Construct a group isomorphism $phi : U(20) to mathbb Z_2 oplus mathbb Z_4$.
Construct a group isomorphism $phi : G_1 to G_2$, with $G_1 = U(20)$ and $G_2 = mathbb Z_2 oplus mathbb Z_4$.
EDIT: removed mapping because it was not an isomorphism
linear-algebra group-theory group-isomorphism
edited Nov 29 '18 at 0:16
Shaun
8,810113680
asked Apr 8 '14 at 14:47
appel
1265
|
show 2 more comments
Construct a group isomorphism $phi : G_1 to G_2$, with $G_1 = U(20)$ and $G_2 = mathbb Z_2 oplus mathbb Z_4$.
EDIT: removed mapping because it was not an isomorphism
linear-algebra group-theory group-isomorphism
edited Nov 29 '18 at 0:16
Shaun
8,810113680
asked Apr 8 '14 at 14:47
appel
1265
1
Why do you want a formula ? Is this really an isomorphism ? What about 3*7 ?
– Denis
Apr 8 '14 at 14:52
It's not even a homomorphism: $3.7 = 1$ in $U(20)$, but their image $(0; 1) + (0; 2) = (0; 3) neq (0; 0)$ in $mathbb{Z}_2 oplusmathbb{Z}_4$. What I'm trying to show is that $f(3.7) neq f(3) + f(7)$. Hence it's not a homomorphism.
– user49685
Apr 8 '14 at 14:54
Oops, you're right. How could I start to find an isomorphism then?
– appel
Apr 8 '14 at 15:01
1
@David: It's the set of all units in the group $(mathbb{Z}_{20}; times)$
– user49685
Apr 8 '14 at 15:10
1
With $U(20)$ I mean the unitary group of degree 20
– appel
Apr 8 '14 at 15:13
|
show 2 more comments
Construct a group isomorphism $phi : G_1 to G_2$, with $G_1 = U(20)$ and $G_2 = mathbb Z_2 oplus mathbb Z_4$.
EDIT: removed mapping because it was not an isomorphism
linear-algebra group-theory group-isomorphism
edited Nov 29 '18 at 0:16
Shaun
8,810113680
asked Apr 8 '14 at 14:47
appel
1265
Construct a group isomorphism $phi : G_1 to G_2$, with $G_1 = U(20)$ and $G_2 = mathbb Z_2 oplus mathbb Z_4$.
EDIT: removed mapping because it was not an isomorphism
linear-algebra group-theory group-isomorphism
linear-algebra group-theory group-isomorphism
edited Nov 29 '18 at 0:16
Shaun
8,810113680
asked Apr 8 '14 at 14:47
appel
1265
edited Nov 29 '18 at 0:16
Shaun
8,810113680
asked Apr 8 '14 at 14:47
appel
1265
edited Nov 29 '18 at 0:16
Shaun
8,810113680
edited Nov 29 '18 at 0:16
Shaun
8,810113680
edited Nov 29 '18 at 0:16
Shaun
8,810113680
8,810113680
asked Apr 8 '14 at 14:47
appel
1265
asked Apr 8 '14 at 14:47
appel
1265
asked Apr 8 '14 at 14:47
appel
1265
1265
1
Why do you want a formula ? Is this really an isomorphism ? What about 3*7 ?
– Denis
Apr 8 '14 at 14:52
It's not even a homomorphism: $3.7 = 1$ in $U(20)$, but their image $(0; 1) + (0; 2) = (0; 3) neq (0; 0)$ in $mathbb{Z}_2 oplusmathbb{Z}_4$. What I'm trying to show is that $f(3.7) neq f(3) + f(7)$. Hence it's not a homomorphism.
– user49685
Apr 8 '14 at 14:54
Oops, you're right. How could I start to find an isomorphism then?
– appel
Apr 8 '14 at 15:01
1
@David: It's the set of all units in the group $(mathbb{Z}_{20}; times)$
– user49685
Apr 8 '14 at 15:10
1
With $U(20)$ I mean the unitary group of degree 20
– appel
Apr 8 '14 at 15:13
|
show 2 more comments
1
Why do you want a formula ? Is this really an isomorphism ? What about 3*7 ?
