Mixing
How do cosets form a group?
$begingroup$
Let G be a group and H be a Subgroup . Then if H is normal then the cosets form a group . I cannot understand what is meant by cosets form a group . To from a group , group operation on two elements must lie inside the group.So when we say that cosets form a group , group operation on any two elements from any two cosets must lie inside the two cosets chosen earlier or the set of cosets (which is group G itself) ?
abstract-algebra group-theory
asked Jan 21 at 17:31
mathnormiemathnormie
184
$endgroup$
add a comment |
$begingroup$
Let G be a group and H be a Subgroup . Then if H is normal then the cosets form a group . I cannot understand what is meant by cosets form a group . To from a group , group operation on two elements must lie inside the group.So when we say that cosets form a group , group operation on any two elements from any two cosets must lie inside the two cosets chosen earlier or the set of cosets (which is group G itself) ?
abstract-algebra group-theory
asked Jan 21 at 17:31
mathnormiemathnormie
184
$endgroup$
$begingroup$
"To from a group" has propagated into the answers...
$endgroup$
– Dietrich Burde
Jan 21 at 19:10
add a comment |
$begingroup$
Let G be a group and H be a Subgroup . Then if H is normal then the cosets form a group . I cannot understand what is meant by cosets form a group . To from a group , group operation on two elements must lie inside the group.So when we say that cosets form a group , group operation on any two elements from any two cosets must lie inside the two cosets chosen earlier or the set of cosets (which is group G itself) ?
abstract-algebra group-theory
asked Jan 21 at 17:31
mathnormiemathnormie
184
$endgroup$
Let G be a group and H be a Subgroup . Then if H is normal then the cosets form a group . I cannot understand what is meant by cosets form a group . To from a group , group operation on two elements must lie inside the group.So when we say that cosets form a group , group operation on any two elements from any two cosets must lie inside the two cosets chosen earlier or the set of cosets (which is group G itself) ?
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 21 at 17:31
mathnormiemathnormie
184
asked Jan 21 at 17:31
mathnormiemathnormie
184
asked Jan 21 at 17:31
mathnormiemathnormie
184
asked Jan 21 at 17:31
mathnormiemathnormie
184
asked Jan 21 at 17:31
mathnormiemathnormie
184
184
$begingroup$
"To from a group" has propagated into the answers...
$endgroup$
– Dietrich Burde
Jan 21 at 19:10
add a comment |
$begingroup$
"To from a group" has propagated into the answers...
$endgroup$
– Dietrich Burde
Jan 21 at 19:10
$begingroup$
"To from a group" has propagated into the answers...
$endgroup$
– Dietrich Burde
Jan 21 at 19:10
$begingroup$
"To from a group" has propagated into the answers...
$endgroup$
– Dietrich Burde
Jan 21 at 19:10
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
To from a group , group operation on two elements must lie inside the group.
Yes, it means that ${aHmid ain G }$ has an operation that makes it a group, namely $aHcdot bH=abH$. This operation does not work unless $H$ is normal.
The way to think of it is that the cosets are "big points" which are globbed together from little points. They are "new points" in a different group, and their multiplication is related to the multiplication in the original group by the simple rule above.
Now, perhaps you're feeling a little disoriented because the new "points" are themselves sets of points. This is a common feeling at the beginning, when one has for so long thought of points and sets of points and nothing "stacked deeper" than that. But if you study a bit of basic set theory, you'll find that almost everything consists of variously stacked sets, and that it's not so unusual. Try to get comfortable of this idea of "big points" and simply forgetting what their internals used to be.
So when we say that cosets form a group , group operation on any two elements from any two cosets must lie inside the two cosets chosen earlier or the set of cosets (which is group G itself) ?
It turns out that yes, this "setwise" multiplication of elements from $aH$ and $bH$ equals $abH$, but in my opinion, this is never useful. In fact, it is counterproductive when one moves on to quotients of rings, where the intuition applied to multiplication in the quotient ring is simply wrong.
answered Jan 21 at 17:37
rschwiebrschwieb
107k12102251
$endgroup$
$begingroup$
Thanks I also got a very beautiful explanation from Abstract algebra book of Dummit where he explained it in terms of fibers and homo-morphism between group G and Subgroup H .
