Mixing
Showing that $Ycong W$ but $X/Ynot cong X/W$.
1
$begingroup$
I was trying to solve the following question:
Let $X=mathbb{Z}_4timesmathbb{Z}_2$, $Y={0,2}times{0}$ and $W={0}times mathbb{Z}_2$. Show that $Ycong W$ but $X/Ynot cong X/W$.
I guess I must verify manually the Isomorphism theorems (link), but how should I do it without those theorems? I don't understand how to verify the statement.
abstract-algebra group-theory
asked Jan 12 at 14:45
BadukBaduk
171
$endgroup$
|
show 1 more comment
1
$begingroup$
I was trying to solve the following question:
Let $X=mathbb{Z}_4timesmathbb{Z}_2$, $Y={0,2}times{0}$ and $W={0}times mathbb{Z}_2$. Show that $Ycong W$ but $X/Ynot cong X/W$.
I guess I must verify manually the Isomorphism theorems (link), but how should I do it without those theorems? I don't understand how to verify the statement.
abstract-algebra group-theory
asked Jan 12 at 14:45
BadukBaduk
171
$endgroup$
$begingroup$
Recall that if $J$ and $K$ are respective subgroups of $G$ and $H$, then $(Gtimes H)/(Jtimes K)cong(G/J)times(H/K)$.
$endgroup$
– Ben W
Jan 12 at 14:52
1
$begingroup$
There are two things you are asked to show. Which of them do you have problems with? The problem is clearly designed such that everything is small enough that you can prove things directly by listing elements one by one.
$endgroup$
– Henning Makholm
Jan 12 at 14:52
$begingroup$
@HenningMakholm I have problem with both of them but lets start with the first one. I just don't understand how to approach to those questions without using those theorems?
$endgroup$
– Baduk
Jan 12 at 14:54
1
$begingroup$
@Baduk: The most basic and direct way to show that two groups are isomorphic is to present an isomorphism between them.
$endgroup$
– Henning Makholm
Jan 12 at 14:54
$begingroup$
@HenningMakholm Yes I know that. Can't think of a good isomorphism to show it
$endgroup$
– Baduk
Jan 12 at 15:33
|
show 1 more comment
1
1
1
1
$begingroup$
I was trying to solve the following question:
Let $X=mathbb{Z}_4timesmathbb{Z}_2$, $Y={0,2}times{0}$ and $W={0}times mathbb{Z}_2$. Show that $Ycong W$ but $X/Ynot cong X/W$.
I guess I must verify manually the Isomorphism theorems (link), but how should I do it without those theorems? I don't understand how to verify the statement.
abstract-algebra group-theory
asked Jan 12 at 14:45
BadukBaduk
171
$endgroup$
I was trying to solve the following question:
Let $X=mathbb{Z}_4timesmathbb{Z}_2$, $Y={0,2}times{0}$ and $W={0}times mathbb{Z}_2$. Show that $Ycong W$ but $X/Ynot cong X/W$.
I guess I must verify manually the Isomorphism theorems (link), but how should I do it without those theorems? I don't understand how to verify the statement.
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 12 at 14:45
BadukBaduk
171
asked Jan 12 at 14:45
BadukBaduk
171
asked Jan 12 at 14:45
BadukBaduk
171
asked Jan 12 at 14:45
BadukBaduk
171
asked Jan 12 at 14:45
BadukBaduk
171
171
$begingroup$
Recall that if $J$ and $K$ are respective subgroups of $G$ and $H$, then $(Gtimes H)/(Jtimes K)cong(G/J)times(H/K)$.
$endgroup$
– Ben W
Jan 12 at 14:52
1
$begingroup$
There are two things you are asked to show. Which of them do you have problems with? The problem is clearly designed such that everything is small enough that you can prove things directly by listing elements one by one.
$endgroup$
– Henning Makholm
Jan 12 at 14:52
$begingroup$
@HenningMakholm I have problem with both of them but lets start with the first one. I just don't understand how to approach to those questions without using those theorems?
$endgroup$
– Baduk
Jan 12 at 14:54
1
$begingroup$
@Baduk: The most basic and direct way to show that two groups are isomorphic is to present an isomorphism between them.