– Denis
Apr 8 '14 at 14:52
It's not even a homomorphism: $3.7 = 1$ in $U(20)$, but their image $(0; 1) + (0; 2) = (0; 3) neq (0; 0)$ in $mathbb{Z}_2 oplusmathbb{Z}_4$. What I'm trying to show is that $f(3.7) neq f(3) + f(7)$. Hence it's not a homomorphism.
– user49685
Apr 8 '14 at 14:54
Oops, you're right. How could I start to find an isomorphism then?
– appel
Apr 8 '14 at 15:01
1
@David: It's the set of all units in the group $(mathbb{Z}_{20}; times)$
– user49685
Apr 8 '14 at 15:10
1
With $U(20)$ I mean the unitary group of degree 20
– appel
Apr 8 '14 at 15:13
1
1
Why do you want a formula ? Is this really an isomorphism ? What about 3*7 ?
– Denis
Apr 8 '14 at 14:52
Why do you want a formula ? Is this really an isomorphism ? What about 3*7 ?
– Denis
Apr 8 '14 at 14:52
It's not even a homomorphism: $3.7 = 1$ in $U(20)$, but their image $(0; 1) + (0; 2) = (0; 3) neq (0; 0)$ in $mathbb{Z}_2 oplusmathbb{Z}_4$. What I'm trying to show is that $f(3.7) neq f(3) + f(7)$. Hence it's not a homomorphism.
– user49685
Apr 8 '14 at 14:54
It's not even a homomorphism: $3.7 = 1$ in $U(20)$, but their image $(0; 1) + (0; 2) = (0; 3) neq (0; 0)$ in $mathbb{Z}_2 oplusmathbb{Z}_4$. What I'm trying to show is that $f(3.7) neq f(3) + f(7)$. Hence it's not a homomorphism.
– user49685
Apr 8 '14 at 14:54
Oops, you're right. How could I start to find an isomorphism then?
– appel
Apr 8 '14 at 15:01
Oops, you're right. How could I start to find an isomorphism then?
– appel
Apr 8 '14 at 15:01
1
1
@David: It's the set of all units in the group $(mathbb{Z}_{20}; times)$
– user49685
Apr 8 '14 at 15:10
@David: It's the set of all units in the group $(mathbb{Z}_{20}; times)$
– user49685
Apr 8 '14 at 15:10
1
1
With $U(20)$ I mean the unitary group of degree 20
– appel
Apr 8 '14 at 15:13
With $U(20)$ I mean the unitary group of degree 20
– appel
Apr 8 '14 at 15:13
|
show 2 more comments
2 Answers
2
active
oldest
votes
Hint: $Bbb Z _2 oplusBbb Z_4$ is generated by two elements, of orders $2$ and $4$.
details:
define
$$
F(1,0)=19, F(0,1)=17
$$
such as $F$ is an homomorphism. This is possible because the orders of
$19$ and $17$ are 2 and 4.
You then get
$$
F(0,2) = 17times 17 = 9\
F(0,3) = 17times 9 = 13\
F(1,1) = 19times 17 = 3\
F(1,2) = 19times 9 = 11\
F(1,3) = 19times 13 = 7
$$ hence $F$ is an isomorphism.
This was the naïve approach.
As soon as you found $a$ of order 4 and $b$ of order 2 and $bneq a^2$,
you know that the subgroup
$langle arangle$ is of order 4 and does not contain $b$.
So the subgroup $langle a,brangle$ has no other choice than being $U(20)$.
Then the morphism has to be onto (without exhaustive check).
edited Nov 29 '18 at 0:17
Shaun
8,810113680
answered Apr 8 '14 at 15:03
mookid
25.5k52447
I understand there should be a mapping $phi(u) = (a,b)$ with $u in U, a in mathbb Z_2, b in mathbb Z_4$, but I don't see what the function could be.
– appel
Apr 8 '14 at 15:10
I wrote the complete details.
– mookid
Apr 8 '14 at 15:13
Thank you, but doesn't 3 also have order 4? and 9 order 2? Why do you take these numbers.
– appel
Apr 8 '14 at 15:16
These are the first I found (see that 19=-1).
– mookid
Apr 8 '14 at 15:18
I added another solution.
