Do homomorphisms preserve subgroups?












1












$begingroup$


Let $(A, cdot), (B, ast)$ be groups, and $phi: A mapsto B$ a homomorphism. Then:
$$ S leq A Rightarrow phi(S) leq B $$



I think this must be a false statement, because there is no guarantee that $textrm{Ker}(phi) subset S$. Yet, I have written the following proof for this statement:



begin{align*}
&quad; S leq A \
&{text{definition of subgroup}} \
&Leftrightarrow forall s, t in S mid s cdot t in S ;wedge; s^{-1} in S \
&{text{definition of set inclusion}} \
&Leftrightarrow forall s, t in S mid (exists u in S mid s cdot t = u) ;wedge; (exists v in S mid v = s^{-1}) \
&{text{(problematic step?) $phi$ is well-defined, so $x = y Rightarrow phi(x) = phi(y)$}} \
&Rightarrow forall phi(s), phi(t) in phi(S) mid (exists phi(u) in phi(S) mid phi(s cdot t) = phi(u)) ;wedge; (exists phi(v) in phi(S) mid phi(v) = phi(s^{-1})) \
&{text{definition of set inclusion}} \
&Leftrightarrow forall phi(s), phi(t) in phi(S) midphi(s cdot t) in phi(S) ;wedge; phi(s^{-1}) in phi(S) \
&{text{$phi$ is a homomorphism, thus preserves the product, and inverse}}\
&Leftrightarrow forall phi(s), phi(t) in phi(S) midphi(s) ast phi(t) in phi(S) ;wedge; {phi(s)}^{-1} in phi(S) \
&{text{definition of subgroup}} \
&Leftrightarrow phi(S) leq B
end{align*}



I think the (weakest) correct statement should be:



$$ textrm{Ker}(phi) subset S ;wedge; S leq A Rightarrow phi(S) leq B $$



Is this the case? If yes, how can I correct my proof to capture this issue? If not, why not?






group-theory proof-verification proof-explanation







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    But it is not a false statement.
    $endgroup$
    – Tobias Kildetoft
    Dec 2 '18 at 21:36










  • $begingroup$
    Your condition in the end will guarantee something strictly better, namely that $S = phi^{-1}(phi(S))$.
    $endgroup$
    – Jef L
    Dec 2 '18 at 21:37










  • $begingroup$
    @TobiasKildetoft So I must be misunderstanding something! Shouldn't we also need the guarantee that $textrm{Ker}(phi) subset S$? There might be subgroups of $A$ which do not contain $textrm{Ker}{phi}$?
    $endgroup$
    – user89
    Dec 2 '18 at 21:56








  • 2




    $begingroup$
    Why do you think that $phi(S)$ being a subgroup of $B$ hinges on whether $mbox{Ker}(phi)subset S$? Suppose that $phi(s)=1$ for all $sin S$. Then $phi(S)={1}$ is definitely a subgroup of $B$.
    $endgroup$
    – Scounged
    Dec 2 '18 at 22:33






  • 1




    $begingroup$
    @user89 Note that homomorphisms always map the identity to the identity due to the property $phi(st)=phi(s)phi(t)$, so the kernel of $phi$ is never empty. And since $S$ is a subgroup it will definitely contain the identity element.
    $endgroup$
    – Scounged
    Dec 2 '18 at 22:46
















1












$begingroup$


Let $(A, cdot), (B, ast)$ be groups, and $phi: A mapsto B$ a homomorphism. Then:
$$ S leq A Rightarrow phi(S) leq B $$



