Mixing
Is the additive group of rationals quotiented out by the integers isomorphic to the additive group of rationals?
Clash Royale CLAN TAG#URR8PPP
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I'm quite new to group theory so please do help me along.
There are some parts of the solution I do not understand.
1) Why is the addition of n terms of (Z+q) defined as equal to Z? Shouldn't it be defined as n(Z+q)?
2) Why is the order of Z+q in Q/Z claimed to be finite? Isn't the order of an element of a group defined to be the smallest number of times an element of a group needs to be composed with itself to get the identity element back again? In the case of (Z+q) it doesn't go back to the identity element again even if you define the addition of (Z+q) n times to be equal to Z.
Thanks in advance. I'm quite puzzled by this question.
EDIT: Thanks for all of your amazing help. Turns out I had the wrong interpretation of what Z/Q represented.
group-theory
asked Jul 28 at 14:42
Yip Jung Hon
18011
add a comment |Â
up vote
0
down vote
favorite
I'm quite new to group theory so please do help me along.
There are some parts of the solution I do not understand.
1) Why is the addition of n terms of (Z+q) defined as equal to Z? Shouldn't it be defined as n(Z+q)?
2) Why is the order of Z+q in Q/Z claimed to be finite? Isn't the order of an element of a group defined to be the smallest number of times an element of a group needs to be composed with itself to get the identity element back again? In the case of (Z+q) it doesn't go back to the identity element again even if you define the addition of (Z+q) n times to be equal to Z.
Thanks in advance. I'm quite puzzled by this question.
EDIT: Thanks for all of your amazing help. Turns out I had the wrong interpretation of what Z/Q represented.
group-theory
asked Jul 28 at 14:42
Yip Jung Hon
18011
Do you understand why the coset $H$ is zero in the quotient group $G/H$?
– Noah Schweber
Jul 28 at 15:04
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm quite new to group theory so please do help me along.
There are some parts of the solution I do not understand.
1) Why is the addition of n terms of (Z+q) defined as equal to Z? Shouldn't it be defined as n(Z+q)?
2) Why is the order of Z+q in Q/Z claimed to be finite? Isn't the order of an element of a group defined to be the smallest number of times an element of a group needs to be composed with itself to get the identity element back again? In the case of (Z+q) it doesn't go back to the identity element again even if you define the addition of (Z+q) n times to be equal to Z.
Thanks in advance. I'm quite puzzled by this question.
EDIT: Thanks for all of your amazing help. Turns out I had the wrong interpretation of what Z/Q represented.
group-theory
asked Jul 28 at 14:42
Yip Jung Hon
18011
I'm quite new to group theory so please do help me along.
There are some parts of the solution I do not understand.
1) Why is the addition of n terms of (Z+q) defined as equal to Z? Shouldn't it be defined as n(Z+q)?
2) Why is the order of Z+q in Q/Z claimed to be finite? Isn't the order of an element of a group defined to be the smallest number of times an element of a group needs to be composed with itself to get the identity element back again? In the case of (Z+q) it doesn't go back to the identity element again even if you define the addition of (Z+q) n times to be equal to Z.
Thanks in advance. I'm quite puzzled by this question.
EDIT: Thanks for all of your amazing help. Turns out I had the wrong interpretation of what Z/Q represented.
group-theory
asked Jul 28 at 14:42
Yip Jung Hon
18011
edited Jul 28 at 15:22
asked Jul 28 at 14:42
Yip Jung Hon
18011
asked Jul 28 at 14:42
Yip Jung Hon
18011
asked Jul 28 at 14:42
Yip Jung Hon
18011
18011
Do you understand why the coset $H$ is zero in the quotient group $G/H$?
– Noah Schweber
Jul 28 at 15:04
add a comment |Â
Do you understand why the coset $H$ is zero in the quotient group $G/H$?
– Noah Schweber
Jul 28 at 15:04
Do you understand why the coset $H$ is zero in the quotient group $G/H$?
– Noah Schweber
Jul 28 at 15:04
Do you understand why the coset $H$ is zero in the quotient group $G/H$?
