Finding Homomorphisms from dihedral groups to cyclical groups
$begingroup$
Ok so there was another question very similar to this on here however it leaves me a little confused.
$bf{Question}$
Let G = $D_{14}$ the Dihedral group order 14 and A = $c_7$ be the cyclical group order 7. Find every homomorphism.
I started out by finding the trivial homomorphism when Im($varphi$)=1 then using the first isomorphism theorem the non trivial homomorphism we get $frac{|G|}{|K|}=7$ so we get |K|=2, however i understand that the answer to this is that there are no homomorphisms as the other subgroups aren't normal. But how do i work that out. And why does that mean there are no homomorphisms.
abstract-algebra group-theory group-isomorphism group-homomorphism dihedral-groups
asked Dec 4 '18 at 19:31
L GL G
248
$endgroup$
add a comment |
$begingroup$
Ok so there was another question very similar to this on here however it leaves me a little confused.
$bf{Question}$
Let G = $D_{14}$ the Dihedral group order 14 and A = $c_7$ be the cyclical group order 7. Find every homomorphism.
I started out by finding the trivial homomorphism when Im($varphi$)=1 then using the first isomorphism theorem the non trivial homomorphism we get $frac{|G|}{|K|}=7$ so we get |K|=2, however i understand that the answer to this is that there are no homomorphisms as the other subgroups aren't normal. But how do i work that out. And why does that mean there are no homomorphisms.
abstract-algebra group-theory group-isomorphism group-homomorphism dihedral-groups
asked Dec 4 '18 at 19:31
L GL G
248
$endgroup$
add a comment |
$begingroup$
Ok so there was another question very similar to this on here however it leaves me a little confused.
$bf{Question}$
Let G = $D_{14}$ the Dihedral group order 14 and A = $c_7$ be the cyclical group order 7. Find every homomorphism.
I started out by finding the trivial homomorphism when Im($varphi$)=1 then using the first isomorphism theorem the non trivial homomorphism we get $frac{|G|}{|K|}=7$ so we get |K|=2, however i understand that the answer to this is that there are no homomorphisms as the other subgroups aren't normal. But how do i work that out. And why does that mean there are no homomorphisms.
abstract-algebra group-theory group-isomorphism group-homomorphism dihedral-groups
asked Dec 4 '18 at 19:31
L GL G
248
$endgroup$
Ok so there was another question very similar to this on here however it leaves me a little confused.
$bf{Question}$
Let G = $D_{14}$ the Dihedral group order 14 and A = $c_7$ be the cyclical group order 7. Find every homomorphism.
I started out by finding the trivial homomorphism when Im($varphi$)=1 then using the first isomorphism theorem the non trivial homomorphism we get $frac{|G|}{|K|}=7$ so we get |K|=2, however i understand that the answer to this is that there are no homomorphisms as the other subgroups aren't normal. But how do i work that out. And why does that mean there are no homomorphisms.
abstract-algebra group-theory group-isomorphism group-homomorphism dihedral-groups
abstract-algebra group-theory group-isomorphism group-homomorphism dihedral-groups
asked Dec 4 '18 at 19:31
L GL G
248
asked Dec 4 '18 at 19:31
L GL G
248
asked Dec 4 '18 at 19:31
L GL G
248
asked Dec 4 '18 at 19:31
L GL G
248
asked Dec 4 '18 at 19:31
L GL G
248
248
add a comment |
add a comment |
2 Answers
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$begingroup$
Here's a productive way to go about this question.
The isomorphism theorems tell us that the image of a homomorphism is determined, up to isomorphism, by the kernel: the image is isomorphic to the quotient modulo the kernel.
Also, the kernel is a normal subgroup.
So, let's start by listing all normal subgroups of $D_{14}$:
- the whole group $D_{14}$;
- its cyclic subgroup of order 7;
- the trivial subgroup.
