Definition of subgroup generated by a set












4












$begingroup$


Yes, I've checked the other similarly titled questions (they seem more on par with abstract algebra for a grad level course). From the Gilbert and Gilbert text we have the following definition



$$langle Arangle ={xin G| x=a_1+a_2+cdots +a_n, quad a_iin Aquad or quad -a_iin A}$$



for an arbitrary $A$ which is a subset of the group $G$. I don't like this definition because it doesn't seem natural; specifically the $n$ is fixed (and comes from apparently nowhere). An exercise I'm currently working on is to show that if $H_1,H_2,ldots H_n$ are subgroups of an abelian group $G$, then $G=H_1+H_2+cdots + H_n$ if and only if $G$ is generated by $bigcup_{j=1}^n H_j$.



I don't want help with this second problem; however, I'm sharing it in hopes that it will clarify why I'm confused. From the definition of $langle A rangle $ it seems that $n$ is arbitrary, that is to say it doesn't "know" that $H$ is the sum of $n$ distinct subgroups. But if that's the case then when trying to the prove the second problem I don't know how to get started because I write something like the following:



If $gin G$ (and $G$ is generated by the union) then $g=a_1+cdots + a_k$ for some $kin mathbb{N}$ and each $a_iin bigcup_{j=1}^n H_j$. I suppose I'm doing this correctly, I just don't trust this definition -- how does $langle A rangle $ know to "go up to" $n$?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You're right that the $n$ in the definition of $<A>$ is arbitrary. It's fixed in the second part of the question. Consider thinking of the definition of $<A>$ as containing those elements $x$ that "for some $m$ are the sum of $m$ elements ..."
    $endgroup$
    – Ethan Bolker
    Nov 3 '18 at 1:08












  • $begingroup$
    The way I am interpreting the definition is " $langle Arangle$ consists of the elements in $G$ that can be expressed as a linear combination of the ones in $A$. " I don't think the $n$'s are related from your first and second questions (as you already figured out and has already been pointed out by Matt and Ethan)
    $endgroup$
    – WaveX
    Nov 3 '18 at 1:23
















4












$begingroup$


Yes, I've checked the other similarly titled questions (they seem more on par with abstract algebra for a grad level course). From the Gilbert and Gilbert text we have the following definition



$$langle Arangle ={xin G| x=a_1+a_2+cdots +a_n, quad a_iin Aquad or quad -a_iin A}$$



for an arbitrary $A$ which is a subset of the group $G$. I don't like this definition because it doesn't seem natural; specifically the $n$ is fixed (and comes from apparently nowhere). An exercise I'm currently working on is to show that if $H_1,H_2,ldots H_n$ are subgroups of an abelian group $G$, then $G=H_1+H_2+cdots + H_n$ if and only if $G$ is generated by $bigcup_{j=1}^n H_j$.



I don't want help with this second problem; however, I'm sharing it in hopes that it will clarify why I'm confused. From the definition of $langle A rangle $ it seems that $n$ is arbitrary, that is to say it doesn't "know" that $H$ is the sum of $n$ distinct subgroups. But if that's the case then when trying to the prove the second problem I don't know how to get started because I write something like the following:



If $gin G$ (and $G$ is generated by the union) then $g=a_1+cdots + a_k$ for some $kin mathbb{N}$ and each $a_iin bigcup_{j=1}^n H_j$. I suppose I'm doing this correctly, I just don't trust this definition -- how does $langle A rangle $ know to "go up to" $n$?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You're right that the $n$ in the definition of $<A>$ is arbitrary. It's fixed in the second part of the question. Consider thinking of the definition of $<A>$ as containing those elements $x$ that "for some $m$ are the sum of $m$ elements ..."
    $endgroup$
    – Ethan Bolker
    Nov 3 '18 at 1:08












  • $begingroup$
    The way I am interpreting the definition is " $langle Arangle$ consists of the elements in $G$ that can be expressed as a linear combination of the ones in $A$. " I don't think the $n$'s are related from your first and second questions (as you already figured out and has already been pointed out by Matt and Ethan)
    $endgroup$
    – WaveX
    Nov 3 '18 at 1:23














