Given certain elements of a proper subgroup $H$ of $mathbbZ$, determine $H$
Clash Royale CLAN TAG#URR8PPP
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Question: Suppose that $H$ is a proper subgroup of $mathbbZ$ under addition and that $H$ contains $18$, $30$, and $40$. Determine $H$.
Can we determine $H$ without knowing the order of $mathbbZ$? Please don't tell the answer; I just need a hint.
abstract-algebra group-theory
edited Sep 9 at 12:07
mrs
58.5k750143
asked Mar 27 '13 at 9:17
TLE
5071718
add a comment |Â
up vote
1
down vote
favorite
Question: Suppose that $H$ is a proper subgroup of $mathbbZ$ under addition and that $H$ contains $18$, $30$, and $40$. Determine $H$.
Can we determine $H$ without knowing the order of $mathbbZ$? Please don't tell the answer; I just need a hint.
abstract-algebra group-theory
edited Sep 9 at 12:07
mrs
58.5k750143
asked Mar 27 '13 at 9:17
TLE
5071718
I am slightly confused by the first sentence in the second paragraph. We already know the order the $mathbbZ$ (it is infinite). Do you perhaps mean the index of $H$ in $mathbbZ$?
– user1729
Mar 27 '13 at 10:54
The wording of the question isn't very good. We can't actually determine $H$ just by knowing that it contains certain elements. I presume what is meant is that $H$ contains only those elements and the elements that it has to contain as a consequence of containing those ones (i.e. those elements generate $H$), but it should say that.
– Tara B
Mar 27 '13 at 12:05
@TaraB: In this question, $H$ is a proper subgroup containing the element $2$, and so it can be determined. But in general, yes, you are right.
– user1729
Mar 27 '13 at 12:47
@user1729: Ah, you are right. I had missed that the word 'proper' gives the necessary extra information.
– Tara B
Mar 27 '13 at 12:49
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Question: Suppose that $H$ is a proper subgroup of $mathbbZ$ under addition and that $H$ contains $18$, $30$, and $40$. Determine $H$.
Can we determine $H$ without knowing the order of $mathbbZ$? Please don't tell the answer; I just need a hint.
abstract-algebra group-theory
edited Sep 9 at 12:07
mrs
58.5k750143
asked Mar 27 '13 at 9:17
TLE
5071718
Question: Suppose that $H$ is a proper subgroup of $mathbbZ$ under addition and that $H$ contains $18$, $30$, and $40$. Determine $H$.
Can we determine $H$ without knowing the order of $mathbbZ$? Please don't tell the answer; I just need a hint.
abstract-algebra group-theory
abstract-algebra group-theory
edited Sep 9 at 12:07
mrs
58.5k750143
asked Mar 27 '13 at 9:17
TLE
5071718
edited Sep 9 at 12:07
mrs
58.5k750143
asked Mar 27 '13 at 9:17
TLE
5071718
edited Sep 9 at 12:07
mrs
58.5k750143
edited Sep 9 at 12:07
mrs
58.5k750143
edited Sep 9 at 12:07
mrs
58.5k750143
58.5k750143
asked Mar 27 '13 at 9:17
TLE
5071718
asked Mar 27 '13 at 9:17
TLE
5071718
asked Mar 27 '13 at 9:17
TLE
5071718
5071718
I am slightly confused by the first sentence in the second paragraph. We already know the order the $mathbbZ$ (it is infinite). Do you perhaps mean the index of $H$ in $mathbbZ$?
– user1729
Mar 27 '13 at 10:54
The wording of the question isn't very good. We can't actually determine $H$ just by knowing that it contains certain elements. I presume what is meant is that $H$ contains only those elements and the elements that it has to contain as a consequence of containing those ones (i.e. those elements generate $H$), but it should say that.
– Tara B
Mar 27 '13 at 12:05
@TaraB: In this question, $H$ is a proper subgroup containing the element $2$, and so it can be determined. But in general, yes, you are right.
– user1729
Mar 27 '13 at 12:47
@user1729: Ah, you are right. I had missed that the word 'proper' gives the necessary extra information.
– Tara B
Mar 27 '13 at 12:49
add a comment |Â
I am slightly confused by the first sentence in the second paragraph. We already know the order the $mathbbZ$ (it is infinite). Do you perhaps mean the index of $H$ in $mathbbZ$?