$endgroup$
– Henning Makholm
Jan 12 at 14:54
$begingroup$
@HenningMakholm Yes I know that. Can't think of a good isomorphism to show it
$endgroup$
– Baduk
Jan 12 at 15:33
|
show 1 more comment
$begingroup$
Recall that if $J$ and $K$ are respective subgroups of $G$ and $H$, then $(Gtimes H)/(Jtimes K)cong(G/J)times(H/K)$.
$endgroup$
– Ben W
Jan 12 at 14:52
1
$begingroup$
There are two things you are asked to show. Which of them do you have problems with? The problem is clearly designed such that everything is small enough that you can prove things directly by listing elements one by one.
$endgroup$
– Henning Makholm
Jan 12 at 14:52
$begingroup$
@HenningMakholm I have problem with both of them but lets start with the first one. I just don't understand how to approach to those questions without using those theorems?
$endgroup$
– Baduk
Jan 12 at 14:54
1
$begingroup$
@Baduk: The most basic and direct way to show that two groups are isomorphic is to present an isomorphism between them.
$endgroup$
– Henning Makholm
Jan 12 at 14:54
$begingroup$
@HenningMakholm Yes I know that. Can't think of a good isomorphism to show it
$endgroup$
– Baduk
Jan 12 at 15:33
$begingroup$
Recall that if $J$ and $K$ are respective subgroups of $G$ and $H$, then $(Gtimes H)/(Jtimes K)cong(G/J)times(H/K)$.
$endgroup$
– Ben W
Jan 12 at 14:52
$begingroup$
Recall that if $J$ and $K$ are respective subgroups of $G$ and $H$, then $(Gtimes H)/(Jtimes K)cong(G/J)times(H/K)$.
$endgroup$
– Ben W
Jan 12 at 14:52
1
1
$begingroup$
There are two things you are asked to show. Which of them do you have problems with? The problem is clearly designed such that everything is small enough that you can prove things directly by listing elements one by one.
$endgroup$
– Henning Makholm
Jan 12 at 14:52
$begingroup$
There are two things you are asked to show. Which of them do you have problems with? The problem is clearly designed such that everything is small enough that you can prove things directly by listing elements one by one.
$endgroup$
– Henning Makholm
Jan 12 at 14:52
$begingroup$
@HenningMakholm I have problem with both of them but lets start with the first one. I just don't understand how to approach to those questions without using those theorems?
$endgroup$
– Baduk
Jan 12 at 14:54
$begingroup$
@HenningMakholm I have problem with both of them but lets start with the first one. I just don't understand how to approach to those questions without using those theorems?
$endgroup$
– Baduk
Jan 12 at 14:54
1
1
$begingroup$
@Baduk: The most basic and direct way to show that two groups are isomorphic is to present an isomorphism between them.
$endgroup$
– Henning Makholm
Jan 12 at 14:54
$begingroup$
@Baduk: The most basic and direct way to show that two groups are isomorphic is to present an isomorphism between them.
$endgroup$
– Henning Makholm
Jan 12 at 14:54
$begingroup$
@HenningMakholm Yes I know that. Can't think of a good isomorphism to show it
$endgroup$
– Baduk
Jan 12 at 15:33
$begingroup$
@HenningMakholm Yes I know that. Can't think of a good isomorphism to show it
$endgroup$
– Baduk
Jan 12 at 15:33
|
show 1 more comment
1 Answer
1
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oldest
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$begingroup$
Hint: Show that $Ycong WcongBbb Z_2$.
Secondly, $X/Ycong V_4=Bbb Z_2×Bbb Z_2$, but $X/WcongBbb Z_4$.
answered Jan 12 at 15:34
Chris CusterChris Custer
13.1k3827
$endgroup$
$begingroup$
I removed all the comments to prevent unnecessary spam. I think I got it. On the second line, did you mean $X/Ycong mathbb{Z}_4$? if so, what good function there are for: $X/Wto mathbb{Z}_2timesmathbb{Z}_2$ and $X/Yto mathbb{Z}_4$?
$endgroup$
– Baduk
Jan 12 at 16:28
$begingroup$
What are the 4 elements of those two gruops?
$endgroup$
– Baduk
Jan 12 at 16:39
$begingroup$
$X/Y={[(0,0)],[(0,1)],[(1,0)],[(1,1)]}$. Each element, other than $[(0,0)]$, has order two. Two elements are equal if they differ by elements of $Y$. So $(0,0)equiv (2,0), (0,1)equiv (2,1),(1,0)equiv (3,0)$ and $(1,1)equiv (3,1)$. $X$ has $8$ elements, but $X/Y$ has $4$.