$endgroup$
– mathnormie
Jan 21 at 19:34
add a comment |
$begingroup$
If group elements $k_1$ and $k_2$ are in the same coset and $k_3$ and $k_4$ are in the same coset, then
$k_1k_2$ has to be in the same coset as $k_3k_4$.
answered Jan 21 at 17:37
J. W. TannerJ. W. Tanner
2,8671217
$endgroup$
add a comment |
$begingroup$
To from a group, group operation on two elements must lie inside the group. So when we say that cosets form a group, group operation on any two elements from any two cosets must lie inside the two cosets chosen earlier or the set of cosets (which is group G itself)?
To say the cosets form a group means that the elements of this group are the cosets, and the group is the set of all cosets. Hence the new group operation acts on the cosets, not on elements of the cosets, and the result must be another coset. This new group operation is:
$$aH*bH=(ab)H.$$
answered Jan 21 at 17:37
kccukccu
10.6k11229
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To from a group , group operation on two elements must lie inside the group.
Yes, it means that ${aHmid ain G }$ has an operation that makes it a group, namely $aHcdot bH=abH$. This operation does not work unless $H$ is normal.
The way to think of it is that the cosets are "big points" which are globbed together from little points. They are "new points" in a different group, and their multiplication is related to the multiplication in the original group by the simple rule above.
Now, perhaps you're feeling a little disoriented because the new "points" are themselves sets of points. This is a common feeling at the beginning, when one has for so long thought of points and sets of points and nothing "stacked deeper" than that. But if you study a bit of basic set theory, you'll find that almost everything consists of variously stacked sets, and that it's not so unusual. Try to get comfortable of this idea of "big points" and simply forgetting what their internals used to be.
So when we say that cosets form a group , group operation on any two elements from any two cosets must lie inside the two cosets chosen earlier or the set of cosets (which is group G itself) ?
It turns out that yes, this "setwise" multiplication of elements from $aH$ and $bH$ equals $abH$, but in my opinion, this is never useful. In fact, it is counterproductive when one moves on to quotients of rings, where the intuition applied to multiplication in the quotient ring is simply wrong.
answered Jan 21 at 17:37
rschwiebrschwieb
107k12102251
$endgroup$
$begingroup$
Thanks I also got a very beautiful explanation from Abstract algebra book of Dummit where he explained it in terms of fibers and homo-morphism between group G and Subgroup H .
$endgroup$
– mathnormie
Jan 21 at 19:34
add a comment |
$begingroup$
To from a group , group operation on two elements must lie inside the group.
Yes, it means that ${aHmid ain G }$ has an operation that makes it a group, namely $aHcdot bH=abH$. This operation does not work unless $H$ is normal.
The way to think of it is that the cosets are "big points" which are globbed together from little points. They are "new points" in a different group, and their multiplication is related to the multiplication in the original group by the simple rule above.
Now, perhaps you're feeling a little disoriented because the new "points" are themselves sets of points. This is a common feeling at the beginning, when one has for so long thought of points and sets of points and nothing "stacked deeper" than that. But if you study a bit of basic set theory, you'll find that almost everything consists of variously stacked sets, and that it's not so unusual. Try to get comfortable of this idea of "big points" and simply forgetting what their internals used to be.
So when we say that cosets form a group , group operation on any two elements from any two cosets must lie inside the two cosets chosen earlier or the set of cosets (which is group G itself) ?
It turns out that yes, this "setwise" multiplication of elements from $aH$ and $bH$ equals $abH$, but in my opinion, this is never useful. In fact, it is counterproductive when one moves on to quotients of rings, where the intuition applied to multiplication in the quotient ring is simply wrong.
answered Jan 21 at 17:37
rschwiebrschwieb
107k12102251
$endgroup$
$begingroup$
Thanks I also got a very beautiful explanation from Abstract algebra book of Dummit where he explained it in terms of fibers and homo-morphism between group G and Subgroup H .