– mookid
Apr 8 '14 at 15:21
|
show 1 more comment
$U(20)={1,3,7,9,11,13,17,19}$ with the group operation multiplication modulo $20$.
Both $U(20)$ and $mathbb Z_2oplusmathbb Z_4$ have presentation $langle a,bmid a^2=e,b^4=e, ab=barangle$.
To see that this is true for $U(20)$ , check the orders of the elements: $o(1)=1,o(3)=o(7)=o(13)=o(17)=4$ and $o(9)=o(11)=o(19)=2$.
Thus to get an isomorphism just match up elements of the same order.
answered Nov 29 '18 at 1:09
Chris Custer
10.9k3824
add a comment |
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2 Answers
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2 Answers
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Hint: $Bbb Z _2 oplusBbb Z_4$ is generated by two elements, of orders $2$ and $4$.
details:
define
$$
F(1,0)=19, F(0,1)=17
$$
such as $F$ is an homomorphism. This is possible because the orders of
$19$ and $17$ are 2 and 4.
You then get
$$
F(0,2) = 17times 17 = 9\
F(0,3) = 17times 9 = 13\
F(1,1) = 19times 17 = 3\
F(1,2) = 19times 9 = 11\
F(1,3) = 19times 13 = 7
$$ hence $F$ is an isomorphism.
This was the naïve approach.
As soon as you found $a$ of order 4 and $b$ of order 2 and $bneq a^2$,
you know that the subgroup
$langle arangle$ is of order 4 and does not contain $b$.
So the subgroup $langle a,brangle$ has no other choice than being $U(20)$.
Then the morphism has to be onto (without exhaustive check).
edited Nov 29 '18 at 0:17
Shaun
8,810113680
answered Apr 8 '14 at 15:03
mookid
25.5k52447
I understand there should be a mapping $phi(u) = (a,b)$ with $u in U, a in mathbb Z_2, b in mathbb Z_4$, but I don't see what the function could be.
– appel
Apr 8 '14 at 15:10
I wrote the complete details.
– mookid
Apr 8 '14 at 15:13
Thank you, but doesn't 3 also have order 4? and 9 order 2? Why do you take these numbers.
– appel
Apr 8 '14 at 15:16
These are the first I found (see that 19=-1).
– mookid
Apr 8 '14 at 15:18
I added another solution.
– mookid
Apr 8 '14 at 15:21
|
show 1 more comment
Hint: $Bbb Z _2 oplusBbb Z_4$ is generated by two elements, of orders $2$ and $4$.
details:
define
$$
F(1,0)=19, F(0,1)=17
$$
such as $F$ is an homomorphism. This is possible because the orders of
$19$ and $17$ are 2 and 4.
You then get
$$
F(0,2) = 17times 17 = 9\
F(0,3) = 17times 9 = 13\
F(1,1) = 19times 17 = 3\
F(1,2) = 19times 9 = 11\
F(1,3) = 19times 13 = 7
$$ hence $F$ is an isomorphism.
This was the naïve approach.
As soon as you found $a$ of order 4 and $b$ of order 2 and $bneq a^2$,
you know that the subgroup
$langle arangle$ is of order 4 and does not contain $b$.
So the subgroup $langle a,brangle$ has no other choice than being $U(20)$.
Then the morphism has to be onto (without exhaustive check).
edited Nov 29 '18 at 0:17
Shaun
8,810113680
answered Apr 8 '14 at 15:03
mookid
25.5k52447
I understand there should be a mapping $phi(u) = (a,b)$ with $u in U, a in mathbb Z_2, b in mathbb Z_4$, but I don't see what the function could be.
– appel
Apr 8 '14 at 15:10
I wrote the complete details.
– mookid
Apr 8 '14 at 15:13
Thank you, but doesn't 3 also have order 4? and 9 order 2? Why do you take these numbers.
– appel
Apr 8 '14 at 15:16
These are the first I found (see that 19=-1).
– mookid
Apr 8 '14 at 15:18
I added another solution.
– mookid
Apr 8 '14 at 15:21
|
show 1 more comment
Hint: $Bbb Z _2 oplusBbb Z_4$ is generated by two elements, of orders $2$ and $4$.
details:
define
$$
F(1,0)=19, F(0,1)=17
$$
such as $F$ is an homomorphism. This is possible because the orders of
$19$ and $17$ are 2 and 4.