– Noah Schweber
Jul 28 at 15:04
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
4
down vote
accepted
Remember that the elements of $mathbbQ/mathbbZ$ are right cosets of $mathbbZ$ which partitions $mathbbQ$. So $mathbbQ/mathbbZ$ gives an equivalence relation on $mathbbQ$ with
beginalign*
q sim p iff mathbbZ + p = mathbbZ + q iff p- q in mathbbZ.
endalign*
Since for any $n in mathbbZ$ we have $n - 0 in mathbbZ$ and so the identity element in $mathbbQ/mathbbZ$ is $mathbbZ + 0 = mathbbZ + n = mathbbZ$. If $q = m/n in mathbbQ$ then
beginalign*
sum_i=1^n(mathbbZ + m/n) = mathbbZ + (sum_i=1^n m/n) = mathbbZ + m = mathbbZ
endalign*
since $m in mathbbZ$. Hence $mathbbZ + q$ has finite order in $mathbbQ/mathbbZ$.
Now ask yourself if any element in $mathbbQ$ has this property.
answered Jul 28 at 15:01
matt stokes
1047
add a comment |Â
up vote
1
down vote
It may help to first consider an intuitive solution to the problem:
Suppose I take a rational $pover q$. Then when I add it to itself $q$-many times, I get an integer (namely $p$). Now this is (probably) nonzero in $mathbbQ$, but think about what role the integers serve in $mathbbQ/mathbbZ$: "modding out" a group $G$ (in this case, $mathbbQ$) by a normal subgroup $H$ (in this case, $mathbbZ$) basically means "making every element of $H$ become zero." So we have the following idea (which isn't quite right - see below - but is on the right track): $$mboxIn $mathbbQ/mathbbZ$, $pover q$ added to itself $q$-many times gives zero.$$
Put another way, the informal idea above suggests that $pover q$ has order at most $q$ in $mathbbQ/mathbbZ$." Now this isn't quite right: $pover q$ isn't literally an element of $mathbbQ/mathbbZ$ in the first place (elements of a quotient group are cosets, not elements, of the original group). But when you go back through the argument above and make it rigorous, you get exactly the solution in the book.
answered Jul 28 at 15:02
Noah Schweber
110k9139260
add a comment |Â
up vote
1
down vote
The following may help the intuition about the group $G:=mathbb Q/mathbb Z$. The map
$$psi:quad mathbb Qto S^1,qquad tmapsto e^2pi i t$$
maps rationals $x$, $yinmathbb Q$ that differ by an integer to the same point of $S^1$, hence maps all representants of an element $[x]in G$ to the same point of $S^1$. It is easily verified that this $G$ is isomorphic to the multiplicative group of all complex numbers $e^ialpha$ with $alpha$ a rational multiple of $2pi$, which is the same thing as the set of all $zin S^1$ that can be an $n^rm th$ root of unity for some $ngeq1$.
answered Jul 28 at 15:17
Christian Blatter
163k7107306
add a comment |Â
up vote
0
down vote
$$(Bbb Z+q)+(Bbb Z+q)+cdots( n textrm terms)cdots+(Bbb Z+q)=Bbb Z+nq=Bbb Z+m=Bbb Z.$$
Since $n(Bbb Z+q)$ is the identity element in $Bbb Q/Bbb Z$, then
$Bbb Z+q$ has finite order in $Bbb Q/Bbb Z$.
answered Jul 28 at 14:44
Lord Shark the Unknown
84.6k950111
Clearly I'm missing something here. Why is n(ℤ+q) the identity?
