If $G=D_{infty}=lbrace x, yvert x^{2}=y^{2}=1rbrace$ then what is $G/langle xrangle$?
$begingroup$
If $G=D_{infty}=lbrace x, yvert x^{2}=y^{2}=1rbrace$ then what is $G/langle xrangle$? I thought that it'd be $langle yrangle$ but I've been told this is incorrect. Am I missing something obvious? And then why does $mathbb{Z}[G/langle xrangle]=mathbb{Z}[langle yrangle]$?
group-theory
edited Jan 9 at 14:00
user1729
17.6k64294
asked Jan 9 at 13:22
RhoswynRhoswyn
388211
$endgroup$
|
show 3 more comments
$begingroup$
If $G=D_{infty}=lbrace x, yvert x^{2}=y^{2}=1rbrace$ then what is $G/langle xrangle$? I thought that it'd be $langle yrangle$ but I've been told this is incorrect. Am I missing something obvious? And then why does $mathbb{Z}[G/langle xrangle]=mathbb{Z}[langle yrangle]$?
group-theory
edited Jan 9 at 14:00
user1729
17.6k64294
asked Jan 9 at 13:22
RhoswynRhoswyn
388211
$endgroup$
1
$begingroup$
There are two things that you are missing, which if you can resolve will help you :-). Firstly, $G/langle xrangle$ isn't a group (why not? So what kind of structure is it?!). Secondly, $G/langle xrangle$ is not going to consist of elements of $x$, so it isn't going to be $langle yrangle$! (Working out the answer to the first point will help you with the second point.)
$endgroup$
– user1729
Jan 9 at 13:57
$begingroup$
Okay, I'm still confused as to what $G/langle xrangle$ looks like. I thought it would be $lbrace 1, yrbrace$...or is it all the elements that aren't actually in $langle xrangle$?
$endgroup$
– Rhoswyn
Jan 9 at 14:22
$begingroup$
The key word is "coset". (Also, $G/langle xrangle$ is an infinite set. You are assuming that $langle xrangle$ is a normal subgroup of $G$, which it isn't)
$endgroup$
– user1729
Jan 9 at 14:35
$begingroup$
$mathbb{Z}[G/langle x rangle]$ does not make sense. Are you sure that the question was not about $G/langle xy rangle$?
$endgroup$
– Derek Holt
Jan 9 at 14:39
1
$begingroup$
The answer is then easy - these two free abelian groups are isomorphic (not equal though!) as both of the sets $langle yrangle$ and $G/langle xrangle$ are countably infinite (why?), and then apply uniqueness of the free abelian group on a countably infinite set.
$endgroup$
– user1729
Jan 9 at 15:26
|
show 3 more comments
$begingroup$
If $G=D_{infty}=lbrace x, yvert x^{2}=y^{2}=1rbrace$ then what is $G/langle xrangle$? I thought that it'd be $langle yrangle$ but I've been told this is incorrect. Am I missing something obvious? And then why does $mathbb{Z}[G/langle xrangle]=mathbb{Z}[langle yrangle]$?
group-theory
edited Jan 9 at 14:00
user1729
17.6k64294
asked Jan 9 at 13:22
RhoswynRhoswyn
388211
$endgroup$
If $G=D_{infty}=lbrace x, yvert x^{2}=y^{2}=1rbrace$ then what is $G/langle xrangle$? I thought that it'd be $langle yrangle$ but I've been told this is incorrect. Am I missing something obvious? And then why does $mathbb{Z}[G/langle xrangle]=mathbb{Z}[langle yrangle]$?
group-theory
group-theory
edited Jan 9 at 14:00
user1729
17.6k64294
asked Jan 9 at 13:22
RhoswynRhoswyn
388211
edited Jan 9 at 14:00
user1729
17.6k64294
asked Jan 9 at 13:22
RhoswynRhoswyn
388211
edited Jan 9 at 14:00
user1729
17.6k64294
edited Jan 9 at 14:00
user1729
17.6k64294
edited Jan 9 at 14:00
user1729
17.6k64294
17.6k64294
asked Jan 9 at 13:22
RhoswynRhoswyn
388211
asked Jan 9 at 13:22
RhoswynRhoswyn
388211
asked Jan 9 at 13:22
RhoswynRhoswyn
388211
388211
1
$begingroup$
There are two things that you are missing, which if you can resolve will help you :-). Firstly, $G/langle xrangle$ isn't a group (why not? So what kind of structure is it?!). Secondly, $G/langle xrangle$ is not going to consist of elements of $x$, so it isn't going to be $langle yrangle$! (Working out the answer to the first point will help you with the second point.)