4












4








4





$begingroup$


Yes, I've checked the other similarly titled questions (they seem more on par with abstract algebra for a grad level course). From the Gilbert and Gilbert text we have the following definition



$$langle Arangle ={xin G| x=a_1+a_2+cdots +a_n, quad a_iin Aquad or quad -a_iin A}$$



for an arbitrary $A$ which is a subset of the group $G$. I don't like this definition because it doesn't seem natural; specifically the $n$ is fixed (and comes from apparently nowhere). An exercise I'm currently working on is to show that if $H_1,H_2,ldots H_n$ are subgroups of an abelian group $G$, then $G=H_1+H_2+cdots + H_n$ if and only if $G$ is generated by $bigcup_{j=1}^n H_j$.



I don't want help with this second problem; however, I'm sharing it in hopes that it will clarify why I'm confused. From the definition of $langle A rangle $ it seems that $n$ is arbitrary, that is to say it doesn't "know" that $H$ is the sum of $n$ distinct subgroups. But if that's the case then when trying to the prove the second problem I don't know how to get started because I write something like the following:



If $gin G$ (and $G$ is generated by the union) then $g=a_1+cdots + a_k$ for some $kin mathbb{N}$ and each $a_iin bigcup_{j=1}^n H_j$. I suppose I'm doing this correctly, I just don't trust this definition -- how does $langle A rangle $ know to "go up to" $n$?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$




Yes, I've checked the other similarly titled questions (they seem more on par with abstract algebra for a grad level course). From the Gilbert and Gilbert text we have the following definition



$$langle Arangle ={xin G| x=a_1+a_2+cdots +a_n, quad a_iin Aquad or quad -a_iin A}$$



for an arbitrary $A$ which is a subset of the group $G$. I don't like this definition because it doesn't seem natural; specifically the $n$ is fixed (and comes from apparently nowhere). An exercise I'm currently working on is to show that if $H_1,H_2,ldots H_n$ are subgroups of an abelian group $G$, then $G=H_1+H_2+cdots + H_n$ if and only if $G$ is generated by $bigcup_{j=1}^n H_j$.



I don't want help with this second problem; however, I'm sharing it in hopes that it will clarify why I'm confused. From the definition of $langle A rangle $ it seems that $n$ is arbitrary, that is to say it doesn't "know" that $H$ is the sum of $n$ distinct subgroups. But if that's the case then when trying to the prove the second problem I don't know how to get started because I write something like the following:



If $gin G$ (and $G$ is generated by the union) then $g=a_1+cdots + a_k$ for some $kin mathbb{N}$ and each $a_iin bigcup_{j=1}^n H_j$. I suppose I'm doing this correctly, I just don't trust this definition -- how does $langle A rangle $ know to "go up to" $n$?






abstract-algebra group-theory




abstract-algebra group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 19:24









Matt Samuel

38k63666




38k63666










asked Nov 3 '18 at 1:03









ClclstdntClclstdnt

468314




468314








  • 2




    $begingroup$
    You're right that the $n$ in the definition of $<A>$ is arbitrary. It's fixed in the second part of the question. Consider thinking of the definition of $<A>$ as containing those elements $x$ that "for some $m$ are the sum of $m$ elements ..."
    $endgroup$
    – Ethan Bolker
    Nov 3 '18 at 1:08












  • $begingroup$
    The way I am interpreting the definition is " $langle Arangle$ consists of the elements in $G$ that can be expressed as a linear combination of the ones in $A$. " I don't think the $n$'s are related from your first and second questions (as you already figured out and has already been pointed out by Matt and Ethan)
    $endgroup$
    – WaveX
    Nov 3 '18 at 1:23














  • 2




    $begingroup$
    You're right that the $n$ in the definition of $<A>$ is arbitrary. It's fixed in the second part of the question. Consider thinking of the definition of $<A>$ as containing those elements $x$ that "for some $m$ are the sum of $m$ elements ..."
    $endgroup$
    – Ethan Bolker
    Nov 3 '18 at 1:08












