Mixing
Normal subgroup involving quotient group
$begingroup$
Let $G$ be a group and $N$ be a normal subgroup of G. Show that there
is a bijection between $Ntimes G/N$ and $G$.
I think I can define a map by $f : Ntimes G/Nto G$ where $(n,gN) mapsto ng$.
Then $f$ is surjective. But how can I show that $f$ is injective? Or is it possible to find its inverse function?
Let $G$ be a group, $Ntriangleleft G$, and $G_{i}$, $G_{i+1}$ be two
subgroups of $G$ such that $G_{i}triangleleft G_{i+1}$.
Then ($G_{i+1}cap N$)/($G_{i}cap N$) $triangleleft$
$G_{i+1}$/$G_{i}$
I only know ($G_{i+1}cap N$)/($G_{i}cap N$) is well-defined. But I don't know what else I can do now.
abstract-algebra group-theory
asked Jan 15 at 3:16
Bill_TsaiBill_Tsai
134
$endgroup$
|
show 2 more comments
$begingroup$
Let $G$ be a group and $N$ be a normal subgroup of G. Show that there
is a bijection between $Ntimes G/N$ and $G$.
I think I can define a map by $f : Ntimes G/Nto G$ where $(n,gN) mapsto ng$.
Then $f$ is surjective. But how can I show that $f$ is injective? Or is it possible to find its inverse function?
Let $G$ be a group, $Ntriangleleft G$, and $G_{i}$, $G_{i+1}$ be two
subgroups of $G$ such that $G_{i}triangleleft G_{i+1}$.
Then ($G_{i+1}cap N$)/($G_{i}cap N$) $triangleleft$
$G_{i+1}$/$G_{i}$
I only know ($G_{i+1}cap N$)/($G_{i}cap N$) is well-defined. But I don't know what else I can do now.
abstract-algebra group-theory
asked Jan 15 at 3:16
Bill_TsaiBill_Tsai
134
$endgroup$
$begingroup$
For 1, you're multiplying on the wrong side.
$endgroup$
– Matt Samuel
Jan 15 at 3:25
$begingroup$
What do you mean, gN? ng?
$endgroup$
– Bill_Tsai
Jan 15 at 3:26
$begingroup$
Is your map well defined?
$endgroup$
– mouthetics
Jan 15 at 3:26
$begingroup$
Should be $gn$. Except that's not a well defined map. You have to single out a specific $g$ in each coset for it to work.
$endgroup$
– Matt Samuel
Jan 15 at 3:27
$begingroup$
oops. I forgot to check.
$endgroup$
– Bill_Tsai
Jan 15 at 3:27
|
show 2 more comments
$begingroup$
Let $G$ be a group and $N$ be a normal subgroup of G. Show that there
is a bijection between $Ntimes G/N$ and $G$.
I think I can define a map by $f : Ntimes G/Nto G$ where $(n,gN) mapsto ng$.
Then $f$ is surjective. But how can I show that $f$ is injective? Or is it possible to find its inverse function?
Let $G$ be a group, $Ntriangleleft G$, and $G_{i}$, $G_{i+1}$ be two
subgroups of $G$ such that $G_{i}triangleleft G_{i+1}$.
Then ($G_{i+1}cap N$)/($G_{i}cap N$) $triangleleft$
$G_{i+1}$/$G_{i}$
I only know ($G_{i+1}cap N$)/($G_{i}cap N$) is well-defined. But I don't know what else I can do now.
abstract-algebra group-theory
asked Jan 15 at 3:16
Bill_TsaiBill_Tsai
134
$endgroup$
Let $G$ be a group and $N$ be a normal subgroup of G. Show that there
is a bijection between $Ntimes G/N$ and $G$.
I think I can define a map by $f : Ntimes G/Nto G$ where $(n,gN) mapsto ng$.
Then $f$ is surjective. But how can I show that $f$ is injective? Or is it possible to find its inverse function?
