Mixing
If $G/Hcong 0$, then $Gcong H$?
$begingroup$
Let $H$ be a subgroup of an abelian group $G$.
If $G/Hcong 0$, then is it true that $Gcong H$?
My attempt of proof:
Consider $psi: Hto G$, where $psi(h)=h$. Then, $kerpsi=0$, hence $psi$ is injective.
Suppose to the contrary $psi$ is not surjective, then there exist $gin G$ such that $gnotin H$. But then that would imply that $g+Hneq H$ which contradicts $G/Hcong 0$.
Is the proof correct?
Additional question: In general, if $M/Ncong 0$, is it always true that $Mcong N$, where $M$,$N$ could be rings/modules/algebras, etc?
Thanks a lot.
abstract-algebra group-theory
asked Jan 28 at 14:50
yoyosteinyoyostein
8,145103973
$endgroup$
add a comment |
$begingroup$
Let $H$ be a subgroup of an abelian group $G$.
If $G/Hcong 0$, then is it true that $Gcong H$?
My attempt of proof:
Consider $psi: Hto G$, where $psi(h)=h$. Then, $kerpsi=0$, hence $psi$ is injective.
Suppose to the contrary $psi$ is not surjective, then there exist $gin G$ such that $gnotin H$. But then that would imply that $g+Hneq H$ which contradicts $G/Hcong 0$.
Is the proof correct?
Additional question: In general, if $M/Ncong 0$, is it always true that $Mcong N$, where $M$,$N$ could be rings/modules/algebras, etc?
Thanks a lot.
abstract-algebra group-theory
asked Jan 28 at 14:50
yoyosteinyoyostein
8,145103973
$endgroup$
8
$begingroup$
Even better: $G=H$.
$endgroup$
– Randall
Jan 28 at 14:52
1
$begingroup$
Answer to additional question: yes, since those all imply the isomorphism as abelian groups, for which you have already seen $M=N$ in other comments.
$endgroup$
– rschwieb
Jan 28 at 14:55
add a comment |
$begingroup$
Let $H$ be a subgroup of an abelian group $G$.
If $G/Hcong 0$, then is it true that $Gcong H$?
My attempt of proof:
Consider $psi: Hto G$, where $psi(h)=h$. Then, $kerpsi=0$, hence $psi$ is injective.
Suppose to the contrary $psi$ is not surjective, then there exist $gin G$ such that $gnotin H$. But then that would imply that $g+Hneq H$ which contradicts $G/Hcong 0$.
Is the proof correct?
Additional question: In general, if $M/Ncong 0$, is it always true that $Mcong N$, where $M$,$N$ could be rings/modules/algebras, etc?
Thanks a lot.
abstract-algebra group-theory
asked Jan 28 at 14:50
yoyosteinyoyostein
8,145103973
$endgroup$
Let $H$ be a subgroup of an abelian group $G$.
If $G/Hcong 0$, then is it true that $Gcong H$?
My attempt of proof:
Consider $psi: Hto G$, where $psi(h)=h$. Then, $kerpsi=0$, hence $psi$ is injective.
Suppose to the contrary $psi$ is not surjective, then there exist $gin G$ such that $gnotin H$. But then that would imply that $g+Hneq H$ which contradicts $G/Hcong 0$.
Is the proof correct?
Additional question: In general, if $M/Ncong 0$, is it always true that $Mcong N$, where $M$,$N$ could be rings/modules/algebras, etc?
Thanks a lot.
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 28 at 14:50
yoyosteinyoyostein
8,145103973
asked Jan 28 at 14:50
yoyosteinyoyostein
8,145103973
asked Jan 28 at 14:50
yoyosteinyoyostein
8,145103973
asked Jan 28 at 14:50
yoyosteinyoyostein
8,145103973
asked Jan 28 at 14:50
yoyosteinyoyostein
8,145103973
8,145103973
8
$begingroup$
Even better: $G=H$.
$endgroup$
– Randall
Jan 28 at 14:52
1
$begingroup$
Answer to additional question: yes, since those all imply the isomorphism as abelian groups, for which you have already seen $M=N$ in other comments.
$endgroup$
– rschwieb
Jan 28 at 14:55
add a comment |
8
$begingroup$
Even better: $G=H$.
$endgroup$
– Randall
Jan 28 at 14:52
1
$begingroup$
Answer to additional question: yes, since those all imply the isomorphism as abelian groups, for which you have already seen $M=N$ in other comments.
$endgroup$
– rschwieb
Jan 28 at 14:55
8
8
$begingroup$
Even better: $G=H$.
$endgroup$
– Randall
Jan 28 at 14:52
$begingroup$
Even better: $G=H$.