I think this must be a false statement, because there is no guarantee that $textrm{Ker}(phi) subset S$. Yet, I have written the following proof for this statement:



begin{align*}
&quad; S leq A \
&{text{definition of subgroup}} \
&Leftrightarrow forall s, t in S mid s cdot t in S ;wedge; s^{-1} in S \
&{text{definition of set inclusion}} \
&Leftrightarrow forall s, t in S mid (exists u in S mid s cdot t = u) ;wedge; (exists v in S mid v = s^{-1}) \
&{text{(problematic step?) $phi$ is well-defined, so $x = y Rightarrow phi(x) = phi(y)$}} \
&Rightarrow forall phi(s), phi(t) in phi(S) mid (exists phi(u) in phi(S) mid phi(s cdot t) = phi(u)) ;wedge; (exists phi(v) in phi(S) mid phi(v) = phi(s^{-1})) \
&{text{definition of set inclusion}} \
&Leftrightarrow forall phi(s), phi(t) in phi(S) midphi(s cdot t) in phi(S) ;wedge; phi(s^{-1}) in phi(S) \
&{text{$phi$ is a homomorphism, thus preserves the product, and inverse}}\
&Leftrightarrow forall phi(s), phi(t) in phi(S) midphi(s) ast phi(t) in phi(S) ;wedge; {phi(s)}^{-1} in phi(S) \
&{text{definition of subgroup}} \
&Leftrightarrow phi(S) leq B
end{align*}



I think the (weakest) correct statement should be:



$$ textrm{Ker}(phi) subset S ;wedge; S leq A Rightarrow phi(S) leq B $$



Is this the case? If yes, how can I correct my proof to capture this issue? If not, why not?






group-theory proof-verification proof-explanation







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    But it is not a false statement.
    $endgroup$
    – Tobias Kildetoft
    Dec 2 '18 at 21:36










  • $begingroup$
    Your condition in the end will guarantee something strictly better, namely that $S = phi^{-1}(phi(S))$.
    $endgroup$
    – Jef L
    Dec 2 '18 at 21:37










  • $begingroup$
    @TobiasKildetoft So I must be misunderstanding something! Shouldn't we also need the guarantee that $textrm{Ker}(phi) subset S$? There might be subgroups of $A$ which do not contain $textrm{Ker}{phi}$?
    $endgroup$
    – user89
    Dec 2 '18 at 21:56








  • 2




    $begingroup$
    Why do you think that $phi(S)$ being a subgroup of $B$ hinges on whether $mbox{Ker}(phi)subset S$? Suppose that $phi(s)=1$ for all $sin S$. Then $phi(S)={1}$ is definitely a subgroup of $B$.
    $endgroup$
    – Scounged
    Dec 2 '18 at 22:33






  • 1




    $begingroup$
    @user89 Note that homomorphisms always map the identity to the identity due to the property $phi(st)=phi(s)phi(t)$, so the kernel of $phi$ is never empty. And since $S$ is a subgroup it will definitely contain the identity element.
    $endgroup$
    – Scounged
    Dec 2 '18 at 22:46














1












1








1





$begingroup$


Let $(A, cdot), (B, ast)$ be groups, and $phi: A mapsto B$ a homomorphism. Then:
$$ S leq A Rightarrow phi(S) leq B $$



I think this must be a false statement, because there is no guarantee that $textrm{Ker}(phi) subset S$. Yet, I have written the following proof for this statement:



begin{align*}
&quad; S leq A \
&{text{definition of subgroup}} \
&Leftrightarrow forall s, t in S mid s cdot t in S ;wedge; s^{-1} in S \
&{text{definition of set inclusion}} \
&Leftrightarrow forall s, t in S mid (exists u in S mid s cdot t = u) ;wedge; (exists v in S mid v = s^{-1}) \
&{text{(problematic step?) $phi$ is well-defined, so $x = y Rightarrow phi(x) = phi(y)$}} \
&Rightarrow forall phi(s), phi(t) in phi(S) mid (exists phi(u) in phi(S) mid phi(s cdot t) = phi(u)) ;wedge; (exists phi(v) in phi(S) mid phi(v) = phi(s^{-1})) \
&{text{definition of set inclusion}} \
&Leftrightarrow forall phi(s), phi(t) in phi(S) midphi(s cdot t) in phi(S) ;wedge; phi(s^{-1}) in phi(S) \
&{text{$phi$ is a homomorphism, thus preserves the product, and inverse}}\
&Leftrightarrow forall phi(s), phi(t) in phi(S) midphi(s) ast phi(t) in phi(S) ;wedge; {phi(s)}^{-1} in phi(S) \
&{text{definition of subgroup}} \
&Leftrightarrow phi(S) leq B
end{align*}