Let $G$ be a group, $Ntriangleleft G$, and $G_{i}$, $G_{i+1}$ be two
subgroups of $G$ such that $G_{i}triangleleft G_{i+1}$.
Then ($G_{i+1}cap N$)/($G_{i}cap N$) $triangleleft$
$G_{i+1}$/$G_{i}$
I only know ($G_{i+1}cap N$)/($G_{i}cap N$) is well-defined. But I don't know what else I can do now.
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 15 at 3:16
Bill_TsaiBill_Tsai
134
asked Jan 15 at 3:16
Bill_TsaiBill_Tsai
134
edited Jan 15 at 13:48
Bill_Tsai
asked Jan 15 at 3:16
Bill_TsaiBill_Tsai
134
asked Jan 15 at 3:16
Bill_TsaiBill_Tsai
134
asked Jan 15 at 3:16
Bill_TsaiBill_Tsai
134
134
$begingroup$
For 1, you're multiplying on the wrong side.
$endgroup$
– Matt Samuel
Jan 15 at 3:25
$begingroup$
What do you mean, gN? ng?
$endgroup$
– Bill_Tsai
Jan 15 at 3:26
$begingroup$
Is your map well defined?
$endgroup$
– mouthetics
Jan 15 at 3:26
$begingroup$
Should be $gn$. Except that's not a well defined map. You have to single out a specific $g$ in each coset for it to work.
$endgroup$
– Matt Samuel
Jan 15 at 3:27
$begingroup$
oops. I forgot to check.
$endgroup$
– Bill_Tsai
Jan 15 at 3:27
|
show 2 more comments
$begingroup$
For 1, you're multiplying on the wrong side.
$endgroup$
– Matt Samuel
Jan 15 at 3:25
$begingroup$
What do you mean, gN? ng?
$endgroup$
– Bill_Tsai
Jan 15 at 3:26
$begingroup$
Is your map well defined?
$endgroup$
– mouthetics
Jan 15 at 3:26
$begingroup$
Should be $gn$. Except that's not a well defined map. You have to single out a specific $g$ in each coset for it to work.
$endgroup$
– Matt Samuel
Jan 15 at 3:27
$begingroup$
oops. I forgot to check.
$endgroup$
– Bill_Tsai
Jan 15 at 3:27
$begingroup$
For 1, you're multiplying on the wrong side.
$endgroup$
– Matt Samuel
Jan 15 at 3:25
$begingroup$
For 1, you're multiplying on the wrong side.
$endgroup$
– Matt Samuel
Jan 15 at 3:25
$begingroup$
What do you mean, gN? ng?
$endgroup$
– Bill_Tsai
Jan 15 at 3:26
$begingroup$
What do you mean, gN? ng?
$endgroup$
– Bill_Tsai
Jan 15 at 3:26
$begingroup$
Is your map well defined?
$endgroup$
– mouthetics
Jan 15 at 3:26
$begingroup$
Is your map well defined?
$endgroup$
– mouthetics
Jan 15 at 3:26
$begingroup$
Should be $gn$. Except that's not a well defined map. You have to single out a specific $g$ in each coset for it to work.
$endgroup$
– Matt Samuel
Jan 15 at 3:27
$begingroup$
Should be $gn$. Except that's not a well defined map. You have to single out a specific $g$ in each coset for it to work.
$endgroup$
– Matt Samuel
Jan 15 at 3:27
$begingroup$
oops. I forgot to check.
$endgroup$
– Bill_Tsai
Jan 15 at 3:27
$begingroup$
oops. I forgot to check.
$endgroup$
– Bill_Tsai
Jan 15 at 3:27
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
I will prove the first statement. The proof of the second one is not short (but it's standard) and it needs to be asked in a separate question because it's irrelevant to the first statement.