– Yip Jung Hon
Jul 28 at 14:52
$Bbb Z$ is the identity of $Bbb Q/Bbb Z$. In general the identity of the quotient group $G/H$ is the coset $H$. @YipJungHon
– Lord Shark the Unknown
Jul 28 at 14:54
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Remember that the elements of $mathbbQ/mathbbZ$ are right cosets of $mathbbZ$ which partitions $mathbbQ$. So $mathbbQ/mathbbZ$ gives an equivalence relation on $mathbbQ$ with
beginalign*
q sim p iff mathbbZ + p = mathbbZ + q iff p- q in mathbbZ.
endalign*
Since for any $n in mathbbZ$ we have $n - 0 in mathbbZ$ and so the identity element in $mathbbQ/mathbbZ$ is $mathbbZ + 0 = mathbbZ + n = mathbbZ$. If $q = m/n in mathbbQ$ then
beginalign*
sum_i=1^n(mathbbZ + m/n) = mathbbZ + (sum_i=1^n m/n) = mathbbZ + m = mathbbZ
endalign*
since $m in mathbbZ$. Hence $mathbbZ + q$ has finite order in $mathbbQ/mathbbZ$.
Now ask yourself if any element in $mathbbQ$ has this property.
answered Jul 28 at 15:01
matt stokes
1047
add a comment |Â
up vote
4
down vote
accepted
Remember that the elements of $mathbbQ/mathbbZ$ are right cosets of $mathbbZ$ which partitions $mathbbQ$. So $mathbbQ/mathbbZ$ gives an equivalence relation on $mathbbQ$ with
beginalign*
q sim p iff mathbbZ + p = mathbbZ + q iff p- q in mathbbZ.
endalign*
Since for any $n in mathbbZ$ we have $n - 0 in mathbbZ$ and so the identity element in $mathbbQ/mathbbZ$ is $mathbbZ + 0 = mathbbZ + n = mathbbZ$. If $q = m/n in mathbbQ$ then
beginalign*
sum_i=1^n(mathbbZ + m/n) = mathbbZ + (sum_i=1^n m/n) = mathbbZ + m = mathbbZ
endalign*
since $m in mathbbZ$. Hence $mathbbZ + q$ has finite order in $mathbbQ/mathbbZ$.
Now ask yourself if any element in $mathbbQ$ has this property.
answered Jul 28 at 15:01
matt stokes
1047
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Remember that the elements of $mathbbQ/mathbbZ$ are right cosets of $mathbbZ$ which partitions $mathbbQ$. So $mathbbQ/mathbbZ$ gives an equivalence relation on $mathbbQ$ with
beginalign*
q sim p iff mathbbZ + p = mathbbZ + q iff p- q in mathbbZ.
endalign*
Since for any $n in mathbbZ$ we have $n - 0 in mathbbZ$ and so the identity element in $mathbbQ/mathbbZ$ is $mathbbZ + 0 = mathbbZ + n = mathbbZ$. If $q = m/n in mathbbQ$ then
beginalign*
sum_i=1^n(mathbbZ + m/n) = mathbbZ + (sum_i=1^n m/n) = mathbbZ + m = mathbbZ
endalign*
since $m in mathbbZ$. Hence $mathbbZ + q$ has finite order in $mathbbQ/mathbbZ$.
Now ask yourself if any element in $mathbbQ$ has this property.
answered Jul 28 at 15:01
matt stokes
1047
Remember that the elements of $mathbbQ/mathbbZ$ are right cosets of $mathbbZ$ which partitions $mathbbQ$. So $mathbbQ/mathbbZ$ gives an equivalence relation on $mathbbQ$ with
beginalign*
q sim p iff mathbbZ + p = mathbbZ + q iff p- q in mathbbZ.
endalign*
Since for any $n in mathbbZ$ we have $n - 0 in mathbbZ$ and so the identity element in $mathbbQ/mathbbZ$ is $mathbbZ + 0 = mathbbZ + n = mathbbZ$. If $q = m/n in mathbbQ$ then
beginalign*
sum_i=1^n(mathbbZ + m/n) = mathbbZ + (sum_i=1^n m/n) = mathbbZ + m = mathbbZ
endalign*
since $m in mathbbZ$. Hence $mathbbZ + q$ has finite order in $mathbbQ/mathbbZ$.
