$mathbb Z^n/langle (a,…,a) rangle cong mathbb Z^{n-1} oplus mathbb Z/langle a rangle$
$begingroup$
I am trying to show the isomorphism
$$mathbb Z^n/langle (a,...,a) rangle cong mathbb Z^{n-1} oplus mathbb Z/langle a rangle.$$
I've tried to define $psi:mathbb Z^n to mathbb Z^{n-1} oplus mathbb Z/langle a rangle$ an epimorphism with $ker(psi)=langle (a,...,a) rangle$ so to apply the first isomorphism theorem, but I couldn't come up with an appropiate morphism. Any hints or suggestions would be greatly appreciated.
abstract-algebra group-theory
edited Dec 1 '14 at 20:38
user26857
39.4k124183
asked Dec 1 '14 at 20:15
user156441user156441
1,4461525
$endgroup$
add a comment |
$begingroup$
I am trying to show the isomorphism
$$mathbb Z^n/langle (a,...,a) rangle cong mathbb Z^{n-1} oplus mathbb Z/langle a rangle.$$
I've tried to define $psi:mathbb Z^n to mathbb Z^{n-1} oplus mathbb Z/langle a rangle$ an epimorphism with $ker(psi)=langle (a,...,a) rangle$ so to apply the first isomorphism theorem, but I couldn't come up with an appropiate morphism. Any hints or suggestions would be greatly appreciated.
abstract-algebra group-theory
edited Dec 1 '14 at 20:38
user26857
39.4k124183
asked Dec 1 '14 at 20:15
user156441user156441
1,4461525
$endgroup$
add a comment |
$begingroup$
I am trying to show the isomorphism
$$mathbb Z^n/langle (a,...,a) rangle cong mathbb Z^{n-1} oplus mathbb Z/langle a rangle.$$
I've tried to define $psi:mathbb Z^n to mathbb Z^{n-1} oplus mathbb Z/langle a rangle$ an epimorphism with $ker(psi)=langle (a,...,a) rangle$ so to apply the first isomorphism theorem, but I couldn't come up with an appropiate morphism. Any hints or suggestions would be greatly appreciated.
abstract-algebra group-theory
edited Dec 1 '14 at 20:38
user26857
39.4k124183
asked Dec 1 '14 at 20:15
user156441user156441
1,4461525
$endgroup$
I am trying to show the isomorphism
$$mathbb Z^n/langle (a,...,a) rangle cong mathbb Z^{n-1} oplus mathbb Z/langle a rangle.$$
I've tried to define $psi:mathbb Z^n to mathbb Z^{n-1} oplus mathbb Z/langle a rangle$ an epimorphism with $ker(psi)=langle (a,...,a) rangle$ so to apply the first isomorphism theorem, but I couldn't come up with an appropiate morphism. Any hints or suggestions would be greatly appreciated.
abstract-algebra group-theory
abstract-algebra group-theory
edited Dec 1 '14 at 20:38
user26857
39.4k124183
asked Dec 1 '14 at 20:15
user156441user156441
1,4461525
edited Dec 1 '14 at 20:38
user26857
39.4k124183
asked Dec 1 '14 at 20:15
user156441user156441
1,4461525
edited Dec 1 '14 at 20:38
user26857
39.4k124183
edited Dec 1 '14 at 20:38
user26857
39.4k124183
edited Dec 1 '14 at 20:38
user26857
39.4k124183
39.4k124183
asked Dec 1 '14 at 20:15
user156441user156441
1,4461525
asked Dec 1 '14 at 20:15
user156441user156441
1,4461525
asked Dec 1 '14 at 20:15
user156441user156441
1,4461525
1,4461525
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If you want to use the first isomorphism theorem, then define $psi:mathbb Z^n to mathbb Z^{n-1} oplus mathbb Z/langle a rangle$ by $psi(x_1,dots,x_n)=(x_2-x_1,dots,x_n-x_1,bar x_1)$. Check that $psi$ is a surjective homomorphism and $kerpsi=langle (a,dots,a) rangle$.
answered Dec 1 '14 at 20:30
user26857user26857
39.4k124183
$endgroup$
add a comment |
$begingroup$
The following: $$(1,0,dots,0), (0,1,0,dots,0), dots, (0,dots,0,1,0), (1,1,dots, 1)$$ is a basis of $mathbb{Z}^n$ and $(a,dots,a)$ corresponds to $(0,dots,0,a)$ in this new basis. Your group is thus the isomorphic to the quotient:
$$mathbb{Z}^n / langle (0, dots, 0, a) rangle cong mathbb{Z}^{n-1} oplus mathbb{Z}/(a).$$
answered Dec 1 '14 at 20:19
Najib IdrissiNajib Idrissi
41.1k471139
$endgroup$
$begingroup$
@lhf : well, this is what I did 2 years ago: math.stackexchange.com/questions/1910883
$endgroup$
– Alphonse
Dec 21 '18 at 16:33
add a comment |
$begingroup$
$mathbb{Z}^n/langle (a,dotsc,a)$ is an abelian group with the presentation $langle e_1,dotsc,e_n : a e_1 + cdots + a e_n = 0}$. Observe that $e_1,dotsc,e_{n-1},e_1+cdots+e_n$ is also a generating system and $a(e_1+cdots+e_n)=0$. Hence, we obtain a homomorphism from $langle f_1,dotsc,f_n : a f_n = 0 rangle$. One then constructs its inverse.
