Mixing
If $K$ is normal and $f$ is not epi, can $f(K)$ be normal?
$begingroup$
I'm reading Robert Ash's Basic Abstract Algebra.
I've read and seen the following proposition with proof:
If $f:Gto H$ is an epimorphism and $K$ is normal, then $f(K)$ is also normal.
I have a further question: Suppose $f$ is not an epimorphism, can $f(K)$ still be normal at least in some specific cases? It seems to me that in some cases, we could have $hf(K)h^{-1}=f(K)$ for all $hin H$ and yet, some $hin H$ won't have a $gin G$ such that $f(g)=h$.
abstract-algebra group-theory
edited Jan 8 at 8:02
j.p.
9031118
asked Jan 8 at 7:51
Billy RubinaBilly Rubina
10.4k1458134
$endgroup$
add a comment |
$begingroup$
I'm reading Robert Ash's Basic Abstract Algebra.
I've read and seen the following proposition with proof:
If $f:Gto H$ is an epimorphism and $K$ is normal, then $f(K)$ is also normal.
I have a further question: Suppose $f$ is not an epimorphism, can $f(K)$ still be normal at least in some specific cases? It seems to me that in some cases, we could have $hf(K)h^{-1}=f(K)$ for all $hin H$ and yet, some $hin H$ won't have a $gin G$ such that $f(g)=h$.
abstract-algebra group-theory
edited Jan 8 at 8:02
j.p.
9031118
asked Jan 8 at 7:51
Billy RubinaBilly Rubina
10.4k1458134
$endgroup$
4
$begingroup$
That's precisely the problem: $f(K)$ will be normal in $f(G)$ but not necessarily in larger subgroups of $H$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:54
7
$begingroup$
The answer to the question that you have asked is clearly yes. Of course it can be normal, for example when each of the groups is abelian. The point is that it is $f(G)$ is not necessarily normal in $H$ when $f$ is not an epimorphism.
$endgroup$
– Derek Holt
Jan 8 at 8:28
add a comment |
$begingroup$
I'm reading Robert Ash's Basic Abstract Algebra.
I've read and seen the following proposition with proof:
If $f:Gto H$ is an epimorphism and $K$ is normal, then $f(K)$ is also normal.
I have a further question: Suppose $f$ is not an epimorphism, can $f(K)$ still be normal at least in some specific cases? It seems to me that in some cases, we could have $hf(K)h^{-1}=f(K)$ for all $hin H$ and yet, some $hin H$ won't have a $gin G$ such that $f(g)=h$.
abstract-algebra group-theory
edited Jan 8 at 8:02
j.p.
9031118
asked Jan 8 at 7:51
Billy RubinaBilly Rubina
10.4k1458134
$endgroup$
I'm reading Robert Ash's Basic Abstract Algebra.
I've read and seen the following proposition with proof:
If $f:Gto H$ is an epimorphism and $K$ is normal, then $f(K)$ is also normal.
I have a further question: Suppose $f$ is not an epimorphism, can $f(K)$ still be normal at least in some specific cases? It seems to me that in some cases, we could have $hf(K)h^{-1}=f(K)$ for all $hin H$ and yet, some $hin H$ won't have a $gin G$ such that $f(g)=h$.
abstract-algebra group-theory
abstract-algebra group-theory
edited Jan 8 at 8:02
j.p.
9031118
asked Jan 8 at 7:51
Billy RubinaBilly Rubina
10.4k1458134
edited Jan 8 at 8:02
j.p.
9031118
asked Jan 8 at 7:51
Billy RubinaBilly Rubina
10.4k1458134
edited Jan 8 at 8:02
j.p.
9031118
edited Jan 8 at 8:02
j.p.
9031118
edited Jan 8 at 8:02
j.p.
9031118
9031118
asked Jan 8 at 7:51
Billy RubinaBilly Rubina
10.4k1458134
asked Jan 8 at 7:51
Billy RubinaBilly Rubina
10.4k1458134
asked Jan 8 at 7:51
Billy RubinaBilly Rubina
10.4k1458134
10.4k1458134
4
$begingroup$
That's precisely the problem: $f(K)$ will be normal in $f(G)$ but not necessarily in larger subgroups of $H$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:54
7
$begingroup$
The answer to the question that you have asked is clearly yes. Of course it can be normal, for example when each of the groups is abelian. The point is that it is $f(G)$ is not necessarily normal in $H$ when $f$ is not an epimorphism.
$endgroup$
– Derek Holt
Jan 8 at 8:28
add a comment |
4
$begingroup$
That's precisely the problem: $f(K)$ will be normal in $f(G)$ but not necessarily in larger subgroups of $H$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:54
7
$begingroup$
The answer to the question that you have asked is clearly yes. Of course it can be normal, for example when each of the groups is abelian. The point is that it is $f(G)$ is not necessarily normal in $H$ when $f$ is not an epimorphism.
$endgroup$
– Derek Holt
Jan 8 at 8:28
4
4
$begingroup$
That's precisely the problem: $f(K)$ will be normal in $f(G)$ but not necessarily in larger subgroups of $H$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:54
$begingroup$
That's precisely the problem: $f(K)$ will be normal in $f(G)$ but not necessarily in larger subgroups of $H$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:54
7
7
$begingroup$
The answer to the question that you have asked is clearly yes. Of course it can be normal, for example when each of the groups is abelian. The point is that it is $f(G)$ is not necessarily normal in $H$ when $f$ is not an epimorphism.
$endgroup$
– Derek Holt
Jan 8 at 8:28
$begingroup$
The answer to the question that you have asked is clearly yes. Of course it can be normal, for example when each of the groups is abelian. The point is that it is $f(G)$ is not necessarily normal in $H$ when $f$ is not an epimorphism.
$endgroup$
– Derek Holt
Jan 8 at 8:28
add a comment |
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4
$begingroup$
That's precisely the problem: $f(K)$ will be normal in $f(G)$ but not necessarily in larger subgroups of $H$.
$endgroup$
– Lord Shark the Unknown
Jan 8 at 7:54
7
$begingroup$
The answer to the question that you have asked is clearly yes. Of course it can be normal, for example when each of the groups is abelian. The point is that it is $f(G)$ is not necessarily normal in $H$ when $f$ is not an epimorphism.
$endgroup$
– Derek Holt
Jan 8 at 8:28