  • $begingroup$
    The way I am interpreting the definition is " $langle Arangle$ consists of the elements in $G$ that can be expressed as a linear combination of the ones in $A$. " I don't think the $n$'s are related from your first and second questions (as you already figured out and has already been pointed out by Matt and Ethan)
    $endgroup$
    – WaveX
    Nov 3 '18 at 1:23








2




2




$begingroup$
You're right that the $n$ in the definition of $<A>$ is arbitrary. It's fixed in the second part of the question. Consider thinking of the definition of $<A>$ as containing those elements $x$ that "for some $m$ are the sum of $m$ elements ..."
$endgroup$
– Ethan Bolker
Nov 3 '18 at 1:08






$begingroup$
You're right that the $n$ in the definition of $<A>$ is arbitrary. It's fixed in the second part of the question. Consider thinking of the definition of $<A>$ as containing those elements $x$ that "for some $m$ are the sum of $m$ elements ..."
$endgroup$
– Ethan Bolker
Nov 3 '18 at 1:08














$begingroup$
The way I am interpreting the definition is " $langle Arangle$ consists of the elements in $G$ that can be expressed as a linear combination of the ones in $A$. " I don't think the $n$'s are related from your first and second questions (as you already figured out and has already been pointed out by Matt and Ethan)
$endgroup$
– WaveX
Nov 3 '18 at 1:23




$begingroup$
The way I am interpreting the definition is " $langle Arangle$ consists of the elements in $G$ that can be expressed as a linear combination of the ones in $A$. " I don't think the $n$'s are related from your first and second questions (as you already figured out and has already been pointed out by Matt and Ethan)
$endgroup$
– WaveX
Nov 3 '18 at 1:23










1 Answer
1






active

oldest

votes


















3












$begingroup$

It's an odd definition, but it's correct as long as $n$ is not fixed. $n$ must be allowed to be any positive integer or else ${1}$ does not generate $Bbb Z$ (which it does).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Righto.... yeah, just in the book it seems that $n$ is fixed because of the unusual juxtaposition. This is how it is written in the book: [centered equation above follow by] "It is left as an exercise to prove that if $H_1,H_2,ldots, H_n$ are subgroups of an abelian group $G$ then $G=H_1+H_2+cdots H_n$ if and only if $G$ is generated by $bigcup_{j=1}^n H_j$." I guess it appeared fixed given the exercise that $immediately$ followed it. Poor wording. Thank you for you clarification!
    $endgroup$
    – Clclstdnt
    Nov 3 '18 at 1:17












  • $begingroup$
    @Clcl No problem.
    $endgroup$
    – Matt Samuel
    Nov 3 '18 at 1:19











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1 Answer
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1 Answer
1






active

oldest

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active

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3












$begingroup$

It's an odd definition, but it's correct as long as $n$ is not fixed. $n$ must be allowed to be any positive integer or else ${1}$ does not generate $Bbb Z$ (which it does).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Righto.... yeah, just in the book it seems that $n$ is fixed because of the unusual juxtaposition. This is how it is written in the book: [centered equation above follow by] "It is left as an exercise to prove that if $H_1,H_2,ldots, H_n$ are subgroups of an abelian group $G$ then $G=H_1+H_2+cdots H_n$ if and only if $G$ is generated by $bigcup_{j=1}^n H_j$." I guess it appeared fixed given the exercise that $immediately$ followed it. Poor wording. Thank you for you clarification!
    $endgroup$
    – Clclstdnt
    Nov 3 '18 at 1:17












  • $begingroup$
    @Clcl No problem.
    $endgroup$
    – Matt Samuel
    Nov 3 '18 at 1:19
















