If all homomorphisms $f:G→H$ are trivial or injective, then G is simple.
$begingroup$
Let $G$ be a nontrivial group. Show that $G$ is simple if and only if, for every group $H$ and homomorphism $f:G→H$, either $f$ is trivial or $f$ is injective.
So I have already proved that if $f$ is trivial, then given any normal subgroup $G'$ of $G$, $G'$={1}. But I have problem proving that if $f$ is injective, then given any normal subgroup $G'$ of G, $G'=G$. Any hint would be appreciated!
abstract-algebra group-theory group-homomorphism simple-groups
edited Dec 3 '18 at 17:41
Arturo Magidin
261k34586906
asked Dec 3 '18 at 17:36
JJW22JJW22
1667
$endgroup$
add a comment |
$begingroup$
Let $G$ be a nontrivial group. Show that $G$ is simple if and only if, for every group $H$ and homomorphism $f:G→H$, either $f$ is trivial or $f$ is injective.
So I have already proved that if $f$ is trivial, then given any normal subgroup $G'$ of $G$, $G'$={1}. But I have problem proving that if $f$ is injective, then given any normal subgroup $G'$ of G, $G'=G$. Any hint would be appreciated!
abstract-algebra group-theory group-homomorphism simple-groups
edited Dec 3 '18 at 17:41
Arturo Magidin
261k34586906
asked Dec 3 '18 at 17:36
JJW22JJW22
1667
$endgroup$
2
$begingroup$
I would prove both sides contrapositively myself. If $G$ is not simple, there exists $N$ such that ${1}triangleleft Ntriangleleft G$. Can you show a homomorphism from $G$ to $G/N$? what properties would it have. For the other direction if there exists a homomorphism $f:Gto H$ that is non-trivial non-injective, then what can we say about its kernel?
$endgroup$
– Melody
Dec 3 '18 at 17:41
1
$begingroup$
Your title was not the same as your question.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 17:42
$begingroup$
@ArturoMagidin Thank you!
$endgroup$
– JJW22
Dec 3 '18 at 17:42
add a comment |
$begingroup$
Let $G$ be a nontrivial group. Show that $G$ is simple if and only if, for every group $H$ and homomorphism $f:G→H$, either $f$ is trivial or $f$ is injective.
So I have already proved that if $f$ is trivial, then given any normal subgroup $G'$ of $G$, $G'$={1}. But I have problem proving that if $f$ is injective, then given any normal subgroup $G'$ of G, $G'=G$. Any hint would be appreciated!
abstract-algebra group-theory group-homomorphism simple-groups
edited Dec 3 '18 at 17:41
Arturo Magidin
261k34586906
asked Dec 3 '18 at 17:36
JJW22JJW22
1667
$endgroup$
Let $G$ be a nontrivial group. Show that $G$ is simple if and only if, for every group $H$ and homomorphism $f:G→H$, either $f$ is trivial or $f$ is injective.
So I have already proved that if $f$ is trivial, then given any normal subgroup $G'$ of $G$, $G'$={1}. But I have problem proving that if $f$ is injective, then given any normal subgroup $G'$ of G, $G'=G$. Any hint would be appreciated!
abstract-algebra group-theory group-homomorphism simple-groups
abstract-algebra group-theory group-homomorphism simple-groups
edited Dec 3 '18 at 17:41
Arturo Magidin
261k34586906
asked Dec 3 '18 at 17:36
JJW22JJW22
1667
edited Dec 3 '18 at 17:41
Arturo Magidin
261k34586906
asked Dec 3 '18 at 17:36
JJW22JJW22
1667
edited Dec 3 '18 at 17:41
Arturo Magidin
261k34586906
edited Dec 3 '18 at 17:41
Arturo Magidin
261k34586906
edited Dec 3 '18 at 17:41
Arturo Magidin
261k34586906
261k34586906
asked Dec 3 '18 at 17:36
JJW22JJW22
1667
asked Dec 3 '18 at 17:36
JJW22JJW22
1667
asked Dec 3 '18 at 17:36
JJW22JJW22
1667
1667
2
$begingroup$
I would prove both sides contrapositively myself. If $G$ is not simple, there exists $N$ such that ${1}triangleleft Ntriangleleft G$. Can you show a homomorphism from $G$ to $G/N$? what properties would it have. For the other direction if there exists a homomorphism $f:Gto H$ that is non-trivial non-injective, then what can we say about its kernel?
$endgroup$
– Melody
Dec 3 '18 at 17:41
1
$begingroup$
Your title was not the same as your question.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 17:42
$begingroup$
@ArturoMagidin Thank you!
$endgroup$
– JJW22
Dec 3 '18 at 17:42
add a comment |
2
$begingroup$
I would prove both sides contrapositively myself. If $G$ is not simple, there exists $N$ such that ${1}triangleleft Ntriangleleft G$. Can you show a homomorphism from $G$ to $G/N$? what properties would it have. For the other direction if there exists a homomorphism $f:Gto H$ that is non-trivial non-injective, then what can we say about its kernel?
$endgroup$
– Melody
Dec 3 '18 at 17:41
1
$begingroup$
Your title was not the same as your question.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 17:42
$begingroup$
@ArturoMagidin Thank you!
$endgroup$
– JJW22
Dec 3 '18 at 17:42
2
2
$begingroup$
I would prove both sides contrapositively myself. If $G$ is not simple, there exists $N$ such that ${1}triangleleft Ntriangleleft G$. Can you show a homomorphism from $G$ to $G/N$? what properties would it have. For the other direction if there exists a homomorphism $f:Gto H$ that is non-trivial non-injective, then what can we say about its kernel?