You then get
$$
F(0,2) = 17times 17 = 9\
F(0,3) = 17times 9 = 13\
F(1,1) = 19times 17 = 3\
F(1,2) = 19times 9 = 11\
F(1,3) = 19times 13 = 7
$$ hence $F$ is an isomorphism.
This was the naïve approach.
As soon as you found $a$ of order 4 and $b$ of order 2 and $bneq a^2$,
you know that the subgroup
$langle arangle$ is of order 4 and does not contain $b$.
So the subgroup $langle a,brangle$ has no other choice than being $U(20)$.
Then the morphism has to be onto (without exhaustive check).
edited Nov 29 '18 at 0:17
Shaun
8,810113680
answered Apr 8 '14 at 15:03
mookid
25.5k52447
Hint: $Bbb Z _2 oplusBbb Z_4$ is generated by two elements, of orders $2$ and $4$.
details:
define
$$
F(1,0)=19, F(0,1)=17
$$
such as $F$ is an homomorphism. This is possible because the orders of
$19$ and $17$ are 2 and 4.
You then get
$$
F(0,2) = 17times 17 = 9\
F(0,3) = 17times 9 = 13\
F(1,1) = 19times 17 = 3\
F(1,2) = 19times 9 = 11\
F(1,3) = 19times 13 = 7
$$ hence $F$ is an isomorphism.
This was the naïve approach.
As soon as you found $a$ of order 4 and $b$ of order 2 and $bneq a^2$,
you know that the subgroup
$langle arangle$ is of order 4 and does not contain $b$.
So the subgroup $langle a,brangle$ has no other choice than being $U(20)$.
Then the morphism has to be onto (without exhaustive check).
edited Nov 29 '18 at 0:17
Shaun
8,810113680
answered Apr 8 '14 at 15:03
mookid
25.5k52447
edited Nov 29 '18 at 0:17
Shaun
8,810113680
edited Nov 29 '18 at 0:17
Shaun
8,810113680
edited Nov 29 '18 at 0:17
Shaun
8,810113680
8,810113680
answered Apr 8 '14 at 15:03
mookid
25.5k52447
answered Apr 8 '14 at 15:03
mookid
25.5k52447
answered Apr 8 '14 at 15:03
mookid
25.5k52447
25.5k52447
I understand there should be a mapping $phi(u) = (a,b)$ with $u in U, a in mathbb Z_2, b in mathbb Z_4$, but I don't see what the function could be.
– appel
Apr 8 '14 at 15:10
I wrote the complete details.
– mookid
Apr 8 '14 at 15:13
Thank you, but doesn't 3 also have order 4? and 9 order 2? Why do you take these numbers.
– appel
Apr 8 '14 at 15:16
These are the first I found (see that 19=-1).
– mookid
Apr 8 '14 at 15:18
I added another solution.
– mookid
Apr 8 '14 at 15:21
|
show 1 more comment
I understand there should be a mapping $phi(u) = (a,b)$ with $u in U, a in mathbb Z_2, b in mathbb Z_4$, but I don't see what the function could be.
– appel
Apr 8 '14 at 15:10
I wrote the complete details.
– mookid
Apr 8 '14 at 15:13
Thank you, but doesn't 3 also have order 4? and 9 order 2? Why do you take these numbers.
– appel
Apr 8 '14 at 15:16
These are the first I found (see that 19=-1).
– mookid
Apr 8 '14 at 15:18
I added another solution.
– mookid
Apr 8 '14 at 15:21
I understand there should be a mapping $phi(u) = (a,b)$ with $u in U, a in mathbb Z_2, b in mathbb Z_4$, but I don't see what the function could be.
– appel
Apr 8 '14 at 15:10
I understand there should be a mapping $phi(u) = (a,b)$ with $u in U, a in mathbb Z_2, b in mathbb Z_4$, but I don't see what the function could be.
– appel
Apr 8 '14 at 15:10
I wrote the complete details.
– mookid
Apr 8 '14 at 15:13
I wrote the complete details.
