Mixing
What is wrong with the followig proof that Z is isomorphic to ZxZ
Clash Royale CLAN TAG#URR8PPP
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2
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Since
$ G =( prod_1^infty mathbbZ )times mathbbZ times mathbbZ cong (prod_1^infty mathbbZ) times mathbbZ $,
by taking quotients we get
$mathbbZ times Z cong G/ prod_1^infty mathbbZ cong mathbbZ$.
Therefore
$mathbbZ cong mathbbZ times Z$.
But $mathbbZ$ is indecomposable! What's wrong with above proof?
group-theory
asked 1 hour ago
student
396
add a comment |Â
up vote
2
down vote
favorite
Since
$ G =( prod_1^infty mathbbZ )times mathbbZ times mathbbZ cong (prod_1^infty mathbbZ) times mathbbZ $,
by taking quotients we get
$mathbbZ times Z cong G/ prod_1^infty mathbbZ cong mathbbZ$.
Therefore
$mathbbZ cong mathbbZ times Z$.
But $mathbbZ$ is indecomposable! What's wrong with above proof?
group-theory
asked 1 hour ago
student
396
You are not working within a field structure. This you can't use cancellation.
– Alephnull
1 hour ago
What's so special about $mathbb Z$? Haven't you just proved $Gcong Gtimes G$ for every group $G$? If $G$ is a group of order $n$, does that mean that $n^2=n$ for every $n$?
– bof
49 mins ago
Possible duplicate of Cancellation Law for Direct Sums - What is wrong with this argument?
– Eric Wofsey
47 mins ago
The isomorphism from $(prod mathbbZ)timesmathbbZtimesmathbbZ$ to $(prodmathbbZ)timesmathbbZ$ does not map $(prodmathbbZ)times0times0$ to $(prodmathbbZ)times0$; so you do not know that you are taking quotients modulo "the same thing"; that is, you have an isomorphism $phicolon Gto K$; and subgroup $Ntriangleleft G$ and $Mtriangleleft K$ with $Ncong M$; but unless $phi(N)=M$, you cannot use $phi$ to conclude $G/Ncong K/M$; yet you are trying to do so here.
– Arturo Magidin
22 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Since
$ G =( prod_1^infty mathbbZ )times mathbbZ times mathbbZ cong (prod_1^infty mathbbZ) times mathbbZ $,
by taking quotients we get
$mathbbZ times Z cong G/ prod_1^infty mathbbZ cong mathbbZ$.
Therefore
$mathbbZ cong mathbbZ times Z$.
But $mathbbZ$ is indecomposable! What's wrong with above proof?
group-theory
asked 1 hour ago
student
396
Since
$ G =( prod_1^infty mathbbZ )times mathbbZ times mathbbZ cong (prod_1^infty mathbbZ) times mathbbZ $,
by taking quotients we get
$mathbbZ times Z cong G/ prod_1^infty mathbbZ cong mathbbZ$.
Therefore
$mathbbZ cong mathbbZ times Z$.
But $mathbbZ$ is indecomposable! What's wrong with above proof?
group-theory
group-theory
asked 1 hour ago
student
396
asked 1 hour ago
student
396
asked 1 hour ago
student
396
asked 1 hour ago
student
396
asked 1 hour ago
student
396
396
You are not working within a field structure. This you can't use cancellation.
– Alephnull
1 hour ago
What's so special about $mathbb Z$? Haven't you just proved $Gcong Gtimes G$ for every group $G$? If $G$ is a group of order $n$, does that mean that $n^2=n$ for every $n$?
– bof
49 mins ago
Possible duplicate of Cancellation Law for Direct Sums - What is wrong with this argument?
– Eric Wofsey
47 mins ago
The isomorphism from $(prod mathbbZ)timesmathbbZtimesmathbbZ$ to $(prodmathbbZ)timesmathbbZ$ does not map $(prodmathbbZ)times0times0$ to $(prodmathbbZ)times0$; so you do not know that you are taking quotients modulo "the same thing"; that is, you have an isomorphism $phicolon Gto K$; and subgroup $Ntriangleleft G$ and $Mtriangleleft K$ with $Ncong M$; but unless $phi(N)=M$, you cannot use $phi$ to conclude $G/Ncong K/M$; yet you are trying to do so here.
– Arturo Magidin
22 mins ago
add a comment |Â
You are not working within a field structure. This you can't use cancellation.
– Alephnull
1 hour ago
What's so special about $mathbb Z$? Haven't you just proved $Gcong Gtimes G$ for every group $G$? If $G$ is a group of order $n$, does that mean that $n^2=n$ for every $n$?