3












$begingroup$

It's an odd definition, but it's correct as long as $n$ is not fixed. $n$ must be allowed to be any positive integer or else ${1}$ does not generate $Bbb Z$ (which it does).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Righto.... yeah, just in the book it seems that $n$ is fixed because of the unusual juxtaposition. This is how it is written in the book: [centered equation above follow by] "It is left as an exercise to prove that if $H_1,H_2,ldots, H_n$ are subgroups of an abelian group $G$ then $G=H_1+H_2+cdots H_n$ if and only if $G$ is generated by $bigcup_{j=1}^n H_j$." I guess it appeared fixed given the exercise that $immediately$ followed it. Poor wording. Thank you for you clarification!
    $endgroup$
    – Clclstdnt
    Nov 3 '18 at 1:17












  • $begingroup$
    @Clcl No problem.
    $endgroup$
    – Matt Samuel
    Nov 3 '18 at 1:19














3












3








3





$begingroup$

It's an odd definition, but it's correct as long as $n$ is not fixed. $n$ must be allowed to be any positive integer or else ${1}$ does not generate $Bbb Z$ (which it does).






share|cite|improve this answer









$endgroup$



It's an odd definition, but it's correct as long as $n$ is not fixed. $n$ must be allowed to be any positive integer or else ${1}$ does not generate $Bbb Z$ (which it does).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 3 '18 at 1:07









Matt SamuelMatt Samuel

38k63666




38k63666












  • $begingroup$
    Righto.... yeah, just in the book it seems that $n$ is fixed because of the unusual juxtaposition. This is how it is written in the book: [centered equation above follow by] "It is left as an exercise to prove that if $H_1,H_2,ldots, H_n$ are subgroups of an abelian group $G$ then $G=H_1+H_2+cdots H_n$ if and only if $G$ is generated by $bigcup_{j=1}^n H_j$." I guess it appeared fixed given the exercise that $immediately$ followed it. Poor wording. Thank you for you clarification!
    $endgroup$
    – Clclstdnt
    Nov 3 '18 at 1:17












  • $begingroup$
    @Clcl No problem.
    $endgroup$
    – Matt Samuel
    Nov 3 '18 at 1:19


















  • $begingroup$
    Righto.... yeah, just in the book it seems that $n$ is fixed because of the unusual juxtaposition. This is how it is written in the book: [centered equation above follow by] "It is left as an exercise to prove that if $H_1,H_2,ldots, H_n$ are subgroups of an abelian group $G$ then $G=H_1+H_2+cdots H_n$ if and only if $G$ is generated by $bigcup_{j=1}^n H_j$." I guess it appeared fixed given the exercise that $immediately$ followed it. Poor wording. Thank you for you clarification!
    $endgroup$
    – Clclstdnt
    Nov 3 '18 at 1:17












  • $begingroup$
    @Clcl No problem.
    $endgroup$
    – Matt Samuel
    Nov 3 '18 at 1:19
















$begingroup$
Righto.... yeah, just in the book it seems that $n$ is fixed because of the unusual juxtaposition. This is how it is written in the book: [centered equation above follow by] "It is left as an exercise to prove that if $H_1,H_2,ldots, H_n$ are subgroups of an abelian group $G$ then $G=H_1+H_2+cdots H_n$ if and only if $G$ is generated by $bigcup_{j=1}^n H_j$." I guess it appeared fixed given the exercise that $immediately$ followed it. Poor wording. Thank you for you clarification!
$endgroup$
– Clclstdnt
Nov 3 '18 at 1:17






$begingroup$
Righto.... yeah, just in the book it seems that $n$ is fixed because of the unusual juxtaposition. This is how it is written in the book: [centered equation above follow by] "It is left as an exercise to prove that if $H_1,H_2,ldots, H_n$ are subgroups of an abelian group $G$ then $G=H_1+H_2+cdots H_n$ if and only if $G$ is generated by $bigcup_{j=1}^n H_j$." I guess it appeared fixed given the exercise that $immediately$ followed it. Poor wording. Thank you for you clarification!
$endgroup$
– Clclstdnt
Nov 3 '18 at 1:17














$begingroup$
@Clcl No problem.
$endgroup$
– Matt Samuel
Nov 3 '18 at 1:19




$begingroup$
@Clcl No problem.
$endgroup$
– Matt Samuel
Nov 3 '18 at 1:19


















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