$endgroup$
– Chris Custer
Jan 12 at 20:13
add a comment |
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$begingroup$
Hint: Show that $Ycong WcongBbb Z_2$.
Secondly, $X/Ycong V_4=Bbb Z_2×Bbb Z_2$, but $X/WcongBbb Z_4$.
answered Jan 12 at 15:34
Chris CusterChris Custer
13.1k3827
$endgroup$
$begingroup$
I removed all the comments to prevent unnecessary spam. I think I got it. On the second line, did you mean $X/Ycong mathbb{Z}_4$? if so, what good function there are for: $X/Wto mathbb{Z}_2timesmathbb{Z}_2$ and $X/Yto mathbb{Z}_4$?
$endgroup$
– Baduk
Jan 12 at 16:28
$begingroup$
What are the 4 elements of those two gruops?
$endgroup$
– Baduk
Jan 12 at 16:39
$begingroup$
$X/Y={[(0,0)],[(0,1)],[(1,0)],[(1,1)]}$. Each element, other than $[(0,0)]$, has order two. Two elements are equal if they differ by elements of $Y$. So $(0,0)equiv (2,0), (0,1)equiv (2,1),(1,0)equiv (3,0)$ and $(1,1)equiv (3,1)$. $X$ has $8$ elements, but $X/Y$ has $4$.
$endgroup$
– Chris Custer
Jan 12 at 20:13
add a comment |
0
$begingroup$
Hint: Show that $Ycong WcongBbb Z_2$.
Secondly, $X/Ycong V_4=Bbb Z_2×Bbb Z_2$, but $X/WcongBbb Z_4$.
answered Jan 12 at 15:34
Chris CusterChris Custer
13.1k3827
$endgroup$
$begingroup$
I removed all the comments to prevent unnecessary spam. I think I got it. On the second line, did you mean $X/Ycong mathbb{Z}_4$? if so, what good function there are for: $X/Wto mathbb{Z}_2timesmathbb{Z}_2$ and $X/Yto mathbb{Z}_4$?
$endgroup$
– Baduk
Jan 12 at 16:28
$begingroup$
What are the 4 elements of those two gruops?
$endgroup$
– Baduk
Jan 12 at 16:39
$begingroup$
$X/Y={[(0,0)],[(0,1)],[(1,0)],[(1,1)]}$. Each element, other than $[(0,0)]$, has order two. Two elements are equal if they differ by elements of $Y$. So $(0,0)equiv (2,0), (0,1)equiv (2,1),(1,0)equiv (3,0)$ and $(1,1)equiv (3,1)$. $X$ has $8$ elements, but $X/Y$ has $4$.
$endgroup$
– Chris Custer
Jan 12 at 20:13
add a comment |
0
0
0
$begingroup$
Hint: Show that $Ycong WcongBbb Z_2$.
Secondly, $X/Ycong V_4=Bbb Z_2×Bbb Z_2$, but $X/WcongBbb Z_4$.
answered Jan 12 at 15:34
Chris CusterChris Custer
13.1k3827
$endgroup$
Hint: Show that $Ycong WcongBbb Z_2$.
Secondly, $X/Ycong V_4=Bbb Z_2×Bbb Z_2$, but $X/WcongBbb Z_4$.
answered Jan 12 at 15:34
Chris CusterChris Custer
13.1k3827
edited Jan 12 at 16:30
answered Jan 12 at 15:34
Chris CusterChris Custer
13.1k3827
answered Jan 12 at 15:34
Chris CusterChris Custer
13.1k3827
answered Jan 12 at 15:34
Chris CusterChris Custer
13.1k3827
13.1k3827
$begingroup$
I removed all the comments to prevent unnecessary spam. I think I got it. On the second line, did you mean $X/Ycong mathbb{Z}_4$? if so, what good function there are for: $X/Wto mathbb{Z}_2timesmathbb{Z}_2$ and $X/Yto mathbb{Z}_4$?
$endgroup$
– Baduk
Jan 12 at 16:28
$begingroup$
What are the 4 elements of those two gruops?
