$G$ be a group where $G= mathbb{Z}_6 oplus mathbb{Z}_8$ and the normal subgroup $H=langle(2,4)rangle$ use...
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$begingroup$
I have the elements of $H$:
$langle(2,4)rangle={(2,4),(4,0),(0,4),(2,0),(4,4),(0,0)}$
where $|G/H|=8$
possible isomorphic classes:
$mathbb{Z}_8$ ,$mathbb{Z}_4oplus mathbb{Z}_2$, $mathbb{Z}_2oplus mathbb{Z}_2 oplus mathbb{Z}_2$
I know
$mathbb{Z}_6 oplus mathbb{Z}_8$ has an element of order $8$ and $mathbb{Z}_2 oplus mathbb{Z}_2 oplus mathbb{Z}_2$ has no element of order $8$, so that leaves $G/H$ is either isomorphic to $mathbb{Z}_8$ or $mathbb{Z}_4 oplus mathbb{Z}_2$.
I don't know how to find which one it is between those two.
I know I need to compare the orders.
abstract-algebra group-theory group-isomorphism
edited Dec 5 '18 at 14:28
amWhy
1
$endgroup$
add a comment |
$begingroup$
I have the elements of $H$:
$langle(2,4)rangle={(2,4),(4,0),(0,4),(2,0),(4,4),(0,0)}$
where $|G/H|=8$
possible isomorphic classes:
$mathbb{Z}_8$ ,$mathbb{Z}_4oplus mathbb{Z}_2$, $mathbb{Z}_2oplus mathbb{Z}_2 oplus mathbb{Z}_2$
I know
$mathbb{Z}_6 oplus mathbb{Z}_8$ has an element of order $8$ and $mathbb{Z}_2 oplus mathbb{Z}_2 oplus mathbb{Z}_2$ has no element of order $8$, so that leaves $G/H$ is either isomorphic to $mathbb{Z}_8$ or $mathbb{Z}_4 oplus mathbb{Z}_2$.
I don't know how to find which one it is between those two.
I know I need to compare the orders.
abstract-algebra group-theory group-isomorphism
edited Dec 5 '18 at 14:28
amWhy
1
$endgroup$
$begingroup$
Possible duplicate of Find an isomorphism
$endgroup$
– user595420
Dec 5 '18 at 12:26
add a comment |
$begingroup$
I have the elements of $H$:
$langle(2,4)rangle={(2,4),(4,0),(0,4),(2,0),(4,4),(0,0)}$
where $|G/H|=8$
possible isomorphic classes:
$mathbb{Z}_8$ ,$mathbb{Z}_4oplus mathbb{Z}_2$, $mathbb{Z}_2oplus mathbb{Z}_2 oplus mathbb{Z}_2$
I know
$mathbb{Z}_6 oplus mathbb{Z}_8$ has an element of order $8$ and $mathbb{Z}_2 oplus mathbb{Z}_2 oplus mathbb{Z}_2$ has no element of order $8$, so that leaves $G/H$ is either isomorphic to $mathbb{Z}_8$ or $mathbb{Z}_4 oplus mathbb{Z}_2$.
I don't know how to find which one it is between those two.
I know I need to compare the orders.
abstract-algebra group-theory group-isomorphism
edited Dec 5 '18 at 14:28
amWhy
1
$endgroup$
I have the elements of $H$:
$langle(2,4)rangle={(2,4),(4,0),(0,4),(2,0),(4,4),(0,0)}$
where $|G/H|=8$
possible isomorphic classes:
$mathbb{Z}_8$ ,$mathbb{Z}_4oplus mathbb{Z}_2$, $mathbb{Z}_2oplus mathbb{Z}_2 oplus mathbb{Z}_2$
I know
$mathbb{Z}_6 oplus mathbb{Z}_8$ has an element of order $8$ and $mathbb{Z}_2 oplus mathbb{Z}_2 oplus mathbb{Z}_2$ has no element of order $8$, so that leaves $G/H$ is either isomorphic to $mathbb{Z}_8$ or $mathbb{Z}_4 oplus mathbb{Z}_2$.
