A subgroup in Group theory












2














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


















  • 2




    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.
    – Axion004
    yesterday


















2














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


















  • 2




    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.
    – Axion004
    yesterday
















2












2








2


1





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













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 yesterday









bladiebla

233




233








  • 2




    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.
    – Axion004
    yesterday
















  • 2




    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.
    – Axion004
    yesterday










2




2




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.
– Axion004
yesterday






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.
– Axion004
yesterday












3 Answers
3






active

oldest

votes


















4














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





















  • 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)
    – Ovi
    yesterday












  • Surely, the ''subgroup criterion''.
    – Wuestenfux
    yesterday










  • Yes ${}{}{}{}{}$
    – Ovi
    yesterday



















1














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










New contributor




pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.














  • 1




    I don't see what this has to do with the question.
    – Matt Samuel
    yesterday










  • 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.
    – pendermath
    yesterday












  • 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.
    – Matt Samuel
    yesterday










  • 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?"
    – Matt Samuel
    yesterday










  • I am not assuming that is a subset of a group. Maybe I missed the question.
    – pendermath
    yesterday



















0














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





















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    3 Answers
    3






    active

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    3 Answers
    3






    active

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    active

    oldest

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    active

    oldest

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    4














    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





















    • 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)
      – Ovi
      yesterday












    • Surely, the ''subgroup criterion''.
      – Wuestenfux
      yesterday










    • Yes ${}{}{}{}{}$
      – Ovi
      yesterday
















    4














    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





















    • 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)
      – Ovi
      yesterday












    • Surely, the ''subgroup criterion''.
      – Wuestenfux
      yesterday










    • Yes ${}{}{}{}{}$
      – Ovi
      yesterday














    4












    4








    4






    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












    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 yesterday









    Wuestenfux

    3,6861411




    3,6861411












    • 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)
      – Ovi
      yesterday












    • Surely, the ''subgroup criterion''.
      – Wuestenfux
      yesterday










    • Yes ${}{}{}{}{}$
      – Ovi
      yesterday


















    • 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)
      – Ovi
      yesterday












    • Surely, the ''subgroup criterion''.
      – Wuestenfux
      yesterday










    • Yes ${}{}{}{}{}$
      – Ovi
      yesterday
















    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)
    – Ovi
    yesterday






    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)
    – Ovi
    yesterday














    Surely, the ''subgroup criterion''.
    – Wuestenfux
    yesterday




    Surely, the ''subgroup criterion''.
    – Wuestenfux
    yesterday












    Yes ${}{}{}{}{}$
    – Ovi
    yesterday




    Yes ${}{}{}{}{}$
    – Ovi
    yesterday











    1














    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










    New contributor




    pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.














    • 1




      I don't see what this has to do with the question.
      – Matt Samuel
      yesterday










    • 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.
      – pendermath
      yesterday












    • 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.
      – Matt Samuel
      yesterday










    • 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?"
      – Matt Samuel
      yesterday










    • I am not assuming that is a subset of a group. Maybe I missed the question.
      – pendermath
      yesterday
















    1














    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










    New contributor




    pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.














    • 1




      I don't see what this has to do with the question.
      – Matt Samuel
      yesterday










    • 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.
      – pendermath
      yesterday












    • 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.
      – Matt Samuel
      yesterday










    • 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?"
      – Matt Samuel
      yesterday










    • I am not assuming that is a subset of a group. Maybe I missed the question.
      – pendermath
      yesterday














    1












    1








    1






    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










    New contributor




    pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    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










    New contributor




    pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    share|cite|improve this answer



    share|cite|improve this answer








    edited yesterday





















    New contributor




    pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.









    answered yesterday









    pendermath

    15310




    15310




    New contributor




    pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    New contributor





    pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    pendermath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.








    • 1




      I don't see what this has to do with the question.
      – Matt Samuel
      yesterday










    • 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.
      – pendermath
      yesterday












    • 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.
      – Matt Samuel
      yesterday










    • 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?"
      – Matt Samuel
      yesterday










    • I am not assuming that is a subset of a group. Maybe I missed the question.
      – pendermath
      yesterday














    • 1




      I don't see what this has to do with the question.
      – Matt Samuel
      yesterday










    • 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.
      – pendermath
      yesterday












    • 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.
      – Matt Samuel
      yesterday










    • 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?"
      – Matt Samuel
      yesterday










    • I am not assuming that is a subset of a group. Maybe I missed the question.
      – pendermath
      yesterday








    1




    1




    I don't see what this has to do with the question.
    – Matt Samuel
    yesterday




    I don't see what this has to do with the question.
    – Matt Samuel
    yesterday












    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.
    – pendermath
    yesterday






    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.
    – pendermath
    yesterday














    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.
    – Matt Samuel
    yesterday




    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.
    – Matt Samuel
    yesterday












    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?"
    – Matt Samuel
    yesterday




    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?"
    – Matt Samuel
    yesterday












    I am not assuming that is a subset of a group. Maybe I missed the question.
    – pendermath
    yesterday




    I am not assuming that is a subset of a group. Maybe I missed the question.
    – pendermath
    yesterday











    0














    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


























      0














      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
























        0












        0








        0






        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












        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 yesterday









        ncmathsadist

        42.3k259102




        42.3k259102






























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