Taking the Homomorphism of a Subgroup [duplicate] The Next CEO of Stack OverflowShowing the...
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Taking the Homomorphism of a Subgroup [duplicate]
The Next CEO of Stack OverflowShowing the image of a subgroup is a subgroup.Image of subgroup and Kernel of homomorphism form subgroupsDoes the image $f(H) subseteq G'$ for $H subseteq G$?Is every quotient of a finite abelian group $G$ isomorphic to some subgroup of $G$?How do I show a mapping is a homomorphism?What would the cosets of this subgroup be?Normality of subgroup of image of homomorphismProving that a homomorphism is abeliannumber of homomorphism from $G$ to $G' times G' times cdots times G'$ ($k$ terms)Group Homomorphism Question from Completed MidtermFor homomorphism $phi: Gto H$, if $G_{1}leq G$ and $|G_{1}|=n$, then $|phi(G_{1})|$ divides $n$.If $H$ and $K$ are normal subgroups of $G$ and $Hbigcap K = {e}$, prove that $G$ is isomorphic to a subgroup of $G/H times G/K$How to find homomorphism of direct product?
$begingroup$
This question already has an answer here:
Showing the image of a subgroup is a subgroup. [duplicate]
1 answer
Let $phi : G rightarrow G'$ be a homomorphism. We want to show that $H leq G$ implies that $phi(H) leq G'$.
This is one of those problems that seems obviously true to the intuition, yet I'm having a devil of a time trying to prove it formally. A sketch of my efforts so far:
Let us define the codomain of $H$ under $phi$ to be $H'$, and $h'$ to be an arbitrary element of $H'$ such that $phi(h)=h'$ (for $h in H$).
To achieve the desired result, we have to show that $h' in G'$.
What I want to say is that, since $H leq G$, then every $h$ in $H$ is also in $G$, and thus clearly $h' in G'$. Is it that simple, or am I begging the question?
group-theory group-homomorphism
asked Mar 16 at 17:37
JustSomeGuy716JustSomeGuy716
94
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marked as duplicate by Dietrich Burde group-theory
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Mar 16 at 17:48
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:
Showing the image of a subgroup is a subgroup. [duplicate]
1 answer
Let $phi : G rightarrow G'$ be a homomorphism. We want to show that $H leq G$ implies that $phi(H) leq G'$.
This is one of those problems that seems obviously true to the intuition, yet I'm having a devil of a time trying to prove it formally. A sketch of my efforts so far:
Let us define the codomain of $H$ under $phi$ to be $H'$, and $h'$ to be an arbitrary element of $H'$ such that $phi(h)=h'$ (for $h in H$).
To achieve the desired result, we have to show that $h' in G'$.
What I want to say is that, since $H leq G$, then every $h$ in $H$ is also in $G$, and thus clearly $h' in G'$. Is it that simple, or am I begging the question?
group-theory group-homomorphism
asked Mar 16 at 17:37
JustSomeGuy716JustSomeGuy716
94
$endgroup$
marked as duplicate by Dietrich Burde group-theory
Users with the group-theory badge can single-handedly close group-theory questions as duplicates and reopen them as needed.
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Mar 16 at 17:48
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.
$begingroup$
If $a, b in H$ then $phi(ab^{-1}) in H'.$ Now use the fact that you have a homomorphism.
$endgroup$
– Sean Roberson
Mar 16 at 17:47
$begingroup$
You have shown that $H’$ is a $textit{subset}$ of $G’$, not that it is a $textit{subgroup}$, which is what the question is asking. So to show that it’s a subgroup, what do you have to show about $H’$? (You will need to use the properties of homomorphisms here.)
$endgroup$
– Alex Sanger
Mar 16 at 17:52
add a comment |
$begingroup$
This question already has an answer here:
Showing the image of a subgroup is a subgroup. [duplicate]
1 answer
Let $phi : G rightarrow G'$ be a homomorphism. We want to show that $H leq G$ implies that $phi(H) leq G'$.
This is one of those problems that seems obviously true to the intuition, yet I'm having a devil of a time trying to prove it formally. A sketch of my efforts so far:
Let us define the codomain of $H$ under $phi$ to be $H'$, and $h'$ to be an arbitrary element of $H'$ such that $phi(h)=h'$ (for $h in H$).
To achieve the desired result, we have to show that $h' in G'$.
What I want to say is that, since $H leq G$, then every $h$ in $H$ is also in $G$, and thus clearly $h' in G'$. Is it that simple, or am I begging the question?
group-theory group-homomorphism
asked Mar 16 at 17:37
JustSomeGuy716JustSomeGuy716
94
$endgroup$
This question already has an answer here:
Showing the image of a subgroup is a subgroup. [duplicate]
1 answer
Let $phi : G rightarrow G'$ be a homomorphism. We want to show that $H leq G$ implies that $phi(H) leq G'$.
This is one of those problems that seems obviously true to the intuition, yet I'm having a devil of a time trying to prove it formally. A sketch of my efforts so far:
Let us define the codomain of $H$ under $phi$ to be $H'$, and $h'$ to be an arbitrary element of $H'$ such that $phi(h)=h'$ (for $h in H$).
To achieve the desired result, we have to show that $h' in G'$.
What I want to say is that, since $H leq G$, then every $h$ in $H$ is also in $G$, and thus clearly $h' in G'$. Is it that simple, or am I begging the question?