$endgroup$
– mathnormie
Jan 21 at 19:34
add a comment |
$begingroup$
To from a group , group operation on two elements must lie inside the group.
Yes, it means that ${aHmid ain G }$ has an operation that makes it a group, namely $aHcdot bH=abH$. This operation does not work unless $H$ is normal.
The way to think of it is that the cosets are "big points" which are globbed together from little points. They are "new points" in a different group, and their multiplication is related to the multiplication in the original group by the simple rule above.
Now, perhaps you're feeling a little disoriented because the new "points" are themselves sets of points. This is a common feeling at the beginning, when one has for so long thought of points and sets of points and nothing "stacked deeper" than that. But if you study a bit of basic set theory, you'll find that almost everything consists of variously stacked sets, and that it's not so unusual. Try to get comfortable of this idea of "big points" and simply forgetting what their internals used to be.
So when we say that cosets form a group , group operation on any two elements from any two cosets must lie inside the two cosets chosen earlier or the set of cosets (which is group G itself) ?
It turns out that yes, this "setwise" multiplication of elements from $aH$ and $bH$ equals $abH$, but in my opinion, this is never useful. In fact, it is counterproductive when one moves on to quotients of rings, where the intuition applied to multiplication in the quotient ring is simply wrong.
answered Jan 21 at 17:37
rschwiebrschwieb
107k12102251
$endgroup$
To from a group , group operation on two elements must lie inside the group.
Yes, it means that ${aHmid ain G }$ has an operation that makes it a group, namely $aHcdot bH=abH$. This operation does not work unless $H$ is normal.
The way to think of it is that the cosets are "big points" which are globbed together from little points. They are "new points" in a different group, and their multiplication is related to the multiplication in the original group by the simple rule above.
Now, perhaps you're feeling a little disoriented because the new "points" are themselves sets of points. This is a common feeling at the beginning, when one has for so long thought of points and sets of points and nothing "stacked deeper" than that. But if you study a bit of basic set theory, you'll find that almost everything consists of variously stacked sets, and that it's not so unusual. Try to get comfortable of this idea of "big points" and simply forgetting what their internals used to be.
So when we say that cosets form a group , group operation on any two elements from any two cosets must lie inside the two cosets chosen earlier or the set of cosets (which is group G itself) ?
It turns out that yes, this "setwise" multiplication of elements from $aH$ and $bH$ equals $abH$, but in my opinion, this is never useful. In fact, it is counterproductive when one moves on to quotients of rings, where the intuition applied to multiplication in the quotient ring is simply wrong.
answered Jan 21 at 17:37
rschwiebrschwieb
107k12102251
answered Jan 21 at 17:37
rschwiebrschwieb
107k12102251
answered Jan 21 at 17:37
rschwiebrschwieb
107k12102251
answered Jan 21 at 17:37
rschwiebrschwieb
107k12102251
107k12102251
$begingroup$
Thanks I also got a very beautiful explanation from Abstract algebra book of Dummit where he explained it in terms of fibers and homo-morphism between group G and Subgroup H .
$endgroup$
– mathnormie
Jan 21 at 19:34
add a comment |
$begingroup$
Thanks I also got a very beautiful explanation from Abstract algebra book of Dummit where he explained it in terms of fibers and homo-morphism between group G and Subgroup H .
$endgroup$
– mathnormie
Jan 21 at 19:34
$begingroup$
Thanks I also got a very beautiful explanation from Abstract algebra book of Dummit where he explained it in terms of fibers and homo-morphism between group G and Subgroup H .
$endgroup$
– mathnormie
Jan 21 at 19:34
$begingroup$
Thanks I also got a very beautiful explanation from Abstract algebra book of Dummit where he explained it in terms of fibers and homo-morphism between group G and Subgroup H .