From basic definitions in group theory, we know that the projection map $pi: G to G/N$ is well-defined and it's surjective. By the axiom of choice, it must have at least one right inverse. Take a right inverse of $pi$ and call it $phi: G/N to G$. Therefore, $pi circ phi=mathbb{1}_{G/N}$. Note that $phi$ has the property that it is constant on each equivalence class because of the way we constructed it. Now, define $f: N times G/N to G$ by
$$f(n,gN)=nphi(gN)$$
By construction, $g_1N=g_2N$ gives that $phi(g_1N)=phi(g_2N)$. So, it's seen that $f$ is well-defined.
$f$ is also surjective because for any $gin G$, there exists some $n_0 in N$ such that $gphi(gN)^{-1}=n_0$ because they belong to the same equivalence class. Hence, $g=f(n_0,gN)$.
To check injectivity, note that $f(n_1,g_1N)=f(n_2,g_2N)$ gives that
$$n_2^{-1}n_1=phi(g_2N)phi(g_1N)^{-1}$$
$$N=pi(n_2^{-1}n_1)=pi(phi(g_2N)phi(g_1N)^{-1})=pi(phi(g_2N))pi(phi(g_1N))^{-1}=(g_2N)(g_1^{-1}N)=g_2g_1^{-1}N$$
Hence, $g_2g_1^{-1}in N$ and they belong to the same equivalence class. In other words, $phi(g_1N) = phi(g_2N)$. Hence, $n_1=n_2$ and $f$ is injective too.
answered Jan 15 at 4:12
stressed outstressed out
5,4431638
$endgroup$
1
$begingroup$
$n_{1}n_{2}^{-1}$? Not $n_{2}^{-1}n_{1}$ instead?
$endgroup$
– Bill_Tsai
Jan 15 at 5:31
$begingroup$
@Bill_Tsai Yes, you're right. I fixed it. Thanks.
$endgroup$
– stressed out
Jan 15 at 5:35
add a comment |
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1 Answer
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$begingroup$
I will prove the first statement. The proof of the second one is not short (but it's standard) and it needs to be asked in a separate question because it's irrelevant to the first statement.
From basic definitions in group theory, we know that the projection map $pi: G to G/N$ is well-defined and it's surjective. By the axiom of choice, it must have at least one right inverse. Take a right inverse of $pi$ and call it $phi: G/N to G$. Therefore, $pi circ phi=mathbb{1}_{G/N}$. Note that $phi$ has the property that it is constant on each equivalence class because of the way we constructed it. Now, define $f: N times G/N to G$ by
$$f(n,gN)=nphi(gN)$$
By construction, $g_1N=g_2N$ gives that $phi(g_1N)=phi(g_2N)$. So, it's seen that $f$ is well-defined.
$f$ is also surjective because for any $gin G$, there exists some $n_0 in N$ such that $gphi(gN)^{-1}=n_0$ because they belong to the same equivalence class. Hence, $g=f(n_0,gN)$.
To check injectivity, note that $f(n_1,g_1N)=f(n_2,g_2N)$ gives that
$$n_2^{-1}n_1=phi(g_2N)phi(g_1N)^{-1}$$
$$N=pi(n_2^{-1}n_1)=pi(phi(g_2N)phi(g_1N)^{-1})=pi(phi(g_2N))pi(phi(g_1N))^{-1}=(g_2N)(g_1^{-1}N)=g_2g_1^{-1}N$$
Hence, $g_2g_1^{-1}in N$ and they belong to the same equivalence class. In other words, $phi(g_1N) = phi(g_2N)$. Hence, $n_1=n_2$ and $f$ is injective too.
answered Jan 15 at 4:12
stressed outstressed out
5,4431638
$endgroup$
1
$begingroup$
$n_{1}n_{2}^{-1}$? Not $n_{2}^{-1}n_{1}$ instead?
$endgroup$
– Bill_Tsai
Jan 15 at 5:31
$begingroup$
@Bill_Tsai Yes, you're right. I fixed it. Thanks.
$endgroup$
– stressed out
Jan 15 at 5:35
add a comment |
$begingroup$
I will prove the first statement. The proof of the second one is not short (but it's standard) and it needs to be asked in a separate question because it's irrelevant to the first statement.