Now ask yourself if any element in $mathbbQ$ has this property.
answered Jul 28 at 15:01
matt stokes
1047
edited Jul 28 at 15:09
answered Jul 28 at 15:01
matt stokes
1047
answered Jul 28 at 15:01
matt stokes
1047
answered Jul 28 at 15:01
matt stokes
1047
1047
add a comment |Â
add a comment |Â
up vote
1
down vote
It may help to first consider an intuitive solution to the problem:
Suppose I take a rational $pover q$. Then when I add it to itself $q$-many times, I get an integer (namely $p$). Now this is (probably) nonzero in $mathbbQ$, but think about what role the integers serve in $mathbbQ/mathbbZ$: "modding out" a group $G$ (in this case, $mathbbQ$) by a normal subgroup $H$ (in this case, $mathbbZ$) basically means "making every element of $H$ become zero." So we have the following idea (which isn't quite right - see below - but is on the right track): $$mboxIn $mathbbQ/mathbbZ$, $pover q$ added to itself $q$-many times gives zero.$$
Put another way, the informal idea above suggests that $pover q$ has order at most $q$ in $mathbbQ/mathbbZ$." Now this isn't quite right: $pover q$ isn't literally an element of $mathbbQ/mathbbZ$ in the first place (elements of a quotient group are cosets, not elements, of the original group). But when you go back through the argument above and make it rigorous, you get exactly the solution in the book.
answered Jul 28 at 15:02
Noah Schweber
110k9139260
add a comment |Â
up vote
1
down vote
It may help to first consider an intuitive solution to the problem:
Suppose I take a rational $pover q$. Then when I add it to itself $q$-many times, I get an integer (namely $p$). Now this is (probably) nonzero in $mathbbQ$, but think about what role the integers serve in $mathbbQ/mathbbZ$: "modding out" a group $G$ (in this case, $mathbbQ$) by a normal subgroup $H$ (in this case, $mathbbZ$) basically means "making every element of $H$ become zero." So we have the following idea (which isn't quite right - see below - but is on the right track): $$mboxIn $mathbbQ/mathbbZ$, $pover q$ added to itself $q$-many times gives zero.$$
Put another way, the informal idea above suggests that $pover q$ has order at most $q$ in $mathbbQ/mathbbZ$." Now this isn't quite right: $pover q$ isn't literally an element of $mathbbQ/mathbbZ$ in the first place (elements of a quotient group are cosets, not elements, of the original group). But when you go back through the argument above and make it rigorous, you get exactly the solution in the book.
answered Jul 28 at 15:02
Noah Schweber
110k9139260
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It may help to first consider an intuitive solution to the problem:
Suppose I take a rational $pover q$. Then when I add it to itself $q$-many times, I get an integer (namely $p$). Now this is (probably) nonzero in $mathbbQ$, but think about what role the integers serve in $mathbbQ/mathbbZ$: "modding out" a group $G$ (in this case, $mathbbQ$) by a normal subgroup $H$ (in this case, $mathbbZ$) basically means "making every element of $H$ become zero." So we have the following idea (which isn't quite right - see below - but is on the right track): $$mboxIn $mathbbQ/mathbbZ$, $pover q$ added to itself $q$-many times gives zero.$$
Put another way, the informal idea above suggests that $pover q$ has order at most $q$ in $mathbbQ/mathbbZ$." Now this isn't quite right: $pover q$ isn't literally an element of $mathbbQ/mathbbZ$ in the first place (elements of a quotient group are cosets, not elements, of the original group). But when you go back through the argument above and make it rigorous, you get exactly the solution in the book.
answered Jul 28 at 15:02
Noah Schweber
110k9139260
It may help to first consider an intuitive solution to the problem:
Suppose I take a rational $pover q$. Then when I add it to itself $q$-many times, I get an integer (namely $p$). Now this is (probably) nonzero in $mathbbQ$, but think about what role the integers serve in $mathbbQ/mathbbZ$: "modding out" a group $G$ (in this case, $mathbbQ$) by a normal subgroup $H$ (in this case, $mathbbZ$) basically means "making every element of $H$ become zero." So we have the following idea (which isn't quite right - see below - but is on the right track): $$mboxIn $mathbbQ/mathbbZ$, $pover q$ added to itself $q$-many times gives zero.$$
Put another way, the informal idea above suggests that $pover q$ has order at most $q$ in $mathbbQ/mathbbZ$." Now this isn't quite right: $pover q$ isn't literally an element of $mathbbQ/mathbbZ$ in the first place (elements of a quotient group are cosets, not elements, of the original group). But when you go back through the argument above and make it rigorous, you get exactly the solution in the book.
