Z-module isomorphism
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Let n,d be positive integer numbers such that d|n. Show that $<frac{n}{d}>$/$<n>$ Is isomorphic as a module to $mathbb{Z}_{d}$
abstract-algebra modules
asked Nov 28 at 1:10
Eduardo
32
add a comment |
up vote
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down vote
favorite
Let n,d be positive integer numbers such that d|n. Show that $<frac{n}{d}>$/$<n>$ Is isomorphic as a module to $mathbb{Z}_{d}$
abstract-algebra modules
asked Nov 28 at 1:10
Eduardo
32
What have you tried? Perhaps you can define a map sending one group to the other?
– Aaron
Nov 28 at 1:18
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let n,d be positive integer numbers such that d|n. Show that $<frac{n}{d}>$/$<n>$ Is isomorphic as a module to $mathbb{Z}_{d}$
abstract-algebra modules
asked Nov 28 at 1:10
Eduardo
32
Let n,d be positive integer numbers such that d|n. Show that $<frac{n}{d}>$/$<n>$ Is isomorphic as a module to $mathbb{Z}_{d}$
abstract-algebra modules
abstract-algebra modules
asked Nov 28 at 1:10
Eduardo
32
asked Nov 28 at 1:10
Eduardo
32
edited Nov 28 at 1:15
asked Nov 28 at 1:10
Eduardo
32
asked Nov 28 at 1:10
Eduardo
32
asked Nov 28 at 1:10
Eduardo
32
32
What have you tried? Perhaps you can define a map sending one group to the other?
– Aaron
Nov 28 at 1:18
add a comment |
What have you tried? Perhaps you can define a map sending one group to the other?
– Aaron
Nov 28 at 1:18
What have you tried? Perhaps you can define a map sending one group to the other?
– Aaron
Nov 28 at 1:18
What have you tried? Perhaps you can define a map sending one group to the other?
– Aaron
Nov 28 at 1:18
add a comment |
1 Answer
1
active
oldest
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0
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accepted
$Bbb Z$-modules are just abelian groups.
You can try to find a homomorphism $<n/d> to mathbb Z_d$ with kernel $<n>$ and apply the First Isomorphism Theorem.
edited Nov 28 at 1:32
Aaron
15.7k22653
answered Nov 28 at 1:30
Lukas Kofler
1,2552519
$<n/d> to mathbb Z_d$ $k=x.n/d |to x$ works. Am l right?
– Eduardo
Nov 30 at 3:09
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
$Bbb Z$-modules are just abelian groups.
You can try to find a homomorphism $<n/d> to mathbb Z_d$ with kernel $<n>$ and apply the First Isomorphism Theorem.
edited Nov 28 at 1:32
Aaron
15.7k22653
answered Nov 28 at 1:30
Lukas Kofler
1,2552519
$<n/d> to mathbb Z_d$ $k=x.n/d |to x$ works. Am l right?
– Eduardo
Nov 30 at 3:09
add a comment |
up vote
0
down vote
accepted
$Bbb Z$-modules are just abelian groups.
You can try to find a homomorphism $<n/d> to mathbb Z_d$ with kernel $<n>$ and apply the First Isomorphism Theorem.
edited Nov 28 at 1:32
Aaron
15.7k22653
answered Nov 28 at 1:30
Lukas Kofler
1,2552519
$<n/d> to mathbb Z_d$ $k=x.n/d |to x$ works. Am l right?
– Eduardo
Nov 30 at 3:09
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
$Bbb Z$-modules are just abelian groups.
You can try to find a homomorphism $<n/d> to mathbb Z_d$ with kernel $<n>$ and apply the First Isomorphism Theorem.
edited Nov 28 at 1:32
Aaron
15.7k22653
answered Nov 28 at 1:30
Lukas Kofler
1,2552519
$Bbb Z$-modules are just abelian groups.
You can try to find a homomorphism $<n/d> to mathbb Z_d$ with kernel $<n>$ and apply the First Isomorphism Theorem.
edited Nov 28 at 1:32
Aaron
15.7k22653
answered Nov 28 at 1:30
Lukas Kofler
1,2552519
edited Nov 28 at 1:32
Aaron
15.7k22653
edited Nov 28 at 1:32
Aaron
15.7k22653
edited Nov 28 at 1:32
Aaron
15.7k22653
15.7k22653
answered Nov 28 at 1:30
Lukas Kofler
1,2552519
answered Nov 28 at 1:30
Lukas Kofler
1,2552519
answered Nov 28 at 1:30
Lukas Kofler
1,2552519
1,2552519
$<n/d> to mathbb Z_d$ $k=x.n/d |to x$ works. Am l right?
– Eduardo
Nov 30 at 3:09
add a comment |
$<n/d> to mathbb Z_d$ $k=x.n/d |to x$ works. Am l right?
– Eduardo
Nov 30 at 3:09
$<n/d> to mathbb Z_d$ $k=x.n/d |to x$ works. Am l right?
– Eduardo
Nov 30 at 3:09
$<n/d> to mathbb Z_d$ $k=x.n/d |to x$ works. Am l right?
– Eduardo
Nov 30 at 3:09
add a comment |
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What have you tried? Perhaps you can define a map sending one group to the other?
– Aaron
Nov 28 at 1:18