Next, let's compute the quotient (up to isomorphism) of $D_{14}$ by each of these normal subgroups:
- the quotient by the whole group is the trivial group;
- the quotient by the cyclic subgroup of order 7 is the order 2 cyclic group $c_2$;
- the quotient by the trivial subgroup is the group $D_{14}$.
Finally, observe that the image of any homomorphism $D_{14} to c_7$ cannot be isomorphic to $c_2$ or to $D_{14}$, so the only possibility is that the image is trivial subgroup of $c_7$.
Thus, the only homomorphism $D_{14} to c_7$ is the trivial homomorphism.
answered Dec 5 '18 at 18:20
Lee MosherLee Mosher
48.5k33681
$endgroup$
add a comment |
$begingroup$
If you found that there were only two possible subgroups, then they would have to be G itself and {e}, meaning there are no homomorphisms.
I'm really confused by some of the stuff you wrote, though. I would think that any of the rotations of the heptagon (1/7, 2/7, ..., 6/7) could be generators.
I might just be completely misunderstanding though, apologies.
answered Dec 5 '18 at 18:04
nessness
375
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
oldest
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active
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$begingroup$
Here's a productive way to go about this question.
The isomorphism theorems tell us that the image of a homomorphism is determined, up to isomorphism, by the kernel: the image is isomorphic to the quotient modulo the kernel.
Also, the kernel is a normal subgroup.
So, let's start by listing all normal subgroups of $D_{14}$:
- the whole group $D_{14}$;
- its cyclic subgroup of order 7;
- the trivial subgroup.
Next, let's compute the quotient (up to isomorphism) of $D_{14}$ by each of these normal subgroups:
- the quotient by the whole group is the trivial group;
- the quotient by the cyclic subgroup of order 7 is the order 2 cyclic group $c_2$;
- the quotient by the trivial subgroup is the group $D_{14}$.
Finally, observe that the image of any homomorphism $D_{14} to c_7$ cannot be isomorphic to $c_2$ or to $D_{14}$, so the only possibility is that the image is trivial subgroup of $c_7$.
Thus, the only homomorphism $D_{14} to c_7$ is the trivial homomorphism.
answered Dec 5 '18 at 18:20
Lee MosherLee Mosher
48.5k33681
$endgroup$
add a comment |
$begingroup$
Here's a productive way to go about this question.
The isomorphism theorems tell us that the image of a homomorphism is determined, up to isomorphism, by the kernel: the image is isomorphic to the quotient modulo the kernel.
Also, the kernel is a normal subgroup.
So, let's start by listing all normal subgroups of $D_{14}$:
- the whole group $D_{14}$;
- its cyclic subgroup of order 7;
- the trivial subgroup.
Next, let's compute the quotient (up to isomorphism) of $D_{14}$ by each of these normal subgroups:
- the quotient by the whole group is the trivial group;
- the quotient by the cyclic subgroup of order 7 is the order 2 cyclic group $c_2$;
- the quotient by the trivial subgroup is the group $D_{14}$.
Finally, observe that the image of any homomorphism $D_{14} to c_7$ cannot be isomorphic to $c_2$ or to $D_{14}$, so the only possibility is that the image is trivial subgroup of $c_7$.
Thus, the only homomorphism $D_{14} to c_7$ is the trivial homomorphism.
answered Dec 5 '18 at 18:20
Lee MosherLee Mosher
48.5k33681
$endgroup$
add a comment |
$begingroup$
Here's a productive way to go about this question.
The isomorphism theorems tell us that the image of a homomorphism is determined, up to isomorphism, by the kernel: the image is isomorphic to the quotient modulo the kernel.
Also, the kernel is a normal subgroup.
So, let's start by listing all normal subgroups of $D_{14}$:
- the whole group $D_{14}$;
- its cyclic subgroup of order 7;
- the trivial subgroup.