answered Jul 28 at 15:02
Noah Schweber
110k9139260
answered Jul 28 at 15:02
Noah Schweber
110k9139260
answered Jul 28 at 15:02
Noah Schweber
110k9139260
answered Jul 28 at 15:02
Noah Schweber
110k9139260
110k9139260
add a comment |Â
add a comment |Â
up vote
1
down vote
The following may help the intuition about the group $G:=mathbb Q/mathbb Z$. The map
$$psi:quad mathbb Qto S^1,qquad tmapsto e^2pi i t$$
maps rationals $x$, $yinmathbb Q$ that differ by an integer to the same point of $S^1$, hence maps all representants of an element $[x]in G$ to the same point of $S^1$. It is easily verified that this $G$ is isomorphic to the multiplicative group of all complex numbers $e^ialpha$ with $alpha$ a rational multiple of $2pi$, which is the same thing as the set of all $zin S^1$ that can be an $n^rm th$ root of unity for some $ngeq1$.
answered Jul 28 at 15:17
Christian Blatter
163k7107306
add a comment |Â
up vote
1
down vote
The following may help the intuition about the group $G:=mathbb Q/mathbb Z$. The map
$$psi:quad mathbb Qto S^1,qquad tmapsto e^2pi i t$$
maps rationals $x$, $yinmathbb Q$ that differ by an integer to the same point of $S^1$, hence maps all representants of an element $[x]in G$ to the same point of $S^1$. It is easily verified that this $G$ is isomorphic to the multiplicative group of all complex numbers $e^ialpha$ with $alpha$ a rational multiple of $2pi$, which is the same thing as the set of all $zin S^1$ that can be an $n^rm th$ root of unity for some $ngeq1$.
answered Jul 28 at 15:17
Christian Blatter
163k7107306
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The following may help the intuition about the group $G:=mathbb Q/mathbb Z$. The map
$$psi:quad mathbb Qto S^1,qquad tmapsto e^2pi i t$$
maps rationals $x$, $yinmathbb Q$ that differ by an integer to the same point of $S^1$, hence maps all representants of an element $[x]in G$ to the same point of $S^1$. It is easily verified that this $G$ is isomorphic to the multiplicative group of all complex numbers $e^ialpha$ with $alpha$ a rational multiple of $2pi$, which is the same thing as the set of all $zin S^1$ that can be an $n^rm th$ root of unity for some $ngeq1$.
answered Jul 28 at 15:17
Christian Blatter
163k7107306
The following may help the intuition about the group $G:=mathbb Q/mathbb Z$. The map
$$psi:quad mathbb Qto S^1,qquad tmapsto e^2pi i t$$
maps rationals $x$, $yinmathbb Q$ that differ by an integer to the same point of $S^1$, hence maps all representants of an element $[x]in G$ to the same point of $S^1$. It is easily verified that this $G$ is isomorphic to the multiplicative group of all complex numbers $e^ialpha$ with $alpha$ a rational multiple of $2pi$, which is the same thing as the set of all $zin S^1$ that can be an $n^rm th$ root of unity for some $ngeq1$.
answered Jul 28 at 15:17
Christian Blatter
163k7107306
answered Jul 28 at 15:17
Christian Blatter
163k7107306
answered Jul 28 at 15:17
Christian Blatter
163k7107306
answered Jul 28 at 15:17
Christian Blatter
163k7107306
163k7107306
add a comment |Â
add a comment |Â
up vote
0
down vote
$$(Bbb Z+q)+(Bbb Z+q)+cdots( n textrm terms)cdots+(Bbb Z+q)=Bbb Z+nq=Bbb Z+m=Bbb Z.$$
Since $n(Bbb Z+q)$ is the identity element in $Bbb Q/Bbb Z$, then
$Bbb Z+q$ has finite order in $Bbb Q/Bbb Z$.
answered Jul 28 at 14:44
Lord Shark the Unknown
84.6k950111
Clearly I'm missing something here. Why is n(ℤ+q) the identity?