$endgroup$
– mathnormie
Jan 21 at 19:34
add a comment |
$begingroup$
If group elements $k_1$ and $k_2$ are in the same coset and $k_3$ and $k_4$ are in the same coset, then
$k_1k_2$ has to be in the same coset as $k_3k_4$.
answered Jan 21 at 17:37
J. W. TannerJ. W. Tanner
2,8671217
$endgroup$
add a comment |
$begingroup$
If group elements $k_1$ and $k_2$ are in the same coset and $k_3$ and $k_4$ are in the same coset, then
$k_1k_2$ has to be in the same coset as $k_3k_4$.
answered Jan 21 at 17:37
J. W. TannerJ. W. Tanner
2,8671217
$endgroup$
add a comment |
$begingroup$
If group elements $k_1$ and $k_2$ are in the same coset and $k_3$ and $k_4$ are in the same coset, then
$k_1k_2$ has to be in the same coset as $k_3k_4$.
answered Jan 21 at 17:37
J. W. TannerJ. W. Tanner
2,8671217
$endgroup$
If group elements $k_1$ and $k_2$ are in the same coset and $k_3$ and $k_4$ are in the same coset, then
$k_1k_2$ has to be in the same coset as $k_3k_4$.
answered Jan 21 at 17:37
J. W. TannerJ. W. Tanner
2,8671217
answered Jan 21 at 17:37
J. W. TannerJ. W. Tanner
2,8671217
answered Jan 21 at 17:37
J. W. TannerJ. W. Tanner
2,8671217
answered Jan 21 at 17:37
J. W. TannerJ. W. Tanner
2,8671217
2,8671217
add a comment |
add a comment |
$begingroup$
To from a group, group operation on two elements must lie inside the group. So when we say that cosets form a group, group operation on any two elements from any two cosets must lie inside the two cosets chosen earlier or the set of cosets (which is group G itself)?
To say the cosets form a group means that the elements of this group are the cosets, and the group is the set of all cosets. Hence the new group operation acts on the cosets, not on elements of the cosets, and the result must be another coset. This new group operation is:
$$aH*bH=(ab)H.$$
answered Jan 21 at 17:37
kccukccu
10.6k11229
$endgroup$
add a comment |
$begingroup$
To from a group, group operation on two elements must lie inside the group. So when we say that cosets form a group, group operation on any two elements from any two cosets must lie inside the two cosets chosen earlier or the set of cosets (which is group G itself)?
To say the cosets form a group means that the elements of this group are the cosets, and the group is the set of all cosets. Hence the new group operation acts on the cosets, not on elements of the cosets, and the result must be another coset. This new group operation is:
$$aH*bH=(ab)H.$$
answered Jan 21 at 17:37
kccukccu
10.6k11229
$endgroup$
add a comment |
$begingroup$
To from a group, group operation on two elements must lie inside the group. So when we say that cosets form a group, group operation on any two elements from any two cosets must lie inside the two cosets chosen earlier or the set of cosets (which is group G itself)?
To say the cosets form a group means that the elements of this group are the cosets, and the group is the set of all cosets. Hence the new group operation acts on the cosets, not on elements of the cosets, and the result must be another coset. This new group operation is:
$$aH*bH=(ab)H.$$
answered Jan 21 at 17:37
kccukccu
10.6k11229
$endgroup$
To from a group, group operation on two elements must lie inside the group. So when we say that cosets form a group, group operation on any two elements from any two cosets must lie inside the two cosets chosen earlier or the set of cosets (which is group G itself)?
To say the cosets form a group means that the elements of this group are the cosets, and the group is the set of all cosets. Hence the new group operation acts on the cosets, not on elements of the cosets, and the result must be another coset. This new group operation is:
$$aH*bH=(ab)H.$$
answered Jan 21 at 17:37
kccukccu
10.6k11229
answered Jan 21 at 17:37
kccukccu
10.6k11229
answered Jan 21 at 17:37
kccukccu
10.6k11229
answered Jan 21 at 17:37
kccukccu
10.6k11229
10.6k11229
add a comment |
add a comment |
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$begingroup$
"To from a group" has propagated into the answers...
$endgroup$
– Dietrich Burde
Jan 21 at 19:10