– user1729
Mar 27 '13 at 10:54
The wording of the question isn't very good. We can't actually determine $H$ just by knowing that it contains certain elements. I presume what is meant is that $H$ contains only those elements and the elements that it has to contain as a consequence of containing those ones (i.e. those elements generate $H$), but it should say that.
– Tara B
Mar 27 '13 at 12:05
@TaraB: In this question, $H$ is a proper subgroup containing the element $2$, and so it can be determined. But in general, yes, you are right.
– user1729
Mar 27 '13 at 12:47
@user1729: Ah, you are right. I had missed that the word 'proper' gives the necessary extra information.
– Tara B
Mar 27 '13 at 12:49
I am slightly confused by the first sentence in the second paragraph. We already know the order the $mathbbZ$ (it is infinite). Do you perhaps mean the index of $H$ in $mathbbZ$?
– user1729
Mar 27 '13 at 10:54
I am slightly confused by the first sentence in the second paragraph. We already know the order the $mathbbZ$ (it is infinite). Do you perhaps mean the index of $H$ in $mathbbZ$?
– user1729
Mar 27 '13 at 10:54
The wording of the question isn't very good. We can't actually determine $H$ just by knowing that it contains certain elements. I presume what is meant is that $H$ contains only those elements and the elements that it has to contain as a consequence of containing those ones (i.e. those elements generate $H$), but it should say that.
– Tara B
Mar 27 '13 at 12:05
The wording of the question isn't very good. We can't actually determine $H$ just by knowing that it contains certain elements. I presume what is meant is that $H$ contains only those elements and the elements that it has to contain as a consequence of containing those ones (i.e. those elements generate $H$), but it should say that.
– Tara B
Mar 27 '13 at 12:05
@TaraB: In this question, $H$ is a proper subgroup containing the element $2$, and so it can be determined. But in general, yes, you are right.
– user1729
Mar 27 '13 at 12:47
@TaraB: In this question, $H$ is a proper subgroup containing the element $2$, and so it can be determined. But in general, yes, you are right.
– user1729
Mar 27 '13 at 12:47
@user1729: Ah, you are right. I had missed that the word 'proper' gives the necessary extra information.
– Tara B
Mar 27 '13 at 12:49
@user1729: Ah, you are right. I had missed that the word 'proper' gives the necessary extra information.
– Tara B
Mar 27 '13 at 12:49
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
Hint $1$: The greatest common devisor of $18$, $30$ and $40$ is which number and why is this relevant?
Hint $2$: To find $H$, think about what happens if $operatornamegcd(18, 30, 40)$ is prime and if it is composite.
answered Mar 27 '13 at 9:34
user1729
17k64083
Thanks for the comment. I was not on here to help the OP. +1
– mrs
Mar 27 '13 at 13:47
@ResidentDementor How does it matter whether gcd is prime or composite though? Just that there could be multiple proper subgroups in case gcd is composite?
– yathish
Dec 28 '17 at 5:50
1
@yathish Yes, exactly. There may be multiple proper subgroups, so $H$ cannot be determined. (I consider the word "determined" to mean "determined uniquely", like in a deterministic automata. I think this definition is standard.)
– user1729
Dec 28 '17 at 11:53
add a comment |Â
up vote
3
down vote
If $Hleqmathbb Z$ so for some $nneq 1,0$, $~~H=nmathbb Z$ (Any proper subgroup of $mathbb Z$ should be of this form).
Now think of this: $$18=nk,~ 30=nk',~ 40=nk'',~~ k,k',k''inmathbb Z$$
edited Dec 28 '17 at 7:14
yathish
16510
answered Mar 27 '13 at 9:23
mrs
58.5k750143
what does $H=nZ$ means???Order of $Z$ is divisible by order of $H$
– TLE
Mar 27 '13 at 13:08
$nmathbbZ$ is shorthand for the set $nk; kinmathbbZ$, which is the subgroup of $mathbbZ$ generated by the element $n$ (why?). Note that all subgroups of a cyclic group are themselves cyclic, so $H=nmathbbZ=langle nrangle$ for some $1neq ninmathbbZ$.