$endgroup$
– Melody
Dec 3 '18 at 17:41
$begingroup$
I would prove both sides contrapositively myself. If $G$ is not simple, there exists $N$ such that ${1}triangleleft Ntriangleleft G$. Can you show a homomorphism from $G$ to $G/N$? what properties would it have. For the other direction if there exists a homomorphism $f:Gto H$ that is non-trivial non-injective, then what can we say about its kernel?
$endgroup$
– Melody
Dec 3 '18 at 17:41
1
1
$begingroup$
Your title was not the same as your question.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 17:42
$begingroup$
Your title was not the same as your question.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 17:42
$begingroup$
@ArturoMagidin Thank you!
$endgroup$
– JJW22
Dec 3 '18 at 17:42
$begingroup$
@ArturoMagidin Thank you!
$endgroup$
– JJW22
Dec 3 '18 at 17:42
add a comment |
1 Answer
1
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oldest
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$begingroup$
Hint: Let $N$ be a normal subgroup of $G$. Take $H=G/N$ and the quotient homomorphism $f: G to G/N$.
Solution:
If $f$ is trivial, then $N=G$.
If $f$ is injective, then $N=ker f=1$.
Therefore, the only normal subgroups of $G$ are $1$ and $G$, and so $G$ is simple.
answered Dec 3 '18 at 17:45
lhflhf
163k10168392
$endgroup$
$begingroup$
Well, does this mean that if $f$ is injective, then the normal subgroup of $G$ is {1}?
$endgroup$
– JJW22
Dec 3 '18 at 18:17
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
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oldest
votes
$begingroup$
Hint: Let $N$ be a normal subgroup of $G$. Take $H=G/N$ and the quotient homomorphism $f: G to G/N$.
Solution:
If $f$ is trivial, then $N=G$.
If $f$ is injective, then $N=ker f=1$.
Therefore, the only normal subgroups of $G$ are $1$ and $G$, and so $G$ is simple.
answered Dec 3 '18 at 17:45
lhflhf
163k10168392
$endgroup$
$begingroup$
Well, does this mean that if $f$ is injective, then the normal subgroup of $G$ is {1}?
$endgroup$
– JJW22
Dec 3 '18 at 18:17
add a comment |
$begingroup$
Hint: Let $N$ be a normal subgroup of $G$. Take $H=G/N$ and the quotient homomorphism $f: G to G/N$.
Solution:
If $f$ is trivial, then $N=G$.
If $f$ is injective, then $N=ker f=1$.
Therefore, the only normal subgroups of $G$ are $1$ and $G$, and so $G$ is simple.
answered Dec 3 '18 at 17:45
lhflhf
163k10168392
$endgroup$
$begingroup$
Well, does this mean that if $f$ is injective, then the normal subgroup of $G$ is {1}?
$endgroup$
– JJW22
Dec 3 '18 at 18:17
add a comment |
$begingroup$
Hint: Let $N$ be a normal subgroup of $G$. Take $H=G/N$ and the quotient homomorphism $f: G to G/N$.
Solution:
If $f$ is trivial, then $N=G$.
If $f$ is injective, then $N=ker f=1$.
Therefore, the only normal subgroups of $G$ are $1$ and $G$, and so $G$ is simple.
answered Dec 3 '18 at 17:45
lhflhf
163k10168392
$endgroup$
Hint: Let $N$ be a normal subgroup of $G$. Take $H=G/N$ and the quotient homomorphism $f: G to G/N$.
Solution:
If $f$ is trivial, then $N=G$.
If $f$ is injective, then $N=ker f=1$.
Therefore, the only normal subgroups of $G$ are $1$ and $G$, and so $G$ is simple.
answered Dec 3 '18 at 17:45
lhflhf
163k10168392
edited Dec 4 '18 at 16:54
answered Dec 3 '18 at 17:45
lhflhf
163k10168392
answered Dec 3 '18 at 17:45
lhflhf
163k10168392
answered Dec 3 '18 at 17:45
lhflhf
163k10168392
163k10168392
$begingroup$
Well, does this mean that if $f$ is injective, then the normal subgroup of $G$ is {1}?
$endgroup$
– JJW22
Dec 3 '18 at 18:17
add a comment |
$begingroup$
Well, does this mean that if $f$ is injective, then the normal subgroup of $G$ is {1}?
$endgroup$
– JJW22
Dec 3 '18 at 18:17
$begingroup$
Well, does this mean that if $f$ is injective, then the normal subgroup of $G$ is {1}?
$endgroup$
– JJW22
Dec 3 '18 at 18:17
$begingroup$
Well, does this mean that if $f$ is injective, then the normal subgroup of $G$ is {1}?
$endgroup$
– JJW22
Dec 3 '18 at 18:17
add a comment |
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$begingroup$
I would prove both sides contrapositively myself. If $G$ is not simple, there exists $N$ such that ${1}triangleleft Ntriangleleft G$. Can you show a homomorphism from $G$ to $G/N$? what properties would it have. For the other direction if there exists a homomorphism $f:Gto H$ that is non-trivial non-injective, then what can we say about its kernel?
$endgroup$
– Melody
Dec 3 '18 at 17:41
1
$begingroup$
Your title was not the same as your question.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 17:42
$begingroup$
@ArturoMagidin Thank you!
$endgroup$
– JJW22
Dec 3 '18 at 17:42