Prove or disprove. Let $G$ be an abelian group. Let $N triangleleft G$. Then, $ Gcong N times G/N. $...
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This question already has an answer here:
if $A$ is Abelian group , $B$ is subgroup of $A$ , Is $B times A/B cong A$? [duplicate]
2 answers
Prove or disprove. Let $G$ be an abelian group. Let $N triangleleft G$. Then,
$$
Gcong N times G/N.
$$
I tried to prove this claim, but then it seems that since $G$ is abelian then every subgroup is normal. So every group can be factored as above. But I did not find a counter example.
Any suggestions?
Thanks for helpers!
group-theory examples-counterexamples abelian-groups normal-subgroups group-isomorphism
edited Jan 22 at 17:15
Martin Sleziak
44.7k10119272
asked Jan 22 at 9:25
Shlomo WaknineShlomo Waknine
92
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marked as duplicate by Shaun, Did, José Carlos Santos, user 170039, Martin R Jan 23 at 8:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
if $A$ is Abelian group , $B$ is subgroup of $A$ , Is $B times A/B cong A$? [duplicate]
2 answers
Prove or disprove. Let $G$ be an abelian group. Let $N triangleleft G$. Then,
$$
Gcong N times G/N.
$$
I tried to prove this claim, but then it seems that since $G$ is abelian then every subgroup is normal. So every group can be factored as above. But I did not find a counter example.
Any suggestions?
Thanks for helpers!
group-theory examples-counterexamples abelian-groups normal-subgroups group-isomorphism
edited Jan 22 at 17:15
Martin Sleziak
44.7k10119272
asked Jan 22 at 9:25
Shlomo WaknineShlomo Waknine
92
$endgroup$
marked as duplicate by Shaun, Did, José Carlos Santos, user 170039, Martin R Jan 23 at 8:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
if $A$ is Abelian group , $B$ is subgroup of $A$ , Is $B times A/B cong A$? [duplicate]
2 answers
Prove or disprove. Let $G$ be an abelian group. Let $N triangleleft G$. Then,
$$
Gcong N times G/N.
$$
I tried to prove this claim, but then it seems that since $G$ is abelian then every subgroup is normal. So every group can be factored as above. But I did not find a counter example.
Any suggestions?
Thanks for helpers!
group-theory examples-counterexamples abelian-groups normal-subgroups group-isomorphism
edited Jan 22 at 17:15
Martin Sleziak
44.7k10119272
asked Jan 22 at 9:25
Shlomo WaknineShlomo Waknine
92
$endgroup$
This question already has an answer here:
if $A$ is Abelian group , $B$ is subgroup of $A$ , Is $B times A/B cong A$? [duplicate]
2 answers
Prove or disprove. Let $G$ be an abelian group. Let $N triangleleft G$. Then,
$$
Gcong N times G/N.
$$
I tried to prove this claim, but then it seems that since $G$ is abelian then every subgroup is normal. So every group can be factored as above. But I did not find a counter example.
Any suggestions?
Thanks for helpers!
This question already has an answer here:
if $A$ is Abelian group , $B$ is subgroup of $A$ , Is $B times A/B cong A$? [duplicate]
2 answers
group-theory examples-counterexamples abelian-groups normal-subgroups group-isomorphism
group-theory examples-counterexamples abelian-groups normal-subgroups group-isomorphism
edited Jan 22 at 17:15
Martin Sleziak
44.7k10119272
asked Jan 22 at 9:25
Shlomo WaknineShlomo Waknine
92
edited Jan 22 at 17:15
Martin Sleziak
44.7k10119272
asked Jan 22 at 9:25
Shlomo WaknineShlomo Waknine
92
edited Jan 22 at 17:15
Martin Sleziak
44.7k10119272
edited Jan 22 at 17:15
Martin Sleziak
44.7k10119272
edited Jan 22 at 17:15
Martin Sleziak
44.7k10119272
44.7k10119272
asked Jan 22 at 9:25
Shlomo WaknineShlomo Waknine
92
asked Jan 22 at 9:25
Shlomo WaknineShlomo Waknine
92
asked Jan 22 at 9:25
Shlomo WaknineShlomo Waknine
92
92
marked as duplicate by Shaun, Did, José Carlos Santos, user 170039, Martin R Jan 23 at 8:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Shaun, Did, José Carlos Santos, user 170039, Martin R Jan 23 at 8:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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Take the cyclic group $G=C_{p^2}$ and $N=C_p$ for $p$ prime. Then it is not true that
$$
G=C_{p^2}cong C_ptimes C_p=Ntimes G/N,
$$
because the group $C_ptimes C_p$ is not cyclic.
