Mixing
The First Isomorphism Theorem, begins with a homomorphism $f$ between two groups $G$ and $H$. Can we assume...
$begingroup$
The first isomorphism theorem states that if $f$ is a homomorphism between two groups $G$ and $H$, then $G/textrm{Ker}(f) cong H$.
We know that $f$ is a homomorphism, but do we also get to assume that it is surjective in order to prove the theorem?
Let $textrm{Ker}(f) = K$. Let $phi: G/K mapsto H$ such that for any $a in G$, $phi(aK) = f(a)$. To prove the first isomorphism theorem, we will need to show that $phi$ is well defined, injective, and surjective.
If $textrm{Im}(f) = H$ (i.e. $f$ is surjective), then we can write any element $h in H$, as $f(a)$ (for some $a in G$), since surjectivity of $f$ guarantees the existence of $a$. Thus, $phi$ being surjective naturally follows from how we use $f$ to define it.
It seems that Rotman is making this assumption:
But according to an answer on this website, it is not necessarily the case that $f$ must be surjective: https://math.stackexchange.com/a/2201729/115703 ?
Is Rotman actually assuming that $f$ is surjective, or am I misunderstanding? If he is assuming it, how can he do so given what the answer above discusses?
group-theory
asked Nov 10 '18 at 5:45
user89user89
5731647
$endgroup$
add a comment |
$begingroup$
The first isomorphism theorem states that if $f$ is a homomorphism between two groups $G$ and $H$, then $G/textrm{Ker}(f) cong H$.
We know that $f$ is a homomorphism, but do we also get to assume that it is surjective in order to prove the theorem?
Let $textrm{Ker}(f) = K$. Let $phi: G/K mapsto H$ such that for any $a in G$, $phi(aK) = f(a)$. To prove the first isomorphism theorem, we will need to show that $phi$ is well defined, injective, and surjective.
If $textrm{Im}(f) = H$ (i.e. $f$ is surjective), then we can write any element $h in H$, as $f(a)$ (for some $a in G$), since surjectivity of $f$ guarantees the existence of $a$. Thus, $phi$ being surjective naturally follows from how we use $f$ to define it.
It seems that Rotman is making this assumption:
But according to an answer on this website, it is not necessarily the case that $f$ must be surjective: https://math.stackexchange.com/a/2201729/115703 ?
Is Rotman actually assuming that $f$ is surjective, or am I misunderstanding? If he is assuming it, how can he do so given what the answer above discusses?
group-theory
asked Nov 10 '18 at 5:45
user89user89
5731647
$endgroup$
$begingroup$
$f(b)^{-1}$ is the group inverse of the element $f(b)$. It has nothing to do with the invertibility of the function $f$, in case that's where you think surjectivity is being used.
$endgroup$
– Cheerful Parsnip
Nov 10 '18 at 5:53
$begingroup$
The theorem does not state anything about $H$. It only involves $color{red}{mathrm{im}f}$. Put it in other words, $f: Gto mathrm{im}f$ is already surjective as you expected.
$endgroup$
– Quang Hoang
Nov 10 '18 at 5:56
add a comment |
$begingroup$
The first isomorphism theorem states that if $f$ is a homomorphism between two groups $G$ and $H$, then $G/textrm{Ker}(f) cong H$.
We know that $f$ is a homomorphism, but do we also get to assume that it is surjective in order to prove the theorem?
Let $textrm{Ker}(f) = K$. Let $phi: G/K mapsto H$ such that for any $a in G$, $phi(aK) = f(a)$. To prove the first isomorphism theorem, we will need to show that $phi$ is well defined, injective, and surjective.
If $textrm{Im}(f) = H$ (i.e. $f$ is surjective), then we can write any element $h in H$, as $f(a)$ (for some $a in G$), since surjectivity of $f$ guarantees the existence of $a$. Thus, $phi$ being surjective naturally follows from how we use $f$ to define it.
It seems that Rotman is making this assumption:
But according to an answer on this website, it is not necessarily the case that $f$ must be surjective: https://math.stackexchange.com/a/2201729/115703 ?
Is Rotman actually assuming that $f$ is surjective, or am I misunderstanding? If he is assuming it, how can he do so given what the answer above discusses?
group-theory
asked Nov 10 '18 at 5:45
user89user89
5731647
$endgroup$
The first isomorphism theorem states that if $f$ is a homomorphism between two groups $G$ and $H$, then $G/textrm{Ker}(f) cong H$.
We know that $f$ is a homomorphism, but do we also get to assume that it is surjective in order to prove the theorem?
Let $textrm{Ker}(f) = K$. Let $phi: G/K mapsto H$ such that for any $a in G$, $phi(aK) = f(a)$. To prove the first isomorphism theorem, we will need to show that $phi$ is well defined, injective, and surjective.
If $textrm{Im}(f) = H$ (i.e. $f$ is surjective), then we can write any element $h in H$, as $f(a)$ (for some $a in G$), since surjectivity of $f$ guarantees the existence of $a$. Thus, $phi$ being surjective naturally follows from how we use $f$ to define it.
It seems that Rotman is making this assumption:
But according to an answer on this website, it is not necessarily the case that $f$ must be surjective: https://math.stackexchange.com/a/2201729/115703 ?
Is Rotman actually assuming that $f$ is surjective, or am I misunderstanding? If he is assuming it, how can he do so given what the answer above discusses?
group-theory
group-theory
asked Nov 10 '18 at 5:45
user89user89
5731647
asked Nov 10 '18 at 5:45
user89user89
5731647
edited Jan 22 at 16:37
user89
asked Nov 10 '18 at 5:45
user89user89
5731647
asked Nov 10 '18 at 5:45
user89user89
5731647
asked Nov 10 '18 at 5:45
user89user89
5731647
5731647
$begingroup$
$f(b)^{-1}$ is the group inverse of the element $f(b)$. It has nothing to do with the invertibility of the function $f$, in case that's where you think surjectivity is being used.