– bof
49 mins ago
Possible duplicate of Cancellation Law for Direct Sums - What is wrong with this argument?
– Eric Wofsey
47 mins ago
The isomorphism from $(prod mathbbZ)timesmathbbZtimesmathbbZ$ to $(prodmathbbZ)timesmathbbZ$ does not map $(prodmathbbZ)times0times0$ to $(prodmathbbZ)times0$; so you do not know that you are taking quotients modulo "the same thing"; that is, you have an isomorphism $phicolon Gto K$; and subgroup $Ntriangleleft G$ and $Mtriangleleft K$ with $Ncong M$; but unless $phi(N)=M$, you cannot use $phi$ to conclude $G/Ncong K/M$; yet you are trying to do so here.
– Arturo Magidin
22 mins ago
You are not working within a field structure. This you can't use cancellation.
– Alephnull
1 hour ago
You are not working within a field structure. This you can't use cancellation.
– Alephnull
1 hour ago
What's so special about $mathbb Z$? Haven't you just proved $Gcong Gtimes G$ for every group $G$? If $G$ is a group of order $n$, does that mean that $n^2=n$ for every $n$?
– bof
49 mins ago
What's so special about $mathbb Z$? Haven't you just proved $Gcong Gtimes G$ for every group $G$? If $G$ is a group of order $n$, does that mean that $n^2=n$ for every $n$?
– bof
49 mins ago
Possible duplicate of Cancellation Law for Direct Sums - What is wrong with this argument?
– Eric Wofsey
47 mins ago
Possible duplicate of Cancellation Law for Direct Sums - What is wrong with this argument?
– Eric Wofsey
47 mins ago
The isomorphism from $(prod mathbbZ)timesmathbbZtimesmathbbZ$ to $(prodmathbbZ)timesmathbbZ$ does not map $(prodmathbbZ)times0times0$ to $(prodmathbbZ)times0$; so you do not know that you are taking quotients modulo "the same thing"; that is, you have an isomorphism $phicolon Gto K$; and subgroup $Ntriangleleft G$ and $Mtriangleleft K$ with $Ncong M$; but unless $phi(N)=M$, you cannot use $phi$ to conclude $G/Ncong K/M$; yet you are trying to do so here.
– Arturo Magidin
22 mins ago
The isomorphism from $(prod mathbbZ)timesmathbbZtimesmathbbZ$ to $(prodmathbbZ)timesmathbbZ$ does not map $(prodmathbbZ)times0times0$ to $(prodmathbbZ)times0$; so you do not know that you are taking quotients modulo "the same thing"; that is, you have an isomorphism $phicolon Gto K$; and subgroup $Ntriangleleft G$ and $Mtriangleleft K$ with $Ncong M$; but unless $phi(N)=M$, you cannot use $phi$ to conclude $G/Ncong K/M$; yet you are trying to do so here.
– Arturo Magidin
22 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
Quotienting by isomorphic copy need not preserve isomorphisms. For a simpler related example: $C_n=mathbbZ/nmathbbZ$ and $mmathbbZcong nmathbbZ$ for all $m,n>0$, but obviously the cyclic groups $C_m$ and $C_n$ are not isomorphic if $mneq n$.
answered 1 hour ago
user10354138
4,523422
Quotient here is not by isomorphic copy. I am using the fact that factors are normal subgroups, and then using the first isomorphism theorem. The example you gave has no such product structure.
– student
50 mins ago
1
@student: If you are not using an isomorphic copy, then it is wrong to write $G/ prod_1^infty mathbbZ cong mathbbZ$, because you only have $G cong (prod_1^infty mathbbZ) times mathbbZ$, not $G = (prod_1^infty mathbbZ) times mathbbZ$
– Eric Wofsey
46 mins ago
add a comment |Â
up vote
0
down vote
Your argument is based on the assumption that $G/N_1cong G/N_2$ follows from $N_1cong N_2$. Here is a simple example to show that this assumption is incorrect.