This question already has an answer here:
Showing the image of a subgroup is a subgroup. [duplicate]
1 answer
group-theory group-homomorphism
group-theory group-homomorphism
asked Mar 16 at 17:37
JustSomeGuy716JustSomeGuy716
94
asked Mar 16 at 17:37
JustSomeGuy716JustSomeGuy716
94
asked Mar 16 at 17:37
JustSomeGuy716JustSomeGuy716
94
asked Mar 16 at 17:37
JustSomeGuy716JustSomeGuy716
94
asked Mar 16 at 17:37
JustSomeGuy716JustSomeGuy716
94
94
marked as duplicate by Dietrich Burde group-theory
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Mar 16 at 17:48
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 Dietrich Burde group-theory
Users with the group-theory badge can single-handedly close group-theory questions as duplicates and reopen them as needed.
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Mar 16 at 17:48
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.
$begingroup$
If $a, b in H$ then $phi(ab^{-1}) in H'.$ Now use the fact that you have a homomorphism.
$endgroup$
– Sean Roberson
Mar 16 at 17:47
$begingroup$
You have shown that $H’$ is a $textit{subset}$ of $G’$, not that it is a $textit{subgroup}$, which is what the question is asking. So to show that it’s a subgroup, what do you have to show about $H’$? (You will need to use the properties of homomorphisms here.)
$endgroup$
– Alex Sanger
Mar 16 at 17:52
add a comment |
$begingroup$
If $a, b in H$ then $phi(ab^{-1}) in H'.$ Now use the fact that you have a homomorphism.
$endgroup$
– Sean Roberson
Mar 16 at 17:47
$begingroup$
You have shown that $H’$ is a $textit{subset}$ of $G’$, not that it is a $textit{subgroup}$, which is what the question is asking. So to show that it’s a subgroup, what do you have to show about $H’$? (You will need to use the properties of homomorphisms here.)
$endgroup$
– Alex Sanger
Mar 16 at 17:52
$begingroup$
If $a, b in H$ then $phi(ab^{-1}) in H'.$ Now use the fact that you have a homomorphism.
$endgroup$
– Sean Roberson
Mar 16 at 17:47
$begingroup$
If $a, b in H$ then $phi(ab^{-1}) in H'.$ Now use the fact that you have a homomorphism.
$endgroup$
– Sean Roberson
Mar 16 at 17:47
$begingroup$
You have shown that $H’$ is a $textit{subset}$ of $G’$, not that it is a $textit{subgroup}$, which is what the question is asking. So to show that it’s a subgroup, what do you have to show about $H’$? (You will need to use the properties of homomorphisms here.)
$endgroup$
– Alex Sanger
Mar 16 at 17:52
$begingroup$
You have shown that $H’$ is a $textit{subset}$ of $G’$, not that it is a $textit{subgroup}$, which is what the question is asking. So to show that it’s a subgroup, what do you have to show about $H’$? (You will need to use the properties of homomorphisms here.)
$endgroup$
– Alex Sanger
Mar 16 at 17:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: You have to show that, for $h_1,h_2$ we have $phi(h_1)phi(h_2)in G'$ and $phi(h_1)^{-1}in G'$. Then $phi(H)$ is a subgroup of $G'$.
Reference: This duplicate:
Image of subgroup and Kernel of homomorphism form subgroups
answered Mar 16 at 17:46
Dietrich BurdeDietrich Burde
81.6k648106
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: You have to show that, for $h_1,h_2$ we have $phi(h_1)phi(h_2)in G'$ and $phi(h_1)^{-1}in G'$. Then $phi(H)$ is a subgroup of $G'$.
Reference: This duplicate:
Image of subgroup and Kernel of homomorphism form subgroups
answered Mar 16 at 17:46
Dietrich BurdeDietrich Burde
81.6k648106
$endgroup$
add a comment |
$begingroup$
Hint: You have to show that, for $h_1,h_2$ we have $phi(h_1)phi(h_2)in G'$ and $phi(h_1)^{-1}in G'$. Then $phi(H)$ is a subgroup of $G'$.
Reference: This duplicate:
Image of subgroup and Kernel of homomorphism form subgroups
answered Mar 16 at 17:46
Dietrich BurdeDietrich Burde
81.6k648106
$endgroup$
add a comment |
$begingroup$
Hint: You have to show that, for $h_1,h_2$ we have $phi(h_1)phi(h_2)in G'$ and $phi(h_1)^{-1}in G'$. Then $phi(H)$ is a subgroup of $G'$.
Reference: This duplicate:
Image of subgroup and Kernel of homomorphism form subgroups
answered Mar 16 at 17:46
Dietrich BurdeDietrich Burde
81.6k648106
$endgroup$
Hint: You have to show that, for $h_1,h_2$ we have $phi(h_1)phi(h_2)in G'$ and $phi(h_1)^{-1}in G'$. Then $phi(H)$ is a subgroup of $G'$.
Reference: This duplicate:
Image of subgroup and Kernel of homomorphism form subgroups
answered Mar 16 at 17:46
Dietrich BurdeDietrich Burde
81.6k648106
answered Mar 16 at 17:46
Dietrich BurdeDietrich Burde
81.6k648106
answered Mar 16 at 17:46
Dietrich BurdeDietrich Burde
81.6k648106
answered Mar 16 at 17:46
Dietrich BurdeDietrich Burde
81.6k648106
81.6k648106
add a comment |
add a comment |
$begingroup$
If $a, b in H$ then $phi(ab^{-1}) in H'.$ Now use the fact that you have a homomorphism.
$endgroup$
– Sean Roberson
Mar 16 at 17:47
$begingroup$
You have shown that $H’$ is a $textit{subset}$ of $G’$, not that it is a $textit{subgroup}$, which is what the question is asking. So to show that it’s a subgroup, what do you have to show about $H’$? (You will need to use the properties of homomorphisms here.)
$endgroup$
– Alex Sanger
Mar 16 at 17:52