– user1729
Mar 27 '13 at 13:22
$ddotsmile$
– amWhy
Mar 27 '13 at 14:16
add a comment |Â
up vote
1
down vote
As an example of the sort of reasoning you might use, you might think that since $18 in H$ and $30in H$, we know that $18 + 18 - 30 = 6 in H$. Using reasoning like this, you can arrive at a unique answer.
answered Mar 27 '13 at 9:29
davidlowryduda♦
72.8k6113244
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Hint $1$: The greatest common devisor of $18$, $30$ and $40$ is which number and why is this relevant?
Hint $2$: To find $H$, think about what happens if $operatornamegcd(18, 30, 40)$ is prime and if it is composite.
answered Mar 27 '13 at 9:34
user1729
17k64083
Thanks for the comment. I was not on here to help the OP. +1
– mrs
Mar 27 '13 at 13:47
@ResidentDementor How does it matter whether gcd is prime or composite though? Just that there could be multiple proper subgroups in case gcd is composite?
– yathish
Dec 28 '17 at 5:50
1
@yathish Yes, exactly. There may be multiple proper subgroups, so $H$ cannot be determined. (I consider the word "determined" to mean "determined uniquely", like in a deterministic automata. I think this definition is standard.)
– user1729
Dec 28 '17 at 11:53
add a comment |Â
up vote
3
down vote
Hint $1$: The greatest common devisor of $18$, $30$ and $40$ is which number and why is this relevant?
Hint $2$: To find $H$, think about what happens if $operatornamegcd(18, 30, 40)$ is prime and if it is composite.
answered Mar 27 '13 at 9:34
user1729
17k64083
Thanks for the comment. I was not on here to help the OP. +1
– mrs
Mar 27 '13 at 13:47
@ResidentDementor How does it matter whether gcd is prime or composite though? Just that there could be multiple proper subgroups in case gcd is composite?
– yathish
Dec 28 '17 at 5:50
1
@yathish Yes, exactly. There may be multiple proper subgroups, so $H$ cannot be determined. (I consider the word "determined" to mean "determined uniquely", like in a deterministic automata. I think this definition is standard.)
– user1729
Dec 28 '17 at 11:53
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Hint $1$: The greatest common devisor of $18$, $30$ and $40$ is which number and why is this relevant?
Hint $2$: To find $H$, think about what happens if $operatornamegcd(18, 30, 40)$ is prime and if it is composite.
answered Mar 27 '13 at 9:34
user1729
17k64083
Hint $1$: The greatest common devisor of $18$, $30$ and $40$ is which number and why is this relevant?
Hint $2$: To find $H$, think about what happens if $operatornamegcd(18, 30, 40)$ is prime and if it is composite.
answered Mar 27 '13 at 9:34
user1729
17k64083
answered Mar 27 '13 at 9:34
user1729
17k64083
answered Mar 27 '13 at 9:34
user1729
17k64083
answered Mar 27 '13 at 9:34
user1729
17k64083
17k64083
Thanks for the comment. I was not on here to help the OP. +1
– mrs
Mar 27 '13 at 13:47
@ResidentDementor How does it matter whether gcd is prime or composite though? Just that there could be multiple proper subgroups in case gcd is composite?
– yathish
Dec 28 '17 at 5:50
1
@yathish Yes, exactly. There may be multiple proper subgroups, so $H$ cannot be determined. (I consider the word "determined" to mean "determined uniquely", like in a deterministic automata. I think this definition is standard.)
– user1729
Dec 28 '17 at 11:53
add a comment |Â
Thanks for the comment. I was not on here to help the OP. +1
– mrs
Mar 27 '13 at 13:47
@ResidentDementor How does it matter whether gcd is prime or composite though? Just that there could be multiple proper subgroups in case gcd is composite?
– yathish
Dec 28 '17 at 5:50
1
@yathish Yes, exactly. There may be multiple proper subgroups, so $H$ cannot be determined. (I consider the word "determined" to mean "determined uniquely", like in a deterministic automata. I think this definition is standard.)
– user1729
Dec 28 '17 at 11:53
Thanks for the comment. I was not on here to help the OP. +1
– mrs
Mar 27 '13 at 13:47
Thanks for the comment. I was not on here to help the OP. +1
– mrs
Mar 27 '13 at 13:47
@ResidentDementor How does it matter whether gcd is prime or composite though? Just that there could be multiple proper subgroups in case gcd is composite?
– yathish
Dec 28 '17 at 5:50
@ResidentDementor How does it matter whether gcd is prime or composite though? Just that there could be multiple proper subgroups in case gcd is composite?