The non-isomorphic groups $mathbb Z_4$ and $mathbb Z_2timesmathbb Z_2$ are both homomorphic images of the group $G=mathbb Z_2timesmathbb Z_4$. Thus there are subgroups $N_1,N_2$ of $G$ such that $G/N_1congmathbb Z_4$ and $G/N_2congmathbb Z_2timesmathbb Z_2$. The subgroups $N_1$ and $N_2$ are isomorphic, as both have order $2$; so $N_1cong N_2$ but $G/N_1notcong G/N_2$.
answered 13 mins ago
bof
48.1k449114
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Quotienting by isomorphic copy need not preserve isomorphisms. For a simpler related example: $C_n=mathbbZ/nmathbbZ$ and $mmathbbZcong nmathbbZ$ for all $m,n>0$, but obviously the cyclic groups $C_m$ and $C_n$ are not isomorphic if $mneq n$.
answered 1 hour ago
user10354138
4,523422
Quotient here is not by isomorphic copy. I am using the fact that factors are normal subgroups, and then using the first isomorphism theorem. The example you gave has no such product structure.
– student
50 mins ago
1
@student: If you are not using an isomorphic copy, then it is wrong to write $G/ prod_1^infty mathbbZ cong mathbbZ$, because you only have $G cong (prod_1^infty mathbbZ) times mathbbZ$, not $G = (prod_1^infty mathbbZ) times mathbbZ$
– Eric Wofsey
46 mins ago
add a comment |Â
up vote
5
down vote
Quotienting by isomorphic copy need not preserve isomorphisms. For a simpler related example: $C_n=mathbbZ/nmathbbZ$ and $mmathbbZcong nmathbbZ$ for all $m,n>0$, but obviously the cyclic groups $C_m$ and $C_n$ are not isomorphic if $mneq n$.
answered 1 hour ago
user10354138
4,523422
Quotient here is not by isomorphic copy. I am using the fact that factors are normal subgroups, and then using the first isomorphism theorem. The example you gave has no such product structure.
– student
50 mins ago
1
@student: If you are not using an isomorphic copy, then it is wrong to write $G/ prod_1^infty mathbbZ cong mathbbZ$, because you only have $G cong (prod_1^infty mathbbZ) times mathbbZ$, not $G = (prod_1^infty mathbbZ) times mathbbZ$
– Eric Wofsey
46 mins ago
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Quotienting by isomorphic copy need not preserve isomorphisms. For a simpler related example: $C_n=mathbbZ/nmathbbZ$ and $mmathbbZcong nmathbbZ$ for all $m,n>0$, but obviously the cyclic groups $C_m$ and $C_n$ are not isomorphic if $mneq n$.
answered 1 hour ago
user10354138
4,523422
Quotienting by isomorphic copy need not preserve isomorphisms. For a simpler related example: $C_n=mathbbZ/nmathbbZ$ and $mmathbbZcong nmathbbZ$ for all $m,n>0$, but obviously the cyclic groups $C_m$ and $C_n$ are not isomorphic if $mneq n$.
answered 1 hour ago
user10354138
4,523422
answered 1 hour ago
user10354138
4,523422
answered 1 hour ago
user10354138
4,523422
answered 1 hour ago
user10354138
4,523422
4,523422
Quotient here is not by isomorphic copy. I am using the fact that factors are normal subgroups, and then using the first isomorphism theorem. The example you gave has no such product structure.
– student
50 mins ago
1
@student: If you are not using an isomorphic copy, then it is wrong to write $G/ prod_1^infty mathbbZ cong mathbbZ$, because you only have $G cong (prod_1^infty mathbbZ) times mathbbZ$, not $G = (prod_1^infty mathbbZ) times mathbbZ$
– Eric Wofsey
46 mins ago
add a comment |Â
Quotient here is not by isomorphic copy. I am using the fact that factors are normal subgroups, and then using the first isomorphism theorem. The example you gave has no such product structure.
– student
50 mins ago
1
@student: If you are not using an isomorphic copy, then it is wrong to write $G/ prod_1^infty mathbbZ cong mathbbZ$, because you only have $G cong (prod_1^infty mathbbZ) times mathbbZ$, not $G = (prod_1^infty mathbbZ) times mathbbZ$
– Eric Wofsey
46 mins ago
Quotient here is not by isomorphic copy. I am using the fact that factors are normal subgroups, and then using the first isomorphism theorem. The example you gave has no such product structure.
– student
50 mins ago
Quotient here is not by isomorphic copy. I am using the fact that factors are normal subgroups, and then using the first isomorphism theorem. The example you gave has no such product structure.
– student
50 mins ago
1
1
@student: If you are not using an isomorphic copy, then it is wrong to write $G/ prod_1^infty mathbbZ cong mathbbZ$, because you only have $G cong (prod_1^infty mathbbZ) times mathbbZ$, not $G = (prod_1^infty mathbbZ) times mathbbZ$
– Eric Wofsey
46 mins ago
@student: If you are not using an isomorphic copy, then it is wrong to write $G/ prod_1^infty mathbbZ cong mathbbZ$, because you only have $G cong (prod_1^infty mathbbZ) times mathbbZ$, not $G = (prod_1^infty mathbbZ) times mathbbZ$
– Eric Wofsey
46 mins ago
add a comment |Â
up vote
0
down vote
Your argument is based on the assumption that $G/N_1cong G/N_2$ follows from $N_1cong N_2$. Here is a simple example to show that this assumption is incorrect.