Next, let's compute the quotient (up to isomorphism) of $D_{14}$ by each of these normal subgroups:
- the quotient by the whole group is the trivial group;
- the quotient by the cyclic subgroup of order 7 is the order 2 cyclic group $c_2$;
- the quotient by the trivial subgroup is the group $D_{14}$.
Finally, observe that the image of any homomorphism $D_{14} to c_7$ cannot be isomorphic to $c_2$ or to $D_{14}$, so the only possibility is that the image is trivial subgroup of $c_7$.
Thus, the only homomorphism $D_{14} to c_7$ is the trivial homomorphism.
answered Dec 5 '18 at 18:20
Lee MosherLee Mosher
48.5k33681
$endgroup$
Here's a productive way to go about this question.
The isomorphism theorems tell us that the image of a homomorphism is determined, up to isomorphism, by the kernel: the image is isomorphic to the quotient modulo the kernel.
Also, the kernel is a normal subgroup.
So, let's start by listing all normal subgroups of $D_{14}$:
- the whole group $D_{14}$;
- its cyclic subgroup of order 7;
- the trivial subgroup.
Next, let's compute the quotient (up to isomorphism) of $D_{14}$ by each of these normal subgroups:
- the quotient by the whole group is the trivial group;
- the quotient by the cyclic subgroup of order 7 is the order 2 cyclic group $c_2$;
- the quotient by the trivial subgroup is the group $D_{14}$.
Finally, observe that the image of any homomorphism $D_{14} to c_7$ cannot be isomorphic to $c_2$ or to $D_{14}$, so the only possibility is that the image is trivial subgroup of $c_7$.
Thus, the only homomorphism $D_{14} to c_7$ is the trivial homomorphism.
answered Dec 5 '18 at 18:20
Lee MosherLee Mosher
48.5k33681
answered Dec 5 '18 at 18:20
Lee MosherLee Mosher
48.5k33681
answered Dec 5 '18 at 18:20
Lee MosherLee Mosher
48.5k33681
answered Dec 5 '18 at 18:20
Lee MosherLee Mosher
48.5k33681
48.5k33681
add a comment |
add a comment |
$begingroup$
If you found that there were only two possible subgroups, then they would have to be G itself and {e}, meaning there are no homomorphisms.
I'm really confused by some of the stuff you wrote, though. I would think that any of the rotations of the heptagon (1/7, 2/7, ..., 6/7) could be generators.
I might just be completely misunderstanding though, apologies.
answered Dec 5 '18 at 18:04
nessness
375
$endgroup$
add a comment |
$begingroup$
If you found that there were only two possible subgroups, then they would have to be G itself and {e}, meaning there are no homomorphisms.
I'm really confused by some of the stuff you wrote, though. I would think that any of the rotations of the heptagon (1/7, 2/7, ..., 6/7) could be generators.
I might just be completely misunderstanding though, apologies.
answered Dec 5 '18 at 18:04
nessness
375
$endgroup$
add a comment |
$begingroup$
If you found that there were only two possible subgroups, then they would have to be G itself and {e}, meaning there are no homomorphisms.
I'm really confused by some of the stuff you wrote, though. I would think that any of the rotations of the heptagon (1/7, 2/7, ..., 6/7) could be generators.
I might just be completely misunderstanding though, apologies.
answered Dec 5 '18 at 18:04
nessness
375
$endgroup$
If you found that there were only two possible subgroups, then they would have to be G itself and {e}, meaning there are no homomorphisms.
I'm really confused by some of the stuff you wrote, though. I would think that any of the rotations of the heptagon (1/7, 2/7, ..., 6/7) could be generators.
I might just be completely misunderstanding though, apologies.
answered Dec 5 '18 at 18:04
nessness
375
answered Dec 5 '18 at 18:04
nessness
375
answered Dec 5 '18 at 18:04
nessness
375
answered Dec 5 '18 at 18:04
nessness
375
375
add a comment |
add a comment |
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