I don't know how to find which one it is between those two.
I know I need to compare the orders.
abstract-algebra group-theory group-isomorphism
abstract-algebra group-theory group-isomorphism
edited Dec 5 '18 at 14:28
amWhy
1
edited Dec 5 '18 at 14:28
amWhy
1
edited Dec 5 '18 at 14:28
amWhy
1
edited Dec 5 '18 at 14:28
amWhy
1
edited Dec 5 '18 at 14:28
amWhy
1
1
asked Dec 5 '18 at 3:31
user595420
$begingroup$
Possible duplicate of Find an isomorphism
$endgroup$
– user595420
Dec 5 '18 at 12:26
add a comment |
$begingroup$
Possible duplicate of Find an isomorphism
$endgroup$
– user595420
Dec 5 '18 at 12:26
$begingroup$
Possible duplicate of Find an isomorphism
$endgroup$
– user595420
Dec 5 '18 at 12:26
$begingroup$
Possible duplicate of Find an isomorphism
$endgroup$
– user595420
Dec 5 '18 at 12:26
add a comment |
1 Answer
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$begingroup$
But $mathbb Z_4oplusmathbb Z_2$ similarly has no element of order $8$.
The question, though, is does $G/H$ have an element of order $8$; or an element of order $4$. It doesn't have an element of order $8$, because the coset $(0,1)+H$ has order $4$. And we can see that no other coset has order $8$.
So there's our answer: $mathbb Z_4oplusmathbb Z_2$.
answered Dec 5 '18 at 4:56
Chris CusterChris Custer
13.9k3827
$endgroup$
add a comment |
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$begingroup$
But $mathbb Z_4oplusmathbb Z_2$ similarly has no element of order $8$.
The question, though, is does $G/H$ have an element of order $8$; or an element of order $4$. It doesn't have an element of order $8$, because the coset $(0,1)+H$ has order $4$. And we can see that no other coset has order $8$.
So there's our answer: $mathbb Z_4oplusmathbb Z_2$.
answered Dec 5 '18 at 4:56
Chris CusterChris Custer
13.9k3827
$endgroup$
add a comment |
$begingroup$
But $mathbb Z_4oplusmathbb Z_2$ similarly has no element of order $8$.
The question, though, is does $G/H$ have an element of order $8$; or an element of order $4$. It doesn't have an element of order $8$, because the coset $(0,1)+H$ has order $4$. And we can see that no other coset has order $8$.
So there's our answer: $mathbb Z_4oplusmathbb Z_2$.
answered Dec 5 '18 at 4:56
Chris CusterChris Custer
13.9k3827
$endgroup$
add a comment |
$begingroup$
But $mathbb Z_4oplusmathbb Z_2$ similarly has no element of order $8$.
The question, though, is does $G/H$ have an element of order $8$; or an element of order $4$. It doesn't have an element of order $8$, because the coset $(0,1)+H$ has order $4$. And we can see that no other coset has order $8$.
So there's our answer: $mathbb Z_4oplusmathbb Z_2$.
answered Dec 5 '18 at 4:56
Chris CusterChris Custer
13.9k3827
$endgroup$
But $mathbb Z_4oplusmathbb Z_2$ similarly has no element of order $8$.
The question, though, is does $G/H$ have an element of order $8$; or an element of order $4$. It doesn't have an element of order $8$, because the coset $(0,1)+H$ has order $4$. And we can see that no other coset has order $8$.
So there's our answer: $mathbb Z_4oplusmathbb Z_2$.
answered Dec 5 '18 at 4:56
Chris CusterChris Custer
13.9k3827
edited Dec 5 '18 at 5:57
answered Dec 5 '18 at 4:56
Chris CusterChris Custer
13.9k3827
answered Dec 5 '18 at 4:56
Chris CusterChris Custer
13.9k3827
answered Dec 5 '18 at 4:56
Chris CusterChris Custer
13.9k3827
13.9k3827
add a comment |
add a comment |
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$begingroup$
Possible duplicate of Find an isomorphism
$endgroup$
– user595420
Dec 5 '18 at 12:26