfies the conditions for being a subgroup (it is closed under * and inverses, and it is nonempty), except for the fact that it is not in a group. But $(S,*)$ is not a group, because associativity fails: $1=a(ba) neq (ab)a=a$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 5 at 15:19

























    answered Jan 5 at 14:31









    pendermathpendermath

    52910




    52910








    • 1




      $begingroup$
      I don't see what this has to do with the question.
      $endgroup$
      – Matt Samuel
      Jan 5 at 19:16










    • $begingroup$
      The question reads: "...to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B...". I provide a counterexample to show that it is not enough. The structure (S,*) satisfies the conditions for a subgroup, but is not a group.
      $endgroup$
      – pendermath
      Jan 5 at 19:37












    • $begingroup$
      But if it's already a subset of a group with the same operation, it's automatically associative. We're not talking about an arbitrary operation.
      $endgroup$
      – Matt Samuel
      Jan 5 at 19:38










    • $begingroup$
      To quote the question, "If we know for sure that B is a group, and we show that some set A satisfies all axioms of being a subgroup of a group B, can we conclude that A is a group?"
      $endgroup$
      – Matt Samuel
      Jan 5 at 19:39










    • $begingroup$
      I am not assuming that is a subset of a group. Maybe I missed the question.
      $endgroup$
      – pendermath
      Jan 5 at 19:39














    • 1




      $begingroup$
      I don't see what this has to do with the question.
      $endgroup$
      – Matt Samuel
      Jan 5 at 19:16










    • $begingroup$
      The question reads: "...to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B...". I provide a counterexample to show that it is not enough. The structure (S,*) satisfies the conditions for a subgroup, but is not a group.
      $endgroup$
      – pendermath
      Jan 5 at 19:37












    • $begingroup$
      But if it's already a subset of a group with the same operation, it's automatically associative. We're not talking about an arbitrary operation.
      $endgroup$
      – Matt Samuel
      Jan 5 at 19:38










    • $begingroup$
      To quote the question, "If we know for sure that B is a group, and we show that some set A satisfies all axioms of being a subgroup of a group B, can we conclude that A is a group?"
      $endgroup$
      – Matt Samuel
      Jan 5 at 19:39










    • $begingroup$
      I am not assuming that is a subset of a group. Maybe I missed the question.
      $endgroup$
      – pendermath
      Jan 5 at 19:39








    1




    1




    $begingroup$
    I don't see what this has to do with the question.
    $endgroup$
    – Matt Samuel
    Jan 5 at 19:16




    $begingroup$
    I don't see what this has to do with the question.
    $endgroup$
    – Matt Samuel
    Jan 5 at 19:16












    $begingroup$
    The question reads: "...to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B...". I provide a counterexample to show that it is not enough. The structure (S,*) satisfies the conditions for a subgroup, but is not a group.
    $endgroup$
    – pendermath
    Jan 5 at 19:37






    $begingroup$
    The question reads: "...to show that some set A is a group, is it sufficient enough to show that that set (A) satisfies all axioms of being a subgroup of a group B...". I provide a counterexample to show that it is not enough. The structure (S,*) satisfies the conditions for a subgroup, but is not a group.
    $endgroup$
    – pendermath
    Jan 5 at 19:37














    $begingroup$
    But if it's already a subset of a group with the same operation, it's automatically associative. We're not talking about an arbitrary operation.
    $endgroup$
    – Matt Samuel
    Jan 5 at 19:38




    $begingroup$
    But if it's already a subset of a group with the same operation, it's automatically associative. We're not talking about an arbitrary operation.
    $endgroup$
    – Matt Samuel
    Jan 5 at 19:38












    $begingroup$
    To quote the question, "If we know for sure that B is a group, and we show that some set A satisfies all axioms of being a subgroup of a group B, can we conclude that A is a group?"
    $endgroup$
    – Matt Samuel
    Jan 5 at 19:39




    $begingroup$
    To quote the question, "If we know for sure that B is a group, and we show that some set A satisfies all axioms of being a subgroup of a group B, can we conclude that A is a group?"
    $endgroup$
    – Matt Samuel
    Jan 5 at 19:39












    $begingroup$
    I am not assuming that is a subset of a group. Maybe I missed the question.
    $endgroup$
    – pendermath
    Jan 5 at 19:39




    $begingroup$
    I am not assuming that is a subset of a group. Maybe I missed the question.
    $endgroup$
    – pendermath
    Jan 5 at 19:39











    0












    $begingroup$

    If $G$ is a group and $Hsubseteq G$, then $H$ is a subgroup of $G$ iff it is nonvoid and closed under multiplication and inversion. You do not need to verify all of the axioms in this case because much is inherited from $G$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If $G$ is a group and $Hsubseteq G$, then $H$ is a subgroup of $G$ iff it is nonvoid and closed under multiplication and inversion. You do not need to verify all of the axioms in this case because much is inherited from $G$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If $G$ is a group and $Hsubseteq G$, then $H$ is a subgroup of $G$ iff it is nonvoid and closed under multiplication and inversion. You do not need to verify all of the axioms in this case because much is inherited from $G$.






        share|cite|improve this answer









        $endgroup$



        If $G$ is a group and $Hsubseteq G$, then $H$ is a subgroup of $G$ iff it is nonvoid and closed under multiplication and inversion. You do not need to verify all of the axioms in this case because much is inherited from $G$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 14:29









        ncmathsadistncmathsadist

        42.6k260103




        42.6k260103






























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