$endgroup$
– Cheerful Parsnip
Nov 10 '18 at 5:53
$begingroup$
The theorem does not state anything about $H$. It only involves $color{red}{mathrm{im}f}$. Put it in other words, $f: Gto mathrm{im}f$ is already surjective as you expected.
$endgroup$
– Quang Hoang
Nov 10 '18 at 5:56
add a comment |
$begingroup$
$f(b)^{-1}$ is the group inverse of the element $f(b)$. It has nothing to do with the invertibility of the function $f$, in case that's where you think surjectivity is being used.
$endgroup$
– Cheerful Parsnip
Nov 10 '18 at 5:53
$begingroup$
The theorem does not state anything about $H$. It only involves $color{red}{mathrm{im}f}$. Put it in other words, $f: Gto mathrm{im}f$ is already surjective as you expected.
$endgroup$
– Quang Hoang
Nov 10 '18 at 5:56
$begingroup$
$f(b)^{-1}$ is the group inverse of the element $f(b)$. It has nothing to do with the invertibility of the function $f$, in case that's where you think surjectivity is being used.
$endgroup$
– Cheerful Parsnip
Nov 10 '18 at 5:53
$begingroup$
$f(b)^{-1}$ is the group inverse of the element $f(b)$. It has nothing to do with the invertibility of the function $f$, in case that's where you think surjectivity is being used.
$endgroup$
– Cheerful Parsnip
Nov 10 '18 at 5:53
$begingroup$
The theorem does not state anything about $H$. It only involves $color{red}{mathrm{im}f}$. Put it in other words, $f: Gto mathrm{im}f$ is already surjective as you expected.
$endgroup$
– Quang Hoang
Nov 10 '18 at 5:56
$begingroup$
The theorem does not state anything about $H$. It only involves $color{red}{mathrm{im}f}$. Put it in other words, $f: Gto mathrm{im}f$ is already surjective as you expected.
$endgroup$
– Quang Hoang
Nov 10 '18 at 5:56
add a comment |
1 Answer
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$begingroup$
The first isomorphism theorem actually states that if $f$ is a homomorphism between two groups $G$ and $H$, then $G/ker(f)cong operatorname{im}(f)$.
Note that every map is surjective onto it's image.
If you want to state that $G/ker(f) cong H$ then you do have to assume that $f$ is surjective.
edited Jan 22 at 16:41
user1729
17.4k64193
answered Nov 10 '18 at 5:59
Dhruv PatelDhruv Patel
913
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The first isomorphism theorem actually states that if $f$ is a homomorphism between two groups $G$ and $H$, then $G/ker(f)cong operatorname{im}(f)$.
Note that every map is surjective onto it's image.
If you want to state that $G/ker(f) cong H$ then you do have to assume that $f$ is surjective.
edited Jan 22 at 16:41
user1729
17.4k64193
answered Nov 10 '18 at 5:59
Dhruv PatelDhruv Patel
913
$endgroup$
add a comment |
$begingroup$
The first isomorphism theorem actually states that if $f$ is a homomorphism between two groups $G$ and $H$, then $G/ker(f)cong operatorname{im}(f)$.
Note that every map is surjective onto it's image.
If you want to state that $G/ker(f) cong H$ then you do have to assume that $f$ is surjective.
edited Jan 22 at 16:41
user1729
17.4k64193
answered Nov 10 '18 at 5:59
Dhruv PatelDhruv Patel
913
$endgroup$
add a comment |
$begingroup$
The first isomorphism theorem actually states that if $f$ is a homomorphism between two groups $G$ and $H$, then $G/ker(f)cong operatorname{im}(f)$.
Note that every map is surjective onto it's image.
If you want to state that $G/ker(f) cong H$ then you do have to assume that $f$ is surjective.
edited Jan 22 at 16:41
user1729
17.4k64193
answered Nov 10 '18 at 5:59
Dhruv PatelDhruv Patel
913
$endgroup$
The first isomorphism theorem actually states that if $f$ is a homomorphism between two groups $G$ and $H$, then $G/ker(f)cong operatorname{im}(f)$.
Note that every map is surjective onto it's image.
If you want to state that $G/ker(f) cong H$ then you do have to assume that $f$ is surjective.
edited Jan 22 at 16:41
user1729
17.4k64193
answered Nov 10 '18 at 5:59
Dhruv PatelDhruv Patel
913
edited Jan 22 at 16:41
user1729
17.4k64193
edited Jan 22 at 16:41
user1729
17.4k64193
edited Jan 22 at 16:41
user1729
17.4k64193
17.4k64193
answered Nov 10 '18 at 5:59
Dhruv PatelDhruv Patel
913
answered Nov 10 '18 at 5:59
Dhruv PatelDhruv Patel
913
answered Nov 10 '18 at 5:59
Dhruv PatelDhruv Patel
913
913
add a comment |
add a comment |
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$begingroup$
$f(b)^{-1}$ is the group inverse of the element $f(b)$. It has nothing to do with the invertibility of the function $f$, in case that's where you think surjectivity is being used.
$endgroup$
– Cheerful Parsnip
Nov 10 '18 at 5:53
$begingroup$
The theorem does not state anything about $H$. It only involves $color{red}{mathrm{im}f}$. Put it in other words, $f: Gto mathrm{im}f$ is already surjective as you expected.
$endgroup$
– Quang Hoang
Nov 10 '18 at 5:56