– yathish
Dec 28 '17 at 5:50
1
1
@yathish Yes, exactly. There may be multiple proper subgroups, so $H$ cannot be determined. (I consider the word "determined" to mean "determined uniquely", like in a deterministic automata. I think this definition is standard.)
– user1729
Dec 28 '17 at 11:53
@yathish Yes, exactly. There may be multiple proper subgroups, so $H$ cannot be determined. (I consider the word "determined" to mean "determined uniquely", like in a deterministic automata. I think this definition is standard.)
– user1729
Dec 28 '17 at 11:53
add a comment |Â
up vote
3
down vote
If $Hleqmathbb Z$ so for some $nneq 1,0$, $~~H=nmathbb Z$ (Any proper subgroup of $mathbb Z$ should be of this form).
Now think of this: $$18=nk,~ 30=nk',~ 40=nk'',~~ k,k',k''inmathbb Z$$
edited Dec 28 '17 at 7:14
yathish
16510
answered Mar 27 '13 at 9:23
mrs
58.5k750143
what does $H=nZ$ means???Order of $Z$ is divisible by order of $H$
– TLE
Mar 27 '13 at 13:08
$nmathbbZ$ is shorthand for the set $nk; kinmathbbZ$, which is the subgroup of $mathbbZ$ generated by the element $n$ (why?). Note that all subgroups of a cyclic group are themselves cyclic, so $H=nmathbbZ=langle nrangle$ for some $1neq ninmathbbZ$.
– user1729
Mar 27 '13 at 13:22
$ddotsmile$
– amWhy
Mar 27 '13 at 14:16
add a comment |Â
up vote
3
down vote
If $Hleqmathbb Z$ so for some $nneq 1,0$, $~~H=nmathbb Z$ (Any proper subgroup of $mathbb Z$ should be of this form).
Now think of this: $$18=nk,~ 30=nk',~ 40=nk'',~~ k,k',k''inmathbb Z$$
edited Dec 28 '17 at 7:14
yathish
16510
answered Mar 27 '13 at 9:23
mrs
58.5k750143
what does $H=nZ$ means???Order of $Z$ is divisible by order of $H$
– TLE
Mar 27 '13 at 13:08
$nmathbbZ$ is shorthand for the set $nk; kinmathbbZ$, which is the subgroup of $mathbbZ$ generated by the element $n$ (why?). Note that all subgroups of a cyclic group are themselves cyclic, so $H=nmathbbZ=langle nrangle$ for some $1neq ninmathbbZ$.
– user1729
Mar 27 '13 at 13:22
$ddotsmile$
– amWhy
Mar 27 '13 at 14:16
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If $Hleqmathbb Z$ so for some $nneq 1,0$, $~~H=nmathbb Z$ (Any proper subgroup of $mathbb Z$ should be of this form).
Now think of this: $$18=nk,~ 30=nk',~ 40=nk'',~~ k,k',k''inmathbb Z$$
edited Dec 28 '17 at 7:14
yathish
16510
answered Mar 27 '13 at 9:23
mrs
58.5k750143
If $Hleqmathbb Z$ so for some $nneq 1,0$, $~~H=nmathbb Z$ (Any proper subgroup of $mathbb Z$ should be of this form).
Now think of this: $$18=nk,~ 30=nk',~ 40=nk'',~~ k,k',k''inmathbb Z$$
edited Dec 28 '17 at 7:14
yathish
16510
answered Mar 27 '13 at 9:23
mrs
58.5k750143
edited Dec 28 '17 at 7:14
yathish
16510
edited Dec 28 '17 at 7:14
yathish
16510
edited Dec 28 '17 at 7:14
yathish
16510
16510
answered Mar 27 '13 at 9:23
mrs
58.5k750143
answered Mar 27 '13 at 9:23
mrs
58.5k750143
answered Mar 27 '13 at 9:23
mrs
58.5k750143
58.5k750143
what does $H=nZ$ means???Order of $Z$ is divisible by order of $H$
– TLE
Mar 27 '13 at 13:08
$nmathbbZ$ is shorthand for the set $nk; kinmathbbZ$, which is the subgroup of $mathbbZ$ generated by the element $n$ (why?). Note that all subgroups of a cyclic group are themselves cyclic, so $H=nmathbbZ=langle nrangle$ for some $1neq ninmathbbZ$.