From basic definitions in group theory, we know that the projection map $pi: G to G/N$ is well-defined and it's surjective. By the axiom of choice, it must have at least one right inverse. Take a right inverse of $pi$ and call it $phi: G/N to G$. Therefore, $pi circ phi=mathbb{1}_{G/N}$. Note that $phi$ has the property that it is constant on each equivalence class because of the way we constructed it. Now, define $f: N times G/N to G$ by
$$f(n,gN)=nphi(gN)$$
By construction, $g_1N=g_2N$ gives that $phi(g_1N)=phi(g_2N)$. So, it's seen that $f$ is well-defined.
$f$ is also surjective because for any $gin G$, there exists some $n_0 in N$ such that $gphi(gN)^{-1}=n_0$ because they belong to the same equivalence class. Hence, $g=f(n_0,gN)$.
To check injectivity, note that $f(n_1,g_1N)=f(n_2,g_2N)$ gives that
$$n_2^{-1}n_1=phi(g_2N)phi(g_1N)^{-1}$$
$$N=pi(n_2^{-1}n_1)=pi(phi(g_2N)phi(g_1N)^{-1})=pi(phi(g_2N))pi(phi(g_1N))^{-1}=(g_2N)(g_1^{-1}N)=g_2g_1^{-1}N$$
Hence, $g_2g_1^{-1}in N$ and they belong to the same equivalence class. In other words, $phi(g_1N) = phi(g_2N)$. Hence, $n_1=n_2$ and $f$ is injective too.
answered Jan 15 at 4:12
stressed outstressed out
5,4431638
$endgroup$
1
$begingroup$
$n_{1}n_{2}^{-1}$? Not $n_{2}^{-1}n_{1}$ instead?
$endgroup$
– Bill_Tsai
Jan 15 at 5:31
$begingroup$
@Bill_Tsai Yes, you're right. I fixed it. Thanks.
$endgroup$
– stressed out
Jan 15 at 5:35
add a comment |
$begingroup$
I will prove the first statement. The proof of the second one is not short (but it's standard) and it needs to be asked in a separate question because it's irrelevant to the first statement.
From basic definitions in group theory, we know that the projection map $pi: G to G/N$ is well-defined and it's surjective. By the axiom of choice, it must have at least one right inverse. Take a right inverse of $pi$ and call it $phi: G/N to G$. Therefore, $pi circ phi=mathbb{1}_{G/N}$. Note that $phi$ has the property that it is constant on each equivalence class because of the way we constructed it. Now, define $f: N times G/N to G$ by
$$f(n,gN)=nphi(gN)$$
By construction, $g_1N=g_2N$ gives that $phi(g_1N)=phi(g_2N)$. So, it's seen that $f$ is well-defined.
$f$ is also surjective because for any $gin G$, there exists some $n_0 in N$ such that $gphi(gN)^{-1}=n_0$ because they belong to the same equivalence class. Hence, $g=f(n_0,gN)$.
To check injectivity, note that $f(n_1,g_1N)=f(n_2,g_2N)$ gives that
$$n_2^{-1}n_1=phi(g_2N)phi(g_1N)^{-1}$$
$$N=pi(n_2^{-1}n_1)=pi(phi(g_2N)phi(g_1N)^{-1})=pi(phi(g_2N))pi(phi(g_1N))^{-1}=(g_2N)(g_1^{-1}N)=g_2g_1^{-1}N$$
Hence, $g_2g_1^{-1}in N$ and they belong to the same equivalence class. In other words, $phi(g_1N) = phi(g_2N)$. Hence, $n_1=n_2$ and $f$ is injective too.
answered Jan 15 at 4:12
stressed outstressed out
5,4431638
$endgroup$
I will prove the first statement. The proof of the second one is not short (but it's standard) and it needs to be asked in a separate question because it's irrelevant to the first statement.