$endgroup$
– Baduk
Jan 12 at 16:39
$begingroup$
$X/Y={[(0,0)],[(0,1)],[(1,0)],[(1,1)]}$. Each element, other than $[(0,0)]$, has order two. Two elements are equal if they differ by elements of $Y$. So $(0,0)equiv (2,0), (0,1)equiv (2,1),(1,0)equiv (3,0)$ and $(1,1)equiv (3,1)$. $X$ has $8$ elements, but $X/Y$ has $4$.
$endgroup$
– Chris Custer
Jan 12 at 20:13
add a comment |
$begingroup$
I removed all the comments to prevent unnecessary spam. I think I got it. On the second line, did you mean $X/Ycong mathbb{Z}_4$? if so, what good function there are for: $X/Wto mathbb{Z}_2timesmathbb{Z}_2$ and $X/Yto mathbb{Z}_4$?
$endgroup$
– Baduk
Jan 12 at 16:28
$begingroup$
What are the 4 elements of those two gruops?
$endgroup$
– Baduk
Jan 12 at 16:39
$begingroup$
$X/Y={[(0,0)],[(0,1)],[(1,0)],[(1,1)]}$. Each element, other than $[(0,0)]$, has order two. Two elements are equal if they differ by elements of $Y$. So $(0,0)equiv (2,0), (0,1)equiv (2,1),(1,0)equiv (3,0)$ and $(1,1)equiv (3,1)$. $X$ has $8$ elements, but $X/Y$ has $4$.
$endgroup$
– Chris Custer
Jan 12 at 20:13
$begingroup$
I removed all the comments to prevent unnecessary spam. I think I got it. On the second line, did you mean $X/Ycong mathbb{Z}_4$? if so, what good function there are for: $X/Wto mathbb{Z}_2timesmathbb{Z}_2$ and $X/Yto mathbb{Z}_4$?
$endgroup$
– Baduk
Jan 12 at 16:28
$begingroup$
I removed all the comments to prevent unnecessary spam. I think I got it. On the second line, did you mean $X/Ycong mathbb{Z}_4$? if so, what good function there are for: $X/Wto mathbb{Z}_2timesmathbb{Z}_2$ and $X/Yto mathbb{Z}_4$?
$endgroup$
– Baduk
Jan 12 at 16:28
$begingroup$
What are the 4 elements of those two gruops?
$endgroup$
– Baduk
Jan 12 at 16:39
$begingroup$
What are the 4 elements of those two gruops?
$endgroup$
– Baduk
Jan 12 at 16:39
$begingroup$
$X/Y={[(0,0)],[(0,1)],[(1,0)],[(1,1)]}$. Each element, other than $[(0,0)]$, has order two. Two elements are equal if they differ by elements of $Y$. So $(0,0)equiv (2,0), (0,1)equiv (2,1),(1,0)equiv (3,0)$ and $(1,1)equiv (3,1)$. $X$ has $8$ elements, but $X/Y$ has $4$.
$endgroup$
– Chris Custer
Jan 12 at 20:13
$begingroup$
$X/Y={[(0,0)],[(0,1)],[(1,0)],[(1,1)]}$. Each element, other than $[(0,0)]$, has order two. Two elements are equal if they differ by elements of $Y$. So $(0,0)equiv (2,0), (0,1)equiv (2,1),(1,0)equiv (3,0)$ and $(1,1)equiv (3,1)$. $X$ has $8$ elements, but $X/Y$ has $4$.
$endgroup$
– Chris Custer
Jan 12 at 20:13
add a comment |
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$begingroup$
Recall that if $J$ and $K$ are respective subgroups of $G$ and $H$, then $(Gtimes H)/(Jtimes K)cong(G/J)times(H/K)$.
$endgroup$
– Ben W
Jan 12 at 14:52
1
$begingroup$
There are two things you are asked to show. Which of them do you have problems with? The problem is clearly designed such that everything is small enough that you can prove things directly by listing elements one by one.
$endgroup$
– Henning Makholm
Jan 12 at 14:52
$begingroup$
@HenningMakholm I have problem with both of them but lets start with the first one. I just don't understand how to approach to those questions without using those theorems?
$endgroup$
– Baduk
Jan 12 at 14:54
1
$begingroup$
@Baduk: The most basic and direct way to show that two groups are isomorphic is to present an isomorphism between them.
$endgroup$
– Henning Makholm
Jan 12 at 14:54
$begingroup$
@HenningMakholm Yes I know that. Can't think of a good isomorphism to show it
$endgroup$
– Baduk
Jan 12 at 15:33