I think the (weakest) correct statement should be:



$$ textrm{Ker}(phi) subset S ;wedge; S leq A Rightarrow phi(S) leq B $$



Is this the case? If yes, how can I correct my proof to capture this issue? If not, why not?






group-theory proof-verification proof-explanation







share|cite|improve this question











$endgroup$




Let $(A, cdot), (B, ast)$ be groups, and $phi: A mapsto B$ a homomorphism. Then:
$$ S leq A Rightarrow phi(S) leq B $$



I think this must be a false statement, because there is no guarantee that $textrm{Ker}(phi) subset S$. Yet, I have written the following proof for this statement:



begin{align*}
&quad; S leq A \
&{text{definition of subgroup}} \
&Leftrightarrow forall s, t in S mid s cdot t in S ;wedge; s^{-1} in S \
&{text{definition of set inclusion}} \
&Leftrightarrow forall s, t in S mid (exists u in S mid s cdot t = u) ;wedge; (exists v in S mid v = s^{-1}) \
&{text{(problematic step?) $phi$ is well-defined, so $x = y Rightarrow phi(x) = phi(y)$}} \
&Rightarrow forall phi(s), phi(t) in phi(S) mid (exists phi(u) in phi(S) mid phi(s cdot t) = phi(u)) ;wedge; (exists phi(v) in phi(S) mid phi(v) = phi(s^{-1})) \
&{text{definition of set inclusion}} \
&Leftrightarrow forall phi(s), phi(t) in phi(S) midphi(s cdot t) in phi(S) ;wedge; phi(s^{-1}) in phi(S) \
&{text{$phi$ is a homomorphism, thus preserves the product, and inverse}}\
&Leftrightarrow forall phi(s), phi(t) in phi(S) midphi(s) ast phi(t) in phi(S) ;wedge; {phi(s)}^{-1} in phi(S) \
&{text{definition of subgroup}} \
&Leftrightarrow phi(S) leq B
end{align*}



I think the (weakest) correct statement should be:



$$ textrm{Ker}(phi) subset S ;wedge; S leq A Rightarrow phi(S) leq B $$



Is this the case? If yes, how can I correct my proof to capture this issue? If not, why not?






group-theory proof-verification proof-explanation




group-theory proof-verification proof-explanation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 22:09







user89

















asked Dec 2 '18 at 21:32









user89user89

6031647




6031647








  • 2




    $begingroup$
    But it is not a false statement.
    $endgroup$
    – Tobias Kildetoft
    Dec 2 '18 at 21:36










  • $begingroup$
    Your condition in the end will guarantee something strictly better, namely that $S = phi^{-1}(phi(S))$.
    $endgroup$
    – Jef L
    Dec 2 '18 at 21:37










  • $begingroup$
    @TobiasKildetoft So I must be misunderstanding something! Shouldn't we also need the guarantee that $textrm{Ker}(phi) subset S$? There might be subgroups of $A$ which do not contain $textrm{Ker}{phi}$?
    $endgroup$
    – user89
    Dec 2 '18 at 21:56








  • 2




    $begingroup$
    Why do you think that $phi(S)$ being a subgroup of $B$ hinges on whether $mbox{Ker}(phi)subset S$? Suppose that $phi(s)=1$ for all $sin S$. Then $phi(S)={1}$ is definitely a subgroup of $B$.
    $endgroup$
    – Scounged
    Dec 2 '18 at 22:33






  • 1




    $begingroup$
    @user89 Note that homomorphisms always map the identity to the identity due to the property $phi(st)=phi(s)phi(t)$, so the kernel of $phi$ is never empty. And since $S$ is a subgroup it will definitely contain the identity element.
    $endgroup$
    – Scounged
    Dec 2 '18 at 22:46