edited Jan 22 at 14:57
Shaun
9,300113684
answered Jan 22 at 9:30
Dietrich BurdeDietrich Burde
79.8k647103
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add a comment |
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Hint: Is $mathbb{Z}_4$ isomorphic to $mathbb{Z}_2timesmathbb{Z}_2$?
answered Jan 22 at 9:29
José Carlos SantosJosé Carlos Santos
164k22131234
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add a comment |
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Take $G=mathbb{Z}timesmathbb{Z}$ and $N=0times 2mathbb{Z}$, $G/N$ has a non-trivial element of finite order, but not $G$.
edited Jan 22 at 17:37
user1729
17.2k64193
answered Jan 22 at 9:30
Tsemo AristideTsemo Aristide
58.7k11445
$endgroup$
$begingroup$
Why $G=mathbb{Z}times mathbb{Z}$ rather than just $G=mathbb{Z}$?!
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– user1729
Jan 22 at 17:36
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take the cyclic group $G=C_{p^2}$ and $N=C_p$ for $p$ prime. Then it is not true that
$$
G=C_{p^2}cong C_ptimes C_p=Ntimes G/N,
$$
because the group $C_ptimes C_p$ is not cyclic.
edited Jan 22 at 14:57
Shaun
9,300113684
answered Jan 22 at 9:30
Dietrich BurdeDietrich Burde
79.8k647103
$endgroup$
add a comment |
$begingroup$
Take the cyclic group $G=C_{p^2}$ and $N=C_p$ for $p$ prime. Then it is not true that
$$
G=C_{p^2}cong C_ptimes C_p=Ntimes G/N,
$$
because the group $C_ptimes C_p$ is not cyclic.
edited Jan 22 at 14:57
Shaun
9,300113684
answered Jan 22 at 9:30
Dietrich BurdeDietrich Burde
79.8k647103
$endgroup$
add a comment |
$begingroup$
Take the cyclic group $G=C_{p^2}$ and $N=C_p$ for $p$ prime. Then it is not true that
$$
G=C_{p^2}cong C_ptimes C_p=Ntimes G/N,
$$
because the group $C_ptimes C_p$ is not cyclic.
edited Jan 22 at 14:57
Shaun
9,300113684
answered Jan 22 at 9:30
Dietrich BurdeDietrich Burde
79.8k647103
$endgroup$
Take the cyclic group $G=C_{p^2}$ and $N=C_p$ for $p$ prime. Then it is not true that
$$
G=C_{p^2}cong C_ptimes C_p=Ntimes G/N,
$$
because the group $C_ptimes C_p$ is not cyclic.
edited Jan 22 at 14:57
Shaun
9,300113684
answered Jan 22 at 9:30
Dietrich BurdeDietrich Burde
79.8k647103
edited Jan 22 at 14:57
Shaun
9,300113684
edited Jan 22 at 14:57
Shaun
9,300113684
edited Jan 22 at 14:57
Shaun
9,300113684
9,300113684
answered Jan 22 at 9:30
Dietrich BurdeDietrich Burde
79.8k647103
answered Jan 22 at 9:30
Dietrich BurdeDietrich Burde
79.8k647103
answered Jan 22 at 9:30
Dietrich BurdeDietrich Burde
79.8k647103
79.8k647103
add a comment |
add a comment |
$begingroup$
Hint: Is $mathbb{Z}_4$ isomorphic to $mathbb{Z}_2timesmathbb{Z}_2$?