The non-isomorphic groups $mathbb Z_4$ and $mathbb Z_2timesmathbb Z_2$ are both homomorphic images of the group $G=mathbb Z_2timesmathbb Z_4$. Thus there are subgroups $N_1,N_2$ of $G$ such that $G/N_1congmathbb Z_4$ and $G/N_2congmathbb Z_2timesmathbb Z_2$. The subgroups $N_1$ and $N_2$ are isomorphic, as both have order $2$; so $N_1cong N_2$ but $G/N_1notcong G/N_2$.
answered 13 mins ago
bof
48.1k449114
add a comment |Â
up vote
0
down vote
Your argument is based on the assumption that $G/N_1cong G/N_2$ follows from $N_1cong N_2$. Here is a simple example to show that this assumption is incorrect.
The non-isomorphic groups $mathbb Z_4$ and $mathbb Z_2timesmathbb Z_2$ are both homomorphic images of the group $G=mathbb Z_2timesmathbb Z_4$. Thus there are subgroups $N_1,N_2$ of $G$ such that $G/N_1congmathbb Z_4$ and $G/N_2congmathbb Z_2timesmathbb Z_2$. The subgroups $N_1$ and $N_2$ are isomorphic, as both have order $2$; so $N_1cong N_2$ but $G/N_1notcong G/N_2$.
answered 13 mins ago
bof
48.1k449114
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your argument is based on the assumption that $G/N_1cong G/N_2$ follows from $N_1cong N_2$. Here is a simple example to show that this assumption is incorrect.
The non-isomorphic groups $mathbb Z_4$ and $mathbb Z_2timesmathbb Z_2$ are both homomorphic images of the group $G=mathbb Z_2timesmathbb Z_4$. Thus there are subgroups $N_1,N_2$ of $G$ such that $G/N_1congmathbb Z_4$ and $G/N_2congmathbb Z_2timesmathbb Z_2$. The subgroups $N_1$ and $N_2$ are isomorphic, as both have order $2$; so $N_1cong N_2$ but $G/N_1notcong G/N_2$.
answered 13 mins ago
bof
48.1k449114
Your argument is based on the assumption that $G/N_1cong G/N_2$ follows from $N_1cong N_2$. Here is a simple example to show that this assumption is incorrect.
The non-isomorphic groups $mathbb Z_4$ and $mathbb Z_2timesmathbb Z_2$ are both homomorphic images of the group $G=mathbb Z_2timesmathbb Z_4$. Thus there are subgroups $N_1,N_2$ of $G$ such that $G/N_1congmathbb Z_4$ and $G/N_2congmathbb Z_2timesmathbb Z_2$. The subgroups $N_1$ and $N_2$ are isomorphic, as both have order $2$; so $N_1cong N_2$ but $G/N_1notcong G/N_2$.
answered 13 mins ago
bof
48.1k449114
answered 13 mins ago
bof
48.1k449114
answered 13 mins ago
bof
48.1k449114
answered 13 mins ago
bof
48.1k449114
48.1k449114
add a comment |Â
add a comment |Â
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You are not working within a field structure. This you can't use cancellation.
– Alephnull
1 hour ago
What's so special about $mathbb Z$? Haven't you just proved $Gcong Gtimes G$ for every group $G$? If $G$ is a group of order $n$, does that mean that $n^2=n$ for every $n$?
– bof
49 mins ago
Possible duplicate of Cancellation Law for Direct Sums - What is wrong with this argument?
– Eric Wofsey
47 mins ago
The isomorphism from $(prod mathbbZ)timesmathbbZtimesmathbbZ$ to $(prodmathbbZ)timesmathbbZ$ does not map $(prodmathbbZ)times0times0$ to $(prodmathbbZ)times0$; so you do not know that you are taking quotients modulo "the same thing"; that is, you have an isomorphism $phicolon Gto K$; and subgroup $Ntriangleleft G$ and $Mtriangleleft K$ with $Ncong M$; but unless $phi(N)=M$, you cannot use $phi$ to conclude $G/Ncong K/M$; yet you are trying to do so here.
– Arturo Magidin
22 mins ago