$endgroup$
– Randall
Jan 28 at 14:52
1
1
$begingroup$
Answer to additional question: yes, since those all imply the isomorphism as abelian groups, for which you have already seen $M=N$ in other comments.
$endgroup$
– rschwieb
Jan 28 at 14:55
$begingroup$
Answer to additional question: yes, since those all imply the isomorphism as abelian groups, for which you have already seen $M=N$ in other comments.
$endgroup$
– rschwieb
Jan 28 at 14:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Think about the canonical projection $pcolon G rightarrow G/H$. We know that $mathrm{ker}, p = H$. But if the image $G/H = 0$, then the kernel has to be everything; i.e. $H = mathrm{ker}, p = G$.
The same approach works for rings, modules, etc. Usually the most efficient way to prove some fact about quotient objects will only involve the canonical projection and its universal property.
answered Jan 28 at 14:53
o.h.o.h.
6917
$endgroup$
add a comment |
$begingroup$
If $[g]in G/H$ then $gsim 0Rightarrow g-0=gin HRightarrow Gsubseteq HRightarrow G=H.$
answered Jan 28 at 15:34
MatematletaMatematleta
11.8k2920
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Think about the canonical projection $pcolon G rightarrow G/H$. We know that $mathrm{ker}, p = H$. But if the image $G/H = 0$, then the kernel has to be everything; i.e. $H = mathrm{ker}, p = G$.
The same approach works for rings, modules, etc. Usually the most efficient way to prove some fact about quotient objects will only involve the canonical projection and its universal property.
answered Jan 28 at 14:53
o.h.o.h.
6917
$endgroup$
add a comment |
$begingroup$
Think about the canonical projection $pcolon G rightarrow G/H$. We know that $mathrm{ker}, p = H$. But if the image $G/H = 0$, then the kernel has to be everything; i.e. $H = mathrm{ker}, p = G$.
The same approach works for rings, modules, etc. Usually the most efficient way to prove some fact about quotient objects will only involve the canonical projection and its universal property.
answered Jan 28 at 14:53
o.h.o.h.
6917
$endgroup$
add a comment |
$begingroup$
Think about the canonical projection $pcolon G rightarrow G/H$. We know that $mathrm{ker}, p = H$. But if the image $G/H = 0$, then the kernel has to be everything; i.e. $H = mathrm{ker}, p = G$.
The same approach works for rings, modules, etc. Usually the most efficient way to prove some fact about quotient objects will only involve the canonical projection and its universal property.
answered Jan 28 at 14:53
o.h.o.h.
6917
$endgroup$
Think about the canonical projection $pcolon G rightarrow G/H$. We know that $mathrm{ker}, p = H$. But if the image $G/H = 0$, then the kernel has to be everything; i.e. $H = mathrm{ker}, p = G$.
The same approach works for rings, modules, etc. Usually the most efficient way to prove some fact about quotient objects will only involve the canonical projection and its universal property.
answered Jan 28 at 14:53
o.h.o.h.
6917
edited Jan 28 at 15:08
answered Jan 28 at 14:53
o.h.o.h.
6917
answered Jan 28 at 14:53
o.h.o.h.
6917
answered Jan 28 at 14:53
o.h.o.h.
6917
6917
add a comment |
add a comment |
$begingroup$
If $[g]in G/H$ then $gsim 0Rightarrow g-0=gin HRightarrow Gsubseteq HRightarrow G=H.$
answered Jan 28 at 15:34
MatematletaMatematleta
11.8k2920
$endgroup$
add a comment |
$begingroup$
If $[g]in G/H$ then $gsim 0Rightarrow g-0=gin HRightarrow Gsubseteq HRightarrow G=H.$
answered Jan 28 at 15:34
MatematletaMatematleta
11.8k2920
$endgroup$
add a comment |
$begingroup$
If $[g]in G/H$ then $gsim 0Rightarrow g-0=gin HRightarrow Gsubseteq HRightarrow G=H.$
answered Jan 28 at 15:34
MatematletaMatematleta
11.8k2920
$endgroup$
If $[g]in G/H$ then $gsim 0Rightarrow g-0=gin HRightarrow Gsubseteq HRightarrow G=H.$
answered Jan 28 at 15:34
MatematletaMatematleta
11.8k2920
answered Jan 28 at 15:34
MatematletaMatematleta
11.8k2920
answered Jan 28 at 15:34
MatematletaMatematleta
11.8k2920
answered Jan 28 at 15:34
MatematletaMatematleta
11.8k2920
11.8k2920
add a comment |
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8
$begingroup$
Even better: $G=H$.
$endgroup$
– Randall
Jan 28 at 14:52
1
$begingroup$
Answer to additional question: yes, since those all imply the isomorphism as abelian groups, for which you have already seen $M=N$ in other comments.
$endgroup$
– rschwieb
Jan 28 at 14:55