answered Jan 22 at 9:29
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
add a comment |
$begingroup$
Hint: Is $mathbb{Z}_4$ isomorphic to $mathbb{Z}_2timesmathbb{Z}_2$?
answered Jan 22 at 9:29
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
add a comment |
$begingroup$
Hint: Is $mathbb{Z}_4$ isomorphic to $mathbb{Z}_2timesmathbb{Z}_2$?
answered Jan 22 at 9:29
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
Hint: Is $mathbb{Z}_4$ isomorphic to $mathbb{Z}_2timesmathbb{Z}_2$?
answered Jan 22 at 9:29
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 22 at 9:29
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 22 at 9:29
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jan 22 at 9:29
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
add a comment |
add a comment |
$begingroup$
Take $G=mathbb{Z}timesmathbb{Z}$ and $N=0times 2mathbb{Z}$, $G/N$ has a non-trivial element of finite order, but not $G$.
edited Jan 22 at 17:37
user1729
17.2k64193
answered Jan 22 at 9:30
Tsemo AristideTsemo Aristide
58.7k11445
$endgroup$
$begingroup$
Why $G=mathbb{Z}times mathbb{Z}$ rather than just $G=mathbb{Z}$?!
$endgroup$
– user1729
Jan 22 at 17:36
add a comment |
$begingroup$
Take $G=mathbb{Z}timesmathbb{Z}$ and $N=0times 2mathbb{Z}$, $G/N$ has a non-trivial element of finite order, but not $G$.
edited Jan 22 at 17:37
user1729
17.2k64193
answered Jan 22 at 9:30
Tsemo AristideTsemo Aristide
58.7k11445
$endgroup$
$begingroup$
Why $G=mathbb{Z}times mathbb{Z}$ rather than just $G=mathbb{Z}$?!
$endgroup$
– user1729
Jan 22 at 17:36
add a comment |
$begingroup$
Take $G=mathbb{Z}timesmathbb{Z}$ and $N=0times 2mathbb{Z}$, $G/N$ has a non-trivial element of finite order, but not $G$.
edited Jan 22 at 17:37
user1729
17.2k64193
answered Jan 22 at 9:30
Tsemo AristideTsemo Aristide
58.7k11445
$endgroup$
Take $G=mathbb{Z}timesmathbb{Z}$ and $N=0times 2mathbb{Z}$, $G/N$ has a non-trivial element of finite order, but not $G$.
edited Jan 22 at 17:37
user1729
17.2k64193
answered Jan 22 at 9:30
Tsemo AristideTsemo Aristide
58.7k11445
edited Jan 22 at 17:37
user1729
17.2k64193
edited Jan 22 at 17:37
user1729
17.2k64193
edited Jan 22 at 17:37
user1729
17.2k64193
17.2k64193
answered Jan 22 at 9:30
Tsemo AristideTsemo Aristide
58.7k11445
answered Jan 22 at 9:30
Tsemo AristideTsemo Aristide
58.7k11445
answered Jan 22 at 9:30
Tsemo AristideTsemo Aristide
58.7k11445
58.7k11445
$begingroup$
Why $G=mathbb{Z}times mathbb{Z}$ rather than just $G=mathbb{Z}$?!
$endgroup$
– user1729
Jan 22 at 17:36
add a comment |
$begingroup$
Why $G=mathbb{Z}times mathbb{Z}$ rather than just $G=mathbb{Z}$?!
$endgroup$
– user1729
Jan 22 at 17:36
$begingroup$
Why $G=mathbb{Z}times mathbb{Z}$ rather than just $G=mathbb{Z}$?!
$endgroup$
– user1729
Jan 22 at 17:36
$begingroup$
Why $G=mathbb{Z}times mathbb{Z}$ rather than just $G=mathbb{Z}$?!
$endgroup$
– user1729
Jan 22 at 17:36
add a comment |
