Hatcher 2.1.14 last part
$begingroup$
This exercise asks to find all abelian groups that fit in the short exact sequence
$0to mathbb{Z}to Ato mathbb{Z}_nto 0$
I've proved that $A$ must be isomorphic to $mathbb{Z}oplus mathbb{Z}_d$ with $d|n$, but I've been unable check that this group fits.
I have to define an injective homomorphism $phi:mathbb{Z}to mathbb{Z}oplus mathbb{Z}_d$. I guess that $phi(1)=(1,x)$, where $x$ must be something that makes $mathrm{coker}(phi)congmathbb{Z}_n$. In the solutions I've read $x=n/d$, so $phi(1)=1cdot (1,0)+n/dcdot(0,1)$ but I can't show that that makes what I want. A presentation of $mathrm{coker}(phi)=langle a,bmid db=0, a+bn/d=0rangle$. If I multiply by $d$, $da=0$, so this is actually a subgroup of $mathbb{Z}_doplusmathbb{Z}_d$, which has no element of order $n$ in general, and therefore cannot be isomorphic to $mathbb{Z}_n$.
What am I doing wrong o what should I take as $x$?
abstract-algebra group-theory homological-algebra abelian-groups exact-sequence
asked Dec 6 '18 at 20:02
JaviJavi
2,6032826
$endgroup$
add a comment |
$begingroup$
This exercise asks to find all abelian groups that fit in the short exact sequence
$0to mathbb{Z}to Ato mathbb{Z}_nto 0$
I've proved that $A$ must be isomorphic to $mathbb{Z}oplus mathbb{Z}_d$ with $d|n$, but I've been unable check that this group fits.
I have to define an injective homomorphism $phi:mathbb{Z}to mathbb{Z}oplus mathbb{Z}_d$. I guess that $phi(1)=(1,x)$, where $x$ must be something that makes $mathrm{coker}(phi)congmathbb{Z}_n$. In the solutions I've read $x=n/d$, so $phi(1)=1cdot (1,0)+n/dcdot(0,1)$ but I can't show that that makes what I want. A presentation of $mathrm{coker}(phi)=langle a,bmid db=0, a+bn/d=0rangle$. If I multiply by $d$, $da=0$, so this is actually a subgroup of $mathbb{Z}_doplusmathbb{Z}_d$, which has no element of order $n$ in general, and therefore cannot be isomorphic to $mathbb{Z}_n$.
What am I doing wrong o what should I take as $x$?
abstract-algebra group-theory homological-algebra abelian-groups exact-sequence
asked Dec 6 '18 at 20:02
JaviJavi
2,6032826
$endgroup$
add a comment |
$begingroup$
This exercise asks to find all abelian groups that fit in the short exact sequence
$0to mathbb{Z}to Ato mathbb{Z}_nto 0$
I've proved that $A$ must be isomorphic to $mathbb{Z}oplus mathbb{Z}_d$ with $d|n$, but I've been unable check that this group fits.
I have to define an injective homomorphism $phi:mathbb{Z}to mathbb{Z}oplus mathbb{Z}_d$. I guess that $phi(1)=(1,x)$, where $x$ must be something that makes $mathrm{coker}(phi)congmathbb{Z}_n$. In the solutions I've read $x=n/d$, so $phi(1)=1cdot (1,0)+n/dcdot(0,1)$ but I can't show that that makes what I want. A presentation of $mathrm{coker}(phi)=langle a,bmid db=0, a+bn/d=0rangle$. If I multiply by $d$, $da=0$, so this is actually a subgroup of $mathbb{Z}_doplusmathbb{Z}_d$, which has no element of order $n$ in general, and therefore cannot be isomorphic to $mathbb{Z}_n$.
What am I doing wrong o what should I take as $x$?
abstract-algebra group-theory homological-algebra abelian-groups exact-sequence
asked Dec 6 '18 at 20:02
JaviJavi
2,6032826
$endgroup$
This exercise asks to find all abelian groups that fit in the short exact sequence
$0to mathbb{Z}to Ato mathbb{Z}_nto 0$
I've proved that $A$ must be isomorphic to $mathbb{Z}oplus mathbb{Z}_d$ with $d|n$, but I've been unable check that this group fits.
I have to define an injective homomorphism $phi:mathbb{Z}to mathbb{Z}oplus mathbb{Z}_d$. I guess that $phi(1)=(1,x)$, where $x$ must be something that makes $mathrm{coker}(phi)congmathbb{Z}_n$. In the solutions I've read $x=n/d$, so $phi(1)=1cdot (1,0)+n/dcdot(0,1)$ but I can't show that that makes what I want. A presentation of $mathrm{coker}(phi)=langle a,bmid db=0, a+bn/d=0rangle$. If I multiply by $d$, $da=0$, so this is actually a subgroup of $mathbb{Z}_doplusmathbb{Z}_d$, which has no element of order $n$ in general, and therefore cannot be isomorphic to $mathbb{Z}_n$.
What am I doing wrong o what should I take as $x$?
abstract-algebra group-theory homological-algebra abelian-groups exact-sequence
abstract-algebra group-theory homological-algebra abelian-groups exact-sequence
asked Dec 6 '18 at 20:02
JaviJavi
2,6032826
asked Dec 6 '18 at 20:02
JaviJavi
2,6032826
asked Dec 6 '18 at 20:02
JaviJavi
2,6032826
asked Dec 6 '18 at 20:02
JaviJavi
2,6032826
asked Dec 6 '18 at 20:02
JaviJavi
2,6032826
2,6032826
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Let's define the surjection $pi:Bbb ZoplusBbb Z_dtoBbb Z_n$ instead.
Write $n=cd$, and define $pi(x,y)=x+cy$. The kernel is generated by
$(c,-1)$ and is isomorphic to $Bbb Z$.
answered Dec 6 '18 at 20:32
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's define the surjection $pi:Bbb ZoplusBbb Z_dtoBbb Z_n$ instead.
Write $n=cd$, and define $pi(x,y)=x+cy$. The kernel is generated by
$(c,-1)$ and is isomorphic to $Bbb Z$.
answered Dec 6 '18 at 20:32
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
add a comment |
$begingroup$
Let's define the surjection $pi:Bbb ZoplusBbb Z_dtoBbb Z_n$ instead.
Write $n=cd$, and define $pi(x,y)=x+cy$. The kernel is generated by
$(c,-1)$ and is isomorphic to $Bbb Z$.
answered Dec 6 '18 at 20:32
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
add a comment |
$begingroup$
Let's define the surjection $pi:Bbb ZoplusBbb Z_dtoBbb Z_n$ instead.
Write $n=cd$, and define $pi(x,y)=x+cy$. The kernel is generated by
$(c,-1)$ and is isomorphic to $Bbb Z$.
answered Dec 6 '18 at 20:32
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
Let's define the surjection $pi:Bbb ZoplusBbb Z_dtoBbb Z_n$ instead.
Write $n=cd$, and define $pi(x,y)=x+cy$. The kernel is generated by
$(c,-1)$ and is isomorphic to $Bbb Z$.
answered Dec 6 '18 at 20:32
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Dec 6 '18 at 20:32
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Dec 6 '18 at 20:32
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Dec 6 '18 at 20:32
Lord Shark the UnknownLord Shark the Unknown
103k1160132
103k1160132
add a comment |
add a comment |
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