– user1729
Mar 27 '13 at 13:22
$ddotsmile$
– amWhy
Mar 27 '13 at 14:16
add a comment |Â
what does $H=nZ$ means???Order of $Z$ is divisible by order of $H$
– TLE
Mar 27 '13 at 13:08
$nmathbbZ$ is shorthand for the set $nk; kinmathbbZ$, which is the subgroup of $mathbbZ$ generated by the element $n$ (why?). Note that all subgroups of a cyclic group are themselves cyclic, so $H=nmathbbZ=langle nrangle$ for some $1neq ninmathbbZ$.
– user1729
Mar 27 '13 at 13:22
$ddotsmile$
– amWhy
Mar 27 '13 at 14:16
what does $H=nZ$ means???Order of $Z$ is divisible by order of $H$
– TLE
Mar 27 '13 at 13:08
what does $H=nZ$ means???Order of $Z$ is divisible by order of $H$
– TLE
Mar 27 '13 at 13:08
$nmathbbZ$ is shorthand for the set $nk; kinmathbbZ$, which is the subgroup of $mathbbZ$ generated by the element $n$ (why?). Note that all subgroups of a cyclic group are themselves cyclic, so $H=nmathbbZ=langle nrangle$ for some $1neq ninmathbbZ$.
– user1729
Mar 27 '13 at 13:22
$nmathbbZ$ is shorthand for the set $nk; kinmathbbZ$, which is the subgroup of $mathbbZ$ generated by the element $n$ (why?). Note that all subgroups of a cyclic group are themselves cyclic, so $H=nmathbbZ=langle nrangle$ for some $1neq ninmathbbZ$.
– user1729
Mar 27 '13 at 13:22
$ddotsmile$
– amWhy
Mar 27 '13 at 14:16
$ddotsmile$
– amWhy
Mar 27 '13 at 14:16
add a comment |Â
up vote
1
down vote
As an example of the sort of reasoning you might use, you might think that since $18 in H$ and $30in H$, we know that $18 + 18 - 30 = 6 in H$. Using reasoning like this, you can arrive at a unique answer.
answered Mar 27 '13 at 9:29
davidlowryduda♦
72.8k6113244
add a comment |Â
up vote
1
down vote
As an example of the sort of reasoning you might use, you might think that since $18 in H$ and $30in H$, we know that $18 + 18 - 30 = 6 in H$. Using reasoning like this, you can arrive at a unique answer.
answered Mar 27 '13 at 9:29
davidlowryduda♦
72.8k6113244
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As an example of the sort of reasoning you might use, you might think that since $18 in H$ and $30in H$, we know that $18 + 18 - 30 = 6 in H$. Using reasoning like this, you can arrive at a unique answer.
answered Mar 27 '13 at 9:29
davidlowryduda♦
72.8k6113244
As an example of the sort of reasoning you might use, you might think that since $18 in H$ and $30in H$, we know that $18 + 18 - 30 = 6 in H$. Using reasoning like this, you can arrive at a unique answer.
answered Mar 27 '13 at 9:29
davidlowryduda♦
72.8k6113244
answered Mar 27 '13 at 9:29
davidlowryduda♦
72.8k6113244
answered Mar 27 '13 at 9:29
davidlowryduda♦
72.8k6113244
answered Mar 27 '13 at 9:29
davidlowryduda♦
72.8k6113244
72.8k6113244
add a comment |Â
add a comment |Â
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I am slightly confused by the first sentence in the second paragraph. We already know the order the $mathbbZ$ (it is infinite). Do you perhaps mean the index of $H$ in $mathbbZ$?
– user1729
Mar 27 '13 at 10:54
The wording of the question isn't very good. We can't actually determine $H$ just by knowing that it contains certain elements. I presume what is meant is that $H$ contains only those elements and the elements that it has to contain as a consequence of containing those ones (i.e. those elements generate $H$), but it should say that.
– Tara B
Mar 27 '13 at 12:05
@TaraB: In this question, $H$ is a proper subgroup containing the element $2$, and so it can be determined. But in general, yes, you are right.
– user1729
Mar 27 '13 at 12:47
@user1729: Ah, you are right. I had missed that the word 'proper' gives the necessary extra information.
– Tara B
Mar 27 '13 at 12:49