Mixing
Why is $Z^n/AZ^n cong Z/d_1Zoplus Z/d_2Z..$ where $d_i$ are the entries of normal form of A?
$begingroup$
In trying to show that $Card(Z^n/AZ^n)=det(A)$ which has been answered earlier here $mathrm{card}(mathbb{Z}^n/Mmathbb{Z}^n) = |det(M)|$? ,
the answer refers to the isomorphism $Z^n/AZ^n cong Z/d_1Zoplus Z/d_2Z..$. Can you point me to a theorem that would lead to this direct sum decomposition? I know that a finitely generated Abelian group can be factored into cyclic groups and I suppose $Z^n/AZ^n$ is finitely generated. But I am not sure what the generator set for $Z^n/AZ^n$ is? A hint is appreciated. Thanks.
abstract-algebra group-theory abelian-groups group-isomorphism finitely-generated
asked Jan 30 at 19:14
manifoldedmanifolded
50219
$endgroup$
add a comment |
$begingroup$
In trying to show that $Card(Z^n/AZ^n)=det(A)$ which has been answered earlier here $mathrm{card}(mathbb{Z}^n/Mmathbb{Z}^n) = |det(M)|$? ,
the answer refers to the isomorphism $Z^n/AZ^n cong Z/d_1Zoplus Z/d_2Z..$. Can you point me to a theorem that would lead to this direct sum decomposition? I know that a finitely generated Abelian group can be factored into cyclic groups and I suppose $Z^n/AZ^n$ is finitely generated. But I am not sure what the generator set for $Z^n/AZ^n$ is? A hint is appreciated. Thanks.
abstract-algebra group-theory abelian-groups group-isomorphism finitely-generated
asked Jan 30 at 19:14
manifoldedmanifolded
50219
$endgroup$
add a comment |
$begingroup$
In trying to show that $Card(Z^n/AZ^n)=det(A)$ which has been answered earlier here $mathrm{card}(mathbb{Z}^n/Mmathbb{Z}^n) = |det(M)|$? ,
the answer refers to the isomorphism $Z^n/AZ^n cong Z/d_1Zoplus Z/d_2Z..$. Can you point me to a theorem that would lead to this direct sum decomposition? I know that a finitely generated Abelian group can be factored into cyclic groups and I suppose $Z^n/AZ^n$ is finitely generated. But I am not sure what the generator set for $Z^n/AZ^n$ is? A hint is appreciated. Thanks.
abstract-algebra group-theory abelian-groups group-isomorphism finitely-generated
asked Jan 30 at 19:14
manifoldedmanifolded
50219
$endgroup$
In trying to show that $Card(Z^n/AZ^n)=det(A)$ which has been answered earlier here $mathrm{card}(mathbb{Z}^n/Mmathbb{Z}^n) = |det(M)|$? ,
the answer refers to the isomorphism $Z^n/AZ^n cong Z/d_1Zoplus Z/d_2Z..$. Can you point me to a theorem that would lead to this direct sum decomposition? I know that a finitely generated Abelian group can be factored into cyclic groups and I suppose $Z^n/AZ^n$ is finitely generated. But I am not sure what the generator set for $Z^n/AZ^n$ is? A hint is appreciated. Thanks.
abstract-algebra group-theory abelian-groups group-isomorphism finitely-generated
abstract-algebra group-theory abelian-groups group-isomorphism finitely-generated
asked Jan 30 at 19:14
manifoldedmanifolded
50219
asked Jan 30 at 19:14
manifoldedmanifolded
50219
edited Jan 30 at 19:25
manifolded
asked Jan 30 at 19:14
manifoldedmanifolded
50219
asked Jan 30 at 19:14
manifoldedmanifolded
50219
asked Jan 30 at 19:14
manifoldedmanifolded
50219
50219
add a comment |
add a comment |
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Suppose $A=P mathrm{diag} (d_1,...,d_n) Q$ with $P,Qin GL_n(mathbb{Z})$.
Then $Amathbb{Z}^n = Pmathrm{diag} (d_1,...,d_n) Qmathbb{Z}^n = Pmathrm{diag} (d_1,...,d_n) mathbb{Z}^n$ because $Q$ is invertible.