– mookid
Apr 8 '14 at 15:13
Thank you, but doesn't 3 also have order 4? and 9 order 2? Why do you take these numbers.
– appel
Apr 8 '14 at 15:16
Thank you, but doesn't 3 also have order 4? and 9 order 2? Why do you take these numbers.
– appel
Apr 8 '14 at 15:16
These are the first I found (see that 19=-1).
– mookid
Apr 8 '14 at 15:18
These are the first I found (see that 19=-1).
– mookid
Apr 8 '14 at 15:18
I added another solution.
– mookid
Apr 8 '14 at 15:21
I added another solution.
– mookid
Apr 8 '14 at 15:21
|
show 1 more comment
$U(20)={1,3,7,9,11,13,17,19}$ with the group operation multiplication modulo $20$.
Both $U(20)$ and $mathbb Z_2oplusmathbb Z_4$ have presentation $langle a,bmid a^2=e,b^4=e, ab=barangle$.
To see that this is true for $U(20)$ , check the orders of the elements: $o(1)=1,o(3)=o(7)=o(13)=o(17)=4$ and $o(9)=o(11)=o(19)=2$.
Thus to get an isomorphism just match up elements of the same order.
answered Nov 29 '18 at 1:09
Chris Custer
10.9k3824
add a comment |
$U(20)={1,3,7,9,11,13,17,19}$ with the group operation multiplication modulo $20$.
Both $U(20)$ and $mathbb Z_2oplusmathbb Z_4$ have presentation $langle a,bmid a^2=e,b^4=e, ab=barangle$.
To see that this is true for $U(20)$ , check the orders of the elements: $o(1)=1,o(3)=o(7)=o(13)=o(17)=4$ and $o(9)=o(11)=o(19)=2$.
Thus to get an isomorphism just match up elements of the same order.
answered Nov 29 '18 at 1:09
Chris Custer
10.9k3824
add a comment |
$U(20)={1,3,7,9,11,13,17,19}$ with the group operation multiplication modulo $20$.
Both $U(20)$ and $mathbb Z_2oplusmathbb Z_4$ have presentation $langle a,bmid a^2=e,b^4=e, ab=barangle$.
To see that this is true for $U(20)$ , check the orders of the elements: $o(1)=1,o(3)=o(7)=o(13)=o(17)=4$ and $o(9)=o(11)=o(19)=2$.
Thus to get an isomorphism just match up elements of the same order.
answered Nov 29 '18 at 1:09
Chris Custer
10.9k3824
$U(20)={1,3,7,9,11,13,17,19}$ with the group operation multiplication modulo $20$.
Both $U(20)$ and $mathbb Z_2oplusmathbb Z_4$ have presentation $langle a,bmid a^2=e,b^4=e, ab=barangle$.
To see that this is true for $U(20)$ , check the orders of the elements: $o(1)=1,o(3)=o(7)=o(13)=o(17)=4$ and $o(9)=o(11)=o(19)=2$.
Thus to get an isomorphism just match up elements of the same order.
answered Nov 29 '18 at 1:09
Chris Custer
10.9k3824
answered Nov 29 '18 at 1:09
Chris Custer
10.9k3824
answered Nov 29 '18 at 1:09
Chris Custer
10.9k3824
answered Nov 29 '18 at 1:09
Chris Custer
10.9k3824
10.9k3824
add a comment |
add a comment |
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1
Why do you want a formula ? Is this really an isomorphism ? What about 3*7 ?
– Denis
Apr 8 '14 at 14:52
It's not even a homomorphism: $3.7 = 1$ in $U(20)$, but their image $(0; 1) + (0; 2) = (0; 3) neq (0; 0)$ in $mathbb{Z}_2 oplusmathbb{Z}_4$. What I'm trying to show is that $f(3.7) neq f(3) + f(7)$. Hence it's not a homomorphism.
– user49685
Apr 8 '14 at 14:54
Oops, you're right. How could I start to find an isomorphism then?
– appel
Apr 8 '14 at 15:01
1
@David: It's the set of all units in the group $(mathbb{Z}_{20}; times)$
– user49685
Apr 8 '14 at 15:10
1
With $U(20)$ I mean the unitary group of degree 20
– appel
Apr 8 '14 at 15:13