– Yip Jung Hon
Jul 28 at 14:52
$Bbb Z$ is the identity of $Bbb Q/Bbb Z$. In general the identity of the quotient group $G/H$ is the coset $H$. @YipJungHon
– Lord Shark the Unknown
Jul 28 at 14:54
add a comment |Â
up vote
0
down vote
$$(Bbb Z+q)+(Bbb Z+q)+cdots( n textrm terms)cdots+(Bbb Z+q)=Bbb Z+nq=Bbb Z+m=Bbb Z.$$
Since $n(Bbb Z+q)$ is the identity element in $Bbb Q/Bbb Z$, then
$Bbb Z+q$ has finite order in $Bbb Q/Bbb Z$.
answered Jul 28 at 14:44
Lord Shark the Unknown
84.6k950111
Clearly I'm missing something here. Why is n(ℤ+q) the identity?
– Yip Jung Hon
Jul 28 at 14:52
$Bbb Z$ is the identity of $Bbb Q/Bbb Z$. In general the identity of the quotient group $G/H$ is the coset $H$. @YipJungHon
– Lord Shark the Unknown
Jul 28 at 14:54
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$(Bbb Z+q)+(Bbb Z+q)+cdots( n textrm terms)cdots+(Bbb Z+q)=Bbb Z+nq=Bbb Z+m=Bbb Z.$$
Since $n(Bbb Z+q)$ is the identity element in $Bbb Q/Bbb Z$, then
$Bbb Z+q$ has finite order in $Bbb Q/Bbb Z$.
answered Jul 28 at 14:44
Lord Shark the Unknown
84.6k950111
$$(Bbb Z+q)+(Bbb Z+q)+cdots( n textrm terms)cdots+(Bbb Z+q)=Bbb Z+nq=Bbb Z+m=Bbb Z.$$
Since $n(Bbb Z+q)$ is the identity element in $Bbb Q/Bbb Z$, then
$Bbb Z+q$ has finite order in $Bbb Q/Bbb Z$.
answered Jul 28 at 14:44
Lord Shark the Unknown
84.6k950111
answered Jul 28 at 14:44
Lord Shark the Unknown
84.6k950111
answered Jul 28 at 14:44
Lord Shark the Unknown
84.6k950111
answered Jul 28 at 14:44
Lord Shark the Unknown
84.6k950111
84.6k950111
Clearly I'm missing something here. Why is n(ℤ+q) the identity?
– Yip Jung Hon
Jul 28 at 14:52
$Bbb Z$ is the identity of $Bbb Q/Bbb Z$. In general the identity of the quotient group $G/H$ is the coset $H$. @YipJungHon
– Lord Shark the Unknown
Jul 28 at 14:54
add a comment |Â
Clearly I'm missing something here. Why is n(ℤ+q) the identity?
– Yip Jung Hon
Jul 28 at 14:52
$Bbb Z$ is the identity of $Bbb Q/Bbb Z$. In general the identity of the quotient group $G/H$ is the coset $H$. @YipJungHon
– Lord Shark the Unknown
Jul 28 at 14:54
Clearly I'm missing something here. Why is n(ℤ+q) the identity?
– Yip Jung Hon
Jul 28 at 14:52
Clearly I'm missing something here. Why is n(ℤ+q) the identity?
– Yip Jung Hon
Jul 28 at 14:52
$Bbb Z$ is the identity of $Bbb Q/Bbb Z$. In general the identity of the quotient group $G/H$ is the coset $H$. @YipJungHon
– Lord Shark the Unknown
Jul 28 at 14:54
$Bbb Z$ is the identity of $Bbb Q/Bbb Z$. In general the identity of the quotient group $G/H$ is the coset $H$. @YipJungHon
– Lord Shark the Unknown
Jul 28 at 14:54
add a comment |Â
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Do you understand why the coset $H$ is zero in the quotient group $G/H$?
– Noah Schweber
Jul 28 at 15:04