edited Dec 21 '18 at 9:51
user26857
39.4k124183
answered Dec 1 '14 at 20:20
Martin BrandenburgMartin Brandenburg
108k13158328
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you want to use the first isomorphism theorem, then define $psi:mathbb Z^n to mathbb Z^{n-1} oplus mathbb Z/langle a rangle$ by $psi(x_1,dots,x_n)=(x_2-x_1,dots,x_n-x_1,bar x_1)$. Check that $psi$ is a surjective homomorphism and $kerpsi=langle (a,dots,a) rangle$.
answered Dec 1 '14 at 20:30
user26857user26857
39.4k124183
$endgroup$
add a comment |
$begingroup$
If you want to use the first isomorphism theorem, then define $psi:mathbb Z^n to mathbb Z^{n-1} oplus mathbb Z/langle a rangle$ by $psi(x_1,dots,x_n)=(x_2-x_1,dots,x_n-x_1,bar x_1)$. Check that $psi$ is a surjective homomorphism and $kerpsi=langle (a,dots,a) rangle$.
answered Dec 1 '14 at 20:30
user26857user26857
39.4k124183
$endgroup$
add a comment |
$begingroup$
If you want to use the first isomorphism theorem, then define $psi:mathbb Z^n to mathbb Z^{n-1} oplus mathbb Z/langle a rangle$ by $psi(x_1,dots,x_n)=(x_2-x_1,dots,x_n-x_1,bar x_1)$. Check that $psi$ is a surjective homomorphism and $kerpsi=langle (a,dots,a) rangle$.
answered Dec 1 '14 at 20:30
user26857user26857
39.4k124183
$endgroup$
If you want to use the first isomorphism theorem, then define $psi:mathbb Z^n to mathbb Z^{n-1} oplus mathbb Z/langle a rangle$ by $psi(x_1,dots,x_n)=(x_2-x_1,dots,x_n-x_1,bar x_1)$. Check that $psi$ is a surjective homomorphism and $kerpsi=langle (a,dots,a) rangle$.
answered Dec 1 '14 at 20:30
user26857user26857
39.4k124183
answered Dec 1 '14 at 20:30
user26857user26857
39.4k124183
answered Dec 1 '14 at 20:30
user26857user26857
39.4k124183
answered Dec 1 '14 at 20:30
user26857user26857
39.4k124183
39.4k124183
add a comment |
add a comment |
$begingroup$
The following: $$(1,0,dots,0), (0,1,0,dots,0), dots, (0,dots,0,1,0), (1,1,dots, 1)$$ is a basis of $mathbb{Z}^n$ and $(a,dots,a)$ corresponds to $(0,dots,0,a)$ in this new basis. Your group is thus the isomorphic to the quotient:
$$mathbb{Z}^n / langle (0, dots, 0, a) rangle cong mathbb{Z}^{n-1} oplus mathbb{Z}/(a).$$
answered Dec 1 '14 at 20:19
Najib IdrissiNajib Idrissi
41.1k471139
$endgroup$
$begingroup$
@lhf : well, this is what I did 2 years ago: math.stackexchange.com/questions/1910883
$endgroup$
– Alphonse
Dec 21 '18 at 16:33
add a comment |
$begingroup$
The following: $$(1,0,dots,0), (0,1,0,dots,0), dots, (0,dots,0,1,0), (1,1,dots, 1)$$ is a basis of $mathbb{Z}^n$ and $(a,dots,a)$ corresponds to $(0,dots,0,a)$ in this new basis. Your group is thus the isomorphic to the quotient:
$$mathbb{Z}^n / langle (0, dots, 0, a) rangle cong mathbb{Z}^{n-1} oplus mathbb{Z}/(a).$$
answered Dec 1 '14 at 20:19
Najib IdrissiNajib Idrissi
41.1k471139
$endgroup$
$begingroup$
@lhf : well, this is what I did 2 years ago: math.stackexchange.com/questions/1910883
$endgroup$
– Alphonse
Dec 21 '18 at 16:33
add a comment |
$begingroup$
The following: $$(1,0,dots,0), (0,1,0,dots,0), dots, (0,dots,0,1,0), (1,1,dots, 1)$$ is a basis of $mathbb{Z}^n$ and $(a,dots,a)$ corresponds to $(0,dots,0,a)$ in this new basis. Your group is thus the isomorphic to the quotient:
$$mathbb{Z}^n / langle (0, dots, 0, a) rangle cong mathbb{Z}^{n-1} oplus mathbb{Z}/(a).$$
answered Dec 1 '14 at 20:19
Najib IdrissiNajib Idrissi
41.1k471139
$endgroup$
The following: $$(1,0,dots,0), (0,1,0,dots,0), dots, (0,dots,0,1,0), (1,1,dots, 1)$$ is a basis of $mathbb{Z}^n$ and $(a,dots,a)$ corresponds to $(0,dots,0,a)$ in this new basis. Your group is thus the isomorphic to the quotient:
$$mathbb{Z}^n / langle (0, dots, 0, a) rangle cong mathbb{Z}^{n-1} oplus mathbb{Z}/(a).$$
answered Dec 1 '14 at 20:19
Najib IdrissiNajib Idrissi
41.1k471139
edited Jan 2 '15 at 11:16
answered Dec 1 '14 at 20:19
Najib IdrissiNajib Idrissi
41.1k471139
answered Dec 1 '14 at 20:19
Najib IdrissiNajib Idrissi
41.1k471139
answered Dec 1 '14 at 20:19
Najib IdrissiNajib Idrissi
41.1k471139
41.1k471139
$begingroup$
@lhf : well, this is what I did 2 years ago: math.stackexchange.com/questions/1910883
$endgroup$
– Alphonse
Dec 21 '18 at 16:33
add a comment |
$begingroup$
@lhf : well, this is what I did 2 years ago: math.stackexchange.com/questions/1910883
$endgroup$
– Alphonse
Dec 21 '18 at 16:33
$begingroup$
@lhf : well, this is what I did 2 years ago: math.stackexchange.com/questions/1910883
$endgroup$
– Alphonse
Dec 21 '18 at 16:33
$begingroup$
@lhf : well, this is what I did 2 years ago: math.stackexchange.com/questions/1910883
$endgroup$
– Alphonse
Dec 21 '18 at 16:33
add a comment |
$begingroup$
$mathbb{Z}^n/langle (a,dotsc,a)$ is an abelian group with the presentation $langle e_1,dotsc,e_n : a e_1 + cdots + a e_n = 0}$. Observe that $e_1,dotsc,e_{n-1},e_1+cdots+e_n$ is also a generating system and $a(e_1+cdots+e_n)=0$. Hence, we obtain a homomorphism from $langle f_1,dotsc,f_n : a f_n = 0 rangle$. One then constructs its inverse.
edited Dec 21 '18 at 9:51
user26857
39.4k124183
answered Dec 1 '14 at 20:20
Martin BrandenburgMartin Brandenburg
108k13158328
$endgroup$
add a comment |
$begingroup$
$mathbb{Z}^n/langle (a,dotsc,a)$ is an abelian group with the presentation $langle e_1,dotsc,e_n : a e_1 + cdots + a e_n = 0}$. Observe that $e_1,dotsc,e_{n-1},e_1+cdots+e_n$ is also a generating system and $a(e_1+cdots+e_n)=0$. Hence, we obtain a homomorphism from $langle f_1,dotsc,f_n : a f_n = 0 rangle$. One then constructs its inverse.
edited Dec 21 '18 at 9:51
user26857
39.4k124183
answered Dec 1 '14 at 20:20
Martin BrandenburgMartin Brandenburg
108k13158328
$endgroup$
add a comment |
$begingroup$
$mathbb{Z}^n/langle (a,dotsc,a)$ is an abelian group with the presentation $langle e_1,dotsc,e_n : a e_1 + cdots + a e_n = 0}$. Observe that $e_1,dotsc,e_{n-1},e_1+cdots+e_n$ is also a generating system and $a(e_1+cdots+e_n)=0$. Hence, we obtain a homomorphism from $langle f_1,dotsc,f_n : a f_n = 0 rangle$. One then constructs its inverse.
edited Dec 21 '18 at 9:51
user26857
39.4k124183
answered Dec 1 '14 at 20:20
Martin BrandenburgMartin Brandenburg
108k13158328
$endgroup$
$mathbb{Z}^n/langle (a,dotsc,a)$ is an abelian group with the presentation $langle e_1,dotsc,e_n : a e_1 + cdots + a e_n = 0}$. Observe that $e_1,dotsc,e_{n-1},e_1+cdots+e_n$ is also a generating system and $a(e_1+cdots+e_n)=0$. Hence, we obtain a homomorphism from $langle f_1,dotsc,f_n : a f_n = 0 rangle$. One then constructs its inverse.
edited Dec 21 '18 at 9:51
user26857
39.4k124183
answered Dec 1 '14 at 20:20
Martin BrandenburgMartin Brandenburg
108k13158328
edited Dec 21 '18 at 9:51
user26857
39.4k124183
edited Dec 21 '18 at 9:51
user26857
39.4k124183
edited Dec 21 '18 at 9:51
user26857
39.4k124183
39.4k124183
answered Dec 1 '14 at 20:20
Martin BrandenburgMartin Brandenburg
108k13158328
answered Dec 1 '14 at 20:20
Martin BrandenburgMartin Brandenburg
108k13158328
answered Dec 1 '14 at 20:20
Martin BrandenburgMartin Brandenburg
108k13158328
108k13158328
add a comment |
add a comment |
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