  • 2




    $begingroup$
    But it is not a false statement.
    $endgroup$
    – Tobias Kildetoft
    Dec 2 '18 at 21:36










  • $begingroup$
    Your condition in the end will guarantee something strictly better, namely that $S = phi^{-1}(phi(S))$.
    $endgroup$
    – Jef L
    Dec 2 '18 at 21:37










  • $begingroup$
    @TobiasKildetoft So I must be misunderstanding something! Shouldn't we also need the guarantee that $textrm{Ker}(phi) subset S$? There might be subgroups of $A$ which do not contain $textrm{Ker}{phi}$?
    $endgroup$
    – user89
    Dec 2 '18 at 21:56








  • 2




    $begingroup$
    Why do you think that $phi(S)$ being a subgroup of $B$ hinges on whether $mbox{Ker}(phi)subset S$? Suppose that $phi(s)=1$ for all $sin S$. Then $phi(S)={1}$ is definitely a subgroup of $B$.
    $endgroup$
    – Scounged
    Dec 2 '18 at 22:33






  • 1




    $begingroup$
    @user89 Note that homomorphisms always map the identity to the identity due to the property $phi(st)=phi(s)phi(t)$, so the kernel of $phi$ is never empty. And since $S$ is a subgroup it will definitely contain the identity element.
    $endgroup$
    – Scounged
    Dec 2 '18 at 22:46








2




2




$begingroup$
But it is not a false statement.
$endgroup$
– Tobias Kildetoft
Dec 2 '18 at 21:36




$begingroup$
But it is not a false statement.
$endgroup$
– Tobias Kildetoft
Dec 2 '18 at 21:36












$begingroup$
Your condition in the end will guarantee something strictly better, namely that $S = phi^{-1}(phi(S))$.
$endgroup$
– Jef L
Dec 2 '18 at 21:37




$begingroup$
Your condition in the end will guarantee something strictly better, namely that $S = phi^{-1}(phi(S))$.
$endgroup$
– Jef L
Dec 2 '18 at 21:37












$begingroup$
@TobiasKildetoft So I must be misunderstanding something! Shouldn't we also need the guarantee that $textrm{Ker}(phi) subset S$? There might be subgroups of $A$ which do not contain $textrm{Ker}{phi}$?
$endgroup$
– user89
Dec 2 '18 at 21:56






$begingroup$
@TobiasKildetoft So I must be misunderstanding something! Shouldn't we also need the guarantee that $textrm{Ker}(phi) subset S$? There might be subgroups of $A$ which do not contain $textrm{Ker}{phi}$?
$endgroup$
– user89
Dec 2 '18 at 21:56






2




2




$begingroup$
Why do you think that $phi(S)$ being a subgroup of $B$ hinges on whether $mbox{Ker}(phi)subset S$? Suppose that $phi(s)=1$ for all $sin S$. Then $phi(S)={1}$ is definitely a subgroup of $B$.
$endgroup$
– Scounged
Dec 2 '18 at 22:33




$begingroup$
Why do you think that $phi(S)$ being a subgroup of $B$ hinges on whether $mbox{Ker}(phi)subset S$? Suppose that $phi(s)=1$ for all $sin S$. Then $phi(S)={1}$ is definitely a subgroup of $B$.
$endgroup$
– Scounged
Dec 2 '18 at 22:33




1




1




$begingroup$
@user89 Note that homomorphisms always map the identity to the identity due to the property $phi(st)=phi(s)phi(t)$, so the kernel of $phi$ is never empty. And since $S$ is a subgroup it will definitely contain the identity element.
$endgroup$
– Scounged
Dec 2 '18 at 22:46




$begingroup$
@user89 Note that homomorphisms always map the identity to the identity due to the property $phi(st)=phi(s)phi(t)$, so the kernel of $phi$ is never empty. And since $S$ is a subgroup it will definitely contain the identity element.
$endgroup$
– Scounged
Dec 2 '18 at 22:46










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