$endgroup$
– user1729
Jan 9 at 13:57
$begingroup$
Okay, I'm still confused as to what $G/langle xrangle$ looks like. I thought it would be $lbrace 1, yrbrace$...or is it all the elements that aren't actually in $langle xrangle$?
$endgroup$
– Rhoswyn
Jan 9 at 14:22
$begingroup$
The key word is "coset". (Also, $G/langle xrangle$ is an infinite set. You are assuming that $langle xrangle$ is a normal subgroup of $G$, which it isn't)
$endgroup$
– user1729
Jan 9 at 14:35
$begingroup$
$mathbb{Z}[G/langle x rangle]$ does not make sense. Are you sure that the question was not about $G/langle xy rangle$?
$endgroup$
– Derek Holt
Jan 9 at 14:39
1
$begingroup$
The answer is then easy - these two free abelian groups are isomorphic (not equal though!) as both of the sets $langle yrangle$ and $G/langle xrangle$ are countably infinite (why?), and then apply uniqueness of the free abelian group on a countably infinite set.
$endgroup$
– user1729
Jan 9 at 15:26
|
show 3 more comments
1
$begingroup$
There are two things that you are missing, which if you can resolve will help you :-). Firstly, $G/langle xrangle$ isn't a group (why not? So what kind of structure is it?!). Secondly, $G/langle xrangle$ is not going to consist of elements of $x$, so it isn't going to be $langle yrangle$! (Working out the answer to the first point will help you with the second point.)
$endgroup$
– user1729
Jan 9 at 13:57
$begingroup$
Okay, I'm still confused as to what $G/langle xrangle$ looks like. I thought it would be $lbrace 1, yrbrace$...or is it all the elements that aren't actually in $langle xrangle$?
$endgroup$
– Rhoswyn
Jan 9 at 14:22
$begingroup$
The key word is "coset". (Also, $G/langle xrangle$ is an infinite set. You are assuming that $langle xrangle$ is a normal subgroup of $G$, which it isn't)
$endgroup$
– user1729
Jan 9 at 14:35
$begingroup$
$mathbb{Z}[G/langle x rangle]$ does not make sense. Are you sure that the question was not about $G/langle xy rangle$?
$endgroup$
– Derek Holt
Jan 9 at 14:39
1
$begingroup$
The answer is then easy - these two free abelian groups are isomorphic (not equal though!) as both of the sets $langle yrangle$ and $G/langle xrangle$ are countably infinite (why?), and then apply uniqueness of the free abelian group on a countably infinite set.
$endgroup$
– user1729
Jan 9 at 15:26
1
1
$begingroup$
There are two things that you are missing, which if you can resolve will help you :-). Firstly, $G/langle xrangle$ isn't a group (why not? So what kind of structure is it?!). Secondly, $G/langle xrangle$ is not going to consist of elements of $x$, so it isn't going to be $langle yrangle$! (Working out the answer to the first point will help you with the second point.)
$endgroup$
– user1729
Jan 9 at 13:57
$begingroup$
There are two things that you are missing, which if you can resolve will help you :-). Firstly, $G/langle xrangle$ isn't a group (why not? So what kind of structure is it?!). Secondly, $G/langle xrangle$ is not going to consist of elements of $x$, so it isn't going to be $langle yrangle$! (Working out the answer to the first point will help you with the second point.)
$endgroup$
– user1729
Jan 9 at 13:57
$begingroup$
Okay, I'm still confused as to what $G/langle xrangle$ looks like. I thought it would be $lbrace 1, yrbrace$...or is it all the elements that aren't actually in $langle xrangle$?
$endgroup$
– Rhoswyn
Jan 9 at 14:22
$begingroup$
Okay, I'm still confused as to what $G/langle xrangle$ looks like. I thought it would be $lbrace 1, yrbrace$...or is it all the elements that aren't actually in $langle xrangle$?