Now it's easy to check that when $f: Mto M$ is an automorphism of an abelian group $M$, and $N$ is a subgroup of $M$, then $M/Nsimeq M/f(N)$. Indeed, consider $f:Mto M$, $p$ the projection onto $M/f(N)$ and $q$ the one onto $N$: then $pcirc f$ is $0$ when restricted to $N$, so that $pcirc f$ factors uniquely through $Q$: we get $overline{f} : M/Nto M/f(N)$. Use $f^{-1}$ to build a map in the opposite direction, and then check (it's easy) that the two maps are invere.
Thus $mathbb{Z}^n/Amathbb{Z}^n = mathbb{Z}^n/P(mathrm{diag}(d_1,...,d_n)mathbb{Z}^n) simeq mathbb{Z}^n/mathrm{diag}(d_1,...,d_n)mathbb{Z}^n$ because $P$ is an automorphism of $mathbb{Z}^n$. Now $mathrm{diag}(d_1,...,d_n)mathbb{Z}^n$ is just $d_1mathbb{Z}oplus ... oplus d_nmathbb{Z}$, and so $mathbb{Z}^n/Amathbb{Z}^n simeq displaystylebigoplus_{i=1}^nmathbb{Z}/d_imathbb{Z}$
answered Jan 30 at 21:39
MaxMax
15.9k11144
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$begingroup$
Suppose $A=P mathrm{diag} (d_1,...,d_n) Q$ with $P,Qin GL_n(mathbb{Z})$.
Then $Amathbb{Z}^n = Pmathrm{diag} (d_1,...,d_n) Qmathbb{Z}^n = Pmathrm{diag} (d_1,...,d_n) mathbb{Z}^n$ because $Q$ is invertible.
Now it's easy to check that when $f: Mto M$ is an automorphism of an abelian group $M$, and $N$ is a subgroup of $M$, then $M/Nsimeq M/f(N)$. Indeed, consider $f:Mto M$, $p$ the projection onto $M/f(N)$ and $q$ the one onto $N$: then $pcirc f$ is $0$ when restricted to $N$, so that $pcirc f$ factors uniquely through $Q$: we get $overline{f} : M/Nto M/f(N)$. Use $f^{-1}$ to build a map in the opposite direction, and then check (it's easy) that the two maps are invere.
Thus $mathbb{Z}^n/Amathbb{Z}^n = mathbb{Z}^n/P(mathrm{diag}(d_1,...,d_n)mathbb{Z}^n) simeq mathbb{Z}^n/mathrm{diag}(d_1,...,d_n)mathbb{Z}^n$ because $P$ is an automorphism of $mathbb{Z}^n$. Now $mathrm{diag}(d_1,...,d_n)mathbb{Z}^n$ is just $d_1mathbb{Z}oplus ... oplus d_nmathbb{Z}$, and so $mathbb{Z}^n/Amathbb{Z}^n simeq displaystylebigoplus_{i=1}^nmathbb{Z}/d_imathbb{Z}$
answered Jan 30 at 21:39
MaxMax
15.9k11144
$endgroup$
add a comment |
$begingroup$
Suppose $A=P mathrm{diag} (d_1,...,d_n) Q$ with $P,Qin GL_n(mathbb{Z})$.
Then $Amathbb{Z}^n = Pmathrm{diag} (d_1,...,d_n) Qmathbb{Z}^n = Pmathrm{diag} (d_1,...,d_n) mathbb{Z}^n$ because $Q$ is invertible.
Now it's easy to check that when $f: Mto M$ is an automorphism of an abelian group $M$, and $N$ is a subgroup of $M$, then $M/Nsimeq M/f(N)$. Indeed, consider $f:Mto M$, $p$ the projection onto $M/f(N)$ and $q$ the one onto $N$: then $pcirc f$ is $0$ when restricted to $N$, so that $pcirc f$ factors uniquely through $Q$: we get $overline{f} : M/Nto M/f(N)$. Use $f^{-1}$ to build a map in the opposite direction, and then check (it's easy) that the two maps are invere.