From basic definitions in group theory, we know that the projection map $pi: G to G/N$ is well-defined and it's surjective. By the axiom of choice, it must have at least one right inverse. Take a right inverse of $pi$ and call it $phi: G/N to G$. Therefore, $pi circ phi=mathbb{1}_{G/N}$. Note that $phi$ has the property that it is constant on each equivalence class because of the way we constructed it. Now, define $f: N times G/N to G$ by
$$f(n,gN)=nphi(gN)$$
By construction, $g_1N=g_2N$ gives that $phi(g_1N)=phi(g_2N)$. So, it's seen that $f$ is well-defined.
$f$ is also surjective because for any $gin G$, there exists some $n_0 in N$ such that $gphi(gN)^{-1}=n_0$ because they belong to the same equivalence class. Hence, $g=f(n_0,gN)$.
To check injectivity, note that $f(n_1,g_1N)=f(n_2,g_2N)$ gives that
$$n_2^{-1}n_1=phi(g_2N)phi(g_1N)^{-1}$$
$$N=pi(n_2^{-1}n_1)=pi(phi(g_2N)phi(g_1N)^{-1})=pi(phi(g_2N))pi(phi(g_1N))^{-1}=(g_2N)(g_1^{-1}N)=g_2g_1^{-1}N$$
Hence, $g_2g_1^{-1}in N$ and they belong to the same equivalence class. In other words, $phi(g_1N) = phi(g_2N)$. Hence, $n_1=n_2$ and $f$ is injective too.
answered Jan 15 at 4:12
stressed outstressed out
5,4431638
edited Jan 15 at 5:37
answered Jan 15 at 4:12
stressed outstressed out
5,4431638
answered Jan 15 at 4:12
stressed outstressed out
5,4431638
answered Jan 15 at 4:12
stressed outstressed out
5,4431638
5,4431638
1
$begingroup$
$n_{1}n_{2}^{-1}$? Not $n_{2}^{-1}n_{1}$ instead?
$endgroup$
– Bill_Tsai
Jan 15 at 5:31
$begingroup$
@Bill_Tsai Yes, you're right. I fixed it. Thanks.
$endgroup$
– stressed out
Jan 15 at 5:35
add a comment |
1
$begingroup$
$n_{1}n_{2}^{-1}$? Not $n_{2}^{-1}n_{1}$ instead?
$endgroup$
– Bill_Tsai
Jan 15 at 5:31
$begingroup$
@Bill_Tsai Yes, you're right. I fixed it. Thanks.
$endgroup$
– stressed out
Jan 15 at 5:35
1
1
$begingroup$
$n_{1}n_{2}^{-1}$? Not $n_{2}^{-1}n_{1}$ instead?
$endgroup$
– Bill_Tsai
Jan 15 at 5:31
$begingroup$
$n_{1}n_{2}^{-1}$? Not $n_{2}^{-1}n_{1}$ instead?
$endgroup$
– Bill_Tsai
Jan 15 at 5:31
$begingroup$
@Bill_Tsai Yes, you're right. I fixed it. Thanks.
$endgroup$
– stressed out
Jan 15 at 5:35
$begingroup$
@Bill_Tsai Yes, you're right. I fixed it. Thanks.
$endgroup$
– stressed out
Jan 15 at 5:35
add a comment |
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$begingroup$
For 1, you're multiplying on the wrong side.
$endgroup$
– Matt Samuel
Jan 15 at 3:25
$begingroup$
What do you mean, gN? ng?
$endgroup$
– Bill_Tsai
Jan 15 at 3:26
$begingroup$
Is your map well defined?
$endgroup$
– mouthetics
Jan 15 at 3:26
$begingroup$
Should be $gn$. Except that's not a well defined map. You have to single out a specific $g$ in each coset for it to work.
$endgroup$
– Matt Samuel
Jan 15 at 3:27
$begingroup$
oops. I forgot to check.
$endgroup$
– Bill_Tsai
Jan 15 at 3:27