$endgroup$
– Rhoswyn
Jan 9 at 14:22
$begingroup$
The key word is "coset". (Also, $G/langle xrangle$ is an infinite set. You are assuming that $langle xrangle$ is a normal subgroup of $G$, which it isn't)
$endgroup$
– user1729
Jan 9 at 14:35
$begingroup$
The key word is "coset". (Also, $G/langle xrangle$ is an infinite set. You are assuming that $langle xrangle$ is a normal subgroup of $G$, which it isn't)
$endgroup$
– user1729
Jan 9 at 14:35
$begingroup$
$mathbb{Z}[G/langle x rangle]$ does not make sense. Are you sure that the question was not about $G/langle xy rangle$?
$endgroup$
– Derek Holt
Jan 9 at 14:39
$begingroup$
$mathbb{Z}[G/langle x rangle]$ does not make sense. Are you sure that the question was not about $G/langle xy rangle$?
$endgroup$
– Derek Holt
Jan 9 at 14:39
1
1
$begingroup$
The answer is then easy - these two free abelian groups are isomorphic (not equal though!) as both of the sets $langle yrangle$ and $G/langle xrangle$ are countably infinite (why?), and then apply uniqueness of the free abelian group on a countably infinite set.
$endgroup$
– user1729
Jan 9 at 15:26
$begingroup$
The answer is then easy - these two free abelian groups are isomorphic (not equal though!) as both of the sets $langle yrangle$ and $G/langle xrangle$ are countably infinite (why?), and then apply uniqueness of the free abelian group on a countably infinite set.
$endgroup$
– user1729
Jan 9 at 15:26
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Note that $x^{-1}=x$, and same for $y$. Therefore, if you conjugate $x$ by $y$ you get
$$x^y = yxy not in langle x rangle$$ which means that $langle x rangle$ is not a normal subgroup. Also note that $X= langle x rangle = {1, x}$.
In particular, this means that the quotient will not necessarily inherit the group structure from $G$, and in this particular case it indeed doesn't. By definition of quotient between groups:
$$G/X = {aX, a in G}$$
the set of cosets, where $aX = {ax, x in X}$ so in our case just ${a, ax}$. In particular, note that $yX neq yxyX$ since $yxy neq y$ and $yxy neq yx$, so it cannot be $langle y rangle$ as there are at least three elements/cosets $(1, y, yxy)$.
In fact, as sets $G/X = G_y$ if we denote by $G_y$ the set of elements of $G$ that "end with $y$" (you can "remove" any $x$ at the end of an element $a in G$ as $ax$ and $a$ are in the same cosets, so the last digit has to be $y$ because $x^2=1$) plus the identity coset or, equivalently, $G/X = G_x$ with similar reasoning.
Be careful though, as this does not inherit the group structure from $G$.
I agree with another comment above: I do not understand your second question.
answered Jan 9 at 14:45
AnalysisStudent0414AnalysisStudent0414
4,408928
$endgroup$
$begingroup$
Ahh yes! That makes total sense! In my second question I meant by $mathbb{Z}[G/langle xrangle]$ the free abelian group with basis the set $G/langle xrangle$.
$endgroup$
– Rhoswyn
Jan 9 at 15:06
$begingroup$
user1729 gave an excellent answer in the comments above then
$endgroup$
– AnalysisStudent0414
Jan 9 at 21:09
add a comment |
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$begingroup$
Note that $x^{-1}=x$, and same for $y$. Therefore, if you conjugate $x$ by $y$ you get
$$x^y = yxy not in langle x rangle$$ which means that $langle x rangle$ is not a normal subgroup. Also note that $X= langle x rangle = {1, x}$.
In particular, this means that the quotient will not necessarily inherit the group structure from $G$, and in this particular case it indeed doesn't. By definition of quotient between groups:
$$G/X = {aX, a in G}$$
the set of cosets, where $aX = {ax, x in X}$ so in our case just ${a, ax}$. In particular, note that $yX neq yxyX$ since $yxy neq y$ and $yxy neq yx$, so it cannot be $langle y rangle$ as there are at least three elements/cosets $(1, y, yxy)$.
In fact, as sets $G/X = G_y$ if we denote by $G_y$ the set of elements of $G$ that "end with $y$" (you can "remove" any $x$ at the end of an element $a in G$ as $ax$ and $a$ are in the same cosets, so the last digit has to be $y$ because $x^2=1$) plus the identity coset or, equivalently, $G/X = G_x$ with similar reasoning.