Thus $mathbb{Z}^n/Amathbb{Z}^n = mathbb{Z}^n/P(mathrm{diag}(d_1,...,d_n)mathbb{Z}^n) simeq mathbb{Z}^n/mathrm{diag}(d_1,...,d_n)mathbb{Z}^n$ because $P$ is an automorphism of $mathbb{Z}^n$. Now $mathrm{diag}(d_1,...,d_n)mathbb{Z}^n$ is just $d_1mathbb{Z}oplus ... oplus d_nmathbb{Z}$, and so $mathbb{Z}^n/Amathbb{Z}^n simeq displaystylebigoplus_{i=1}^nmathbb{Z}/d_imathbb{Z}$
answered Jan 30 at 21:39
MaxMax
15.9k11144
$endgroup$
add a comment |
$begingroup$
Suppose $A=P mathrm{diag} (d_1,...,d_n) Q$ with $P,Qin GL_n(mathbb{Z})$.
Then $Amathbb{Z}^n = Pmathrm{diag} (d_1,...,d_n) Qmathbb{Z}^n = Pmathrm{diag} (d_1,...,d_n) mathbb{Z}^n$ because $Q$ is invertible.
Now it's easy to check that when $f: Mto M$ is an automorphism of an abelian group $M$, and $N$ is a subgroup of $M$, then $M/Nsimeq M/f(N)$. Indeed, consider $f:Mto M$, $p$ the projection onto $M/f(N)$ and $q$ the one onto $N$: then $pcirc f$ is $0$ when restricted to $N$, so that $pcirc f$ factors uniquely through $Q$: we get $overline{f} : M/Nto M/f(N)$. Use $f^{-1}$ to build a map in the opposite direction, and then check (it's easy) that the two maps are invere.
Thus $mathbb{Z}^n/Amathbb{Z}^n = mathbb{Z}^n/P(mathrm{diag}(d_1,...,d_n)mathbb{Z}^n) simeq mathbb{Z}^n/mathrm{diag}(d_1,...,d_n)mathbb{Z}^n$ because $P$ is an automorphism of $mathbb{Z}^n$. Now $mathrm{diag}(d_1,...,d_n)mathbb{Z}^n$ is just $d_1mathbb{Z}oplus ... oplus d_nmathbb{Z}$, and so $mathbb{Z}^n/Amathbb{Z}^n simeq displaystylebigoplus_{i=1}^nmathbb{Z}/d_imathbb{Z}$
answered Jan 30 at 21:39
MaxMax
15.9k11144
$endgroup$
Suppose $A=P mathrm{diag} (d_1,...,d_n) Q$ with $P,Qin GL_n(mathbb{Z})$.
Then $Amathbb{Z}^n = Pmathrm{diag} (d_1,...,d_n) Qmathbb{Z}^n = Pmathrm{diag} (d_1,...,d_n) mathbb{Z}^n$ because $Q$ is invertible.
Now it's easy to check that when $f: Mto M$ is an automorphism of an abelian group $M$, and $N$ is a subgroup of $M$, then $M/Nsimeq M/f(N)$. Indeed, consider $f:Mto M$, $p$ the projection onto $M/f(N)$ and $q$ the one onto $N$: then $pcirc f$ is $0$ when restricted to $N$, so that $pcirc f$ factors uniquely through $Q$: we get $overline{f} : M/Nto M/f(N)$. Use $f^{-1}$ to build a map in the opposite direction, and then check (it's easy) that the two maps are invere.
Thus $mathbb{Z}^n/Amathbb{Z}^n = mathbb{Z}^n/P(mathrm{diag}(d_1,...,d_n)mathbb{Z}^n) simeq mathbb{Z}^n/mathrm{diag}(d_1,...,d_n)mathbb{Z}^n$ because $P$ is an automorphism of $mathbb{Z}^n$. Now $mathrm{diag}(d_1,...,d_n)mathbb{Z}^n$ is just $d_1mathbb{Z}oplus ... oplus d_nmathbb{Z}$, and so $mathbb{Z}^n/Amathbb{Z}^n simeq displaystylebigoplus_{i=1}^nmathbb{Z}/d_imathbb{Z}$
answered Jan 30 at 21:39
MaxMax
15.9k11144
answered Jan 30 at 21:39
MaxMax
15.9k11144
answered Jan 30 at 21:39
MaxMax
15.9k11144
answered Jan 30 at 21:39
MaxMax
15.9k11144
15.9k11144
add a comment |
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