Be careful though, as this does not inherit the group structure from $G$.
I agree with another comment above: I do not understand your second question.
answered Jan 9 at 14:45
AnalysisStudent0414AnalysisStudent0414
4,408928
$endgroup$
$begingroup$
Ahh yes! That makes total sense! In my second question I meant by $mathbb{Z}[G/langle xrangle]$ the free abelian group with basis the set $G/langle xrangle$.
$endgroup$
– Rhoswyn
Jan 9 at 15:06
$begingroup$
user1729 gave an excellent answer in the comments above then
$endgroup$
– AnalysisStudent0414
Jan 9 at 21:09
add a comment |
$begingroup$
Note that $x^{-1}=x$, and same for $y$. Therefore, if you conjugate $x$ by $y$ you get
$$x^y = yxy not in langle x rangle$$ which means that $langle x rangle$ is not a normal subgroup. Also note that $X= langle x rangle = {1, x}$.
In particular, this means that the quotient will not necessarily inherit the group structure from $G$, and in this particular case it indeed doesn't. By definition of quotient between groups:
$$G/X = {aX, a in G}$$
the set of cosets, where $aX = {ax, x in X}$ so in our case just ${a, ax}$. In particular, note that $yX neq yxyX$ since $yxy neq y$ and $yxy neq yx$, so it cannot be $langle y rangle$ as there are at least three elements/cosets $(1, y, yxy)$.
In fact, as sets $G/X = G_y$ if we denote by $G_y$ the set of elements of $G$ that "end with $y$" (you can "remove" any $x$ at the end of an element $a in G$ as $ax$ and $a$ are in the same cosets, so the last digit has to be $y$ because $x^2=1$) plus the identity coset or, equivalently, $G/X = G_x$ with similar reasoning.
Be careful though, as this does not inherit the group structure from $G$.
I agree with another comment above: I do not understand your second question.
answered Jan 9 at 14:45
AnalysisStudent0414AnalysisStudent0414
4,408928
$endgroup$
$begingroup$
Ahh yes! That makes total sense! In my second question I meant by $mathbb{Z}[G/langle xrangle]$ the free abelian group with basis the set $G/langle xrangle$.
$endgroup$
– Rhoswyn
Jan 9 at 15:06
$begingroup$
user1729 gave an excellent answer in the comments above then
$endgroup$
– AnalysisStudent0414
Jan 9 at 21:09
add a comment |
$begingroup$
Note that $x^{-1}=x$, and same for $y$. Therefore, if you conjugate $x$ by $y$ you get
$$x^y = yxy not in langle x rangle$$ which means that $langle x rangle$ is not a normal subgroup. Also note that $X= langle x rangle = {1, x}$.
In particular, this means that the quotient will not necessarily inherit the group structure from $G$, and in this particular case it indeed doesn't. By definition of quotient between groups:
$$G/X = {aX, a in G}$$
the set of cosets, where $aX = {ax, x in X}$ so in our case just ${a, ax}$. In particular, note that $yX neq yxyX$ since $yxy neq y$ and $yxy neq yx$, so it cannot be $langle y rangle$ as there are at least three elements/cosets $(1, y, yxy)$.
In fact, as sets $G/X = G_y$ if we denote by $G_y$ the set of elements of $G$ that "end with $y$" (you can "remove" any $x$ at the end of an element $a in G$ as $ax$ and $a$ are in the same cosets, so the last digit has to be $y$ because $x^2=1$) plus the identity coset or, equivalently, $G/X = G_x$ with similar reasoning.
Be careful though, as this does not inherit the group structure from $G$.
I agree with another comment above: I do not understand your second question.
answered Jan 9 at 14:45
AnalysisStudent0414AnalysisStudent0414
4,408928
$endgroup$
Note that $x^{-1}=x$, and same for $y$. Therefore, if you conjugate $x$ by $y$ you get
$$x^y = yxy not in langle x rangle$$ which means that $langle x rangle$ is not a normal subgroup. Also note that $X= langle x rangle = {1, x}$.
In particular, this means that the quotient will not necessarily inherit the group structure from $G$, and in this particular case it indeed doesn't. By definition of quotient between groups:
$$G/X = {aX, a in G}$$
the set of cosets, where $aX = {ax, x in X}$ so in our case just ${a, ax}$. In particular, note that $yX neq yxyX$ since $yxy neq y$ and $yxy neq yx$, so it cannot be $langle y rangle$ as there are at least three elements/cosets $(1, y, yxy)$.
In fact, as sets $G/X = G_y$ if we denote by $G_y$ the set of elements of $G$ that "end with $y$" (you can "remove" any $x$ at the end of an element $a in G$ as $ax$ and $a$ are in the same cosets, so the last digit has to be $y$ because $x^2=1$) plus the identity coset or, equivalently, $G/X = G_x$ with similar reasoning.
Be careful though, as this does not inherit the group structure from $G$.
I agree with another comment above: I do not understand your second question.
answered Jan 9 at 14:45
AnalysisStudent0414AnalysisStudent0414
4,408928
answered Jan 9 at 14:45
AnalysisStudent0414AnalysisStudent0414
4,408928
answered Jan 9 at 14:45
AnalysisStudent0414AnalysisStudent0414
4,408928
answered Jan 9 at 14:45
AnalysisStudent0414AnalysisStudent0414
4,408928
4,408928
$begingroup$
Ahh yes! That makes total sense! In my second question I meant by $mathbb{Z}[G/langle xrangle]$ the free abelian group with basis the set $G/langle xrangle$.
$endgroup$
– Rhoswyn
Jan 9 at 15:06
$begingroup$
user1729 gave an excellent answer in the comments above then
$endgroup$
– AnalysisStudent0414
Jan 9 at 21:09
add a comment |
$begingroup$
Ahh yes! That makes total sense! In my second question I meant by $mathbb{Z}[G/langle xrangle]$ the free abelian group with basis the set $G/langle xrangle$.
$endgroup$
– Rhoswyn
Jan 9 at 15:06
$begingroup$
user1729 gave an excellent answer in the comments above then
$endgroup$
– AnalysisStudent0414
Jan 9 at 21:09
$begingroup$
Ahh yes! That makes total sense! In my second question I meant by $mathbb{Z}[G/langle xrangle]$ the free abelian group with basis the set $G/langle xrangle$.
$endgroup$
– Rhoswyn
Jan 9 at 15:06
$begingroup$
Ahh yes! That makes total sense! In my second question I meant by $mathbb{Z}[G/langle xrangle]$ the free abelian group with basis the set $G/langle xrangle$.
$endgroup$
– Rhoswyn
Jan 9 at 15:06
$begingroup$
user1729 gave an excellent answer in the comments above then
$endgroup$
– AnalysisStudent0414
Jan 9 at 21:09
$begingroup$
user1729 gave an excellent answer in the comments above then
$endgroup$
– AnalysisStudent0414
Jan 9 at 21:09
add a comment |
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$begingroup$
There are two things that you are missing, which if you can resolve will help you :-). Firstly, $G/langle xrangle$ isn't a group (why not? So what kind of structure is it?!). Secondly, $G/langle xrangle$ is not going to consist of elements of $x$, so it isn't going to be $langle yrangle$! (Working out the answer to the first point will help you with the second point.)
$endgroup$
– user1729
Jan 9 at 13:57
$begingroup$
Okay, I'm still confused as to what $G/langle xrangle$ looks like. I thought it would be $lbrace 1, yrbrace$...or is it all the elements that aren't actually in $langle xrangle$?
$endgroup$
– Rhoswyn
Jan 9 at 14:22
$begingroup$
The key word is "coset". (Also, $G/langle xrangle$ is an infinite set. You are assuming that $langle xrangle$ is a normal subgroup of $G$, which it isn't)
$endgroup$
– user1729
Jan 9 at 14:35
$begingroup$
$mathbb{Z}[G/langle x rangle]$ does not make sense. Are you sure that the question was not about $G/langle xy rangle$?
$endgroup$
– Derek Holt
Jan 9 at 14:39
1
$begingroup$
The answer is then easy - these two free abelian groups are isomorphic (not equal though!) as both of the sets $langle yrangle$ and $G/langle xrangle$ are countably infinite (why?), and then apply uniqueness of the free abelian group on a countably infinite set.
$endgroup$
– user1729
Jan 9 at 15:26