Constructing homomorphisms between a group and its subgroup Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Quotient of an Abelian group with its torsion subgroupConstructing groups with a given subgroupDifficulty understanding subgroups for certain simple groupsFinding all groups H (up to isomorphism) such that there is a surjective homomorphism from D8 to Hsubgroup and group homomorphismSubgroup of the free group on 3 generatorsDetailed explanation wanted about group homomorphismsHomomorphism group of homomorphisms between a homomorphism group, and codomain of that groupDoes every group have a finite index subgroup?Finding all group homomorphisms from $mathbbZ_7$ to $mathbbZ_12$
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Constructing homomorphisms between a group and its subgroup
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Quotient of an Abelian group with its torsion subgroupConstructing groups with a given subgroupDifficulty understanding subgroups for certain simple groupsFinding all groups H (up to isomorphism) such that there is a surjective homomorphism from D8 to Hsubgroup and group homomorphismSubgroup of the free group on 3 generatorsDetailed explanation wanted about group homomorphismsHomomorphism group of homomorphisms between a homomorphism group, and codomain of that groupDoes every group have a finite index subgroup?Finding all group homomorphisms from $mathbbZ_7$ to $mathbbZ_12$
$begingroup$
How would I go about constructing a homomorphism between a group and its subgroup. For example, the group multiplicative group $mathbbZ/15mathbbZ$ and the subgroup of the squares of its elements $1,4$.
Also, in general how would I construct a homomorphism between two groups without generators.
(I'm just at A-level, so I would like a simple explanation)
group-theory normal-subgroups group-homomorphism
edited Apr 8 at 17:51
qweruiop
213111
asked Dec 31 '16 at 1:38
learning_mathematicslearning_mathematics
215
$endgroup$
|
show 3 more comments
$begingroup$
How would I go about constructing a homomorphism between a group and its subgroup. For example, the group multiplicative group $mathbbZ/15mathbbZ$ and the subgroup of the squares of its elements $1,4$.
Also, in general how would I construct a homomorphism between two groups without generators.
(I'm just at A-level, so I would like a simple explanation)
group-theory normal-subgroups group-homomorphism
edited Apr 8 at 17:51
qweruiop
213111
asked Dec 31 '16 at 1:38
learning_mathematicslearning_mathematics
215
$endgroup$
$begingroup$
Are you familiar with quotient maps/cosets?
$endgroup$
– πr8
Dec 31 '16 at 1:40
1
$begingroup$
I don't think the subgroup of squares of the elements of Z/15Z is 1,4, did you mean Z/5Z? To get a homomorphism from an abelian group to the subgroup of squares, the map x-> x^2 would work. Also, going the other way, the identity gives a homomorphism from a subgroup to a group. I don't know what you mean by 'between two groups without generators'
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 1:52
$begingroup$
In general, I don't think there's a natural homomorphism between a group and a subgroup. A quotient group, yes, but that's not necessarily a subgroup. The image of a homomorphism G->G is a subgroup but not all subgroups can be expressed this way.
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 2:00
$begingroup$
For the groups with no generators I meant non cyclic groups. And what do u mean by the identity giving a homomorphism from a subgroup to a group?
$endgroup$
– learning_mathematics
Dec 31 '16 at 2:26
$begingroup$
I guess the more proper term is 'inclusion', not identity. If H is a subgroup of G, there is a map H->G that just takes h->h. It's a homomorphism.
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 11:13
|
show 3 more comments
$begingroup$
How would I go about constructing a homomorphism between a group and its subgroup. For example, the group multiplicative group $mathbbZ/15mathbbZ$ and the subgroup of the squares of its elements $1,4$.
Also, in general how would I construct a homomorphism between two groups without generators.
(I'm just at A-level, so I would like a simple explanation)
group-theory normal-subgroups group-homomorphism
edited Apr 8 at 17:51
qweruiop
213111
asked Dec 31 '16 at 1:38
learning_mathematicslearning_mathematics
215
$endgroup$
How would I go about constructing a homomorphism between a group and its subgroup. For example, the group multiplicative group $mathbbZ/15mathbbZ$ and the subgroup of the squares of its elements $1,4$.
Also, in general how would I construct a homomorphism between two groups without generators.
(I'm just at A-level, so I would like a simple explanation)
group-theory normal-subgroups group-homomorphism
group-theory normal-subgroups group-homomorphism
edited Apr 8 at 17:51
qweruiop
213111
asked Dec 31 '16 at 1:38
learning_mathematicslearning_mathematics
215
edited Apr 8 at 17:51
qweruiop
213111
asked Dec 31 '16 at 1:38
learning_mathematicslearning_mathematics
215
edited Apr 8 at 17:51
qweruiop
213111
edited Apr 8 at 17:51
qweruiop
213111
edited Apr 8 at 17:51
qweruiop
213111
213111
asked Dec 31 '16 at 1:38
learning_mathematicslearning_mathematics
215
asked Dec 31 '16 at 1:38
learning_mathematicslearning_mathematics
215
asked Dec 31 '16 at 1:38
learning_mathematicslearning_mathematics
215
215
$begingroup$
Are you familiar with quotient maps/cosets?
$endgroup$
– πr8
Dec 31 '16 at 1:40
1
$begingroup$
I don't think the subgroup of squares of the elements of Z/15Z is 1,4, did you mean Z/5Z? To get a homomorphism from an abelian group to the subgroup of squares, the map x-> x^2 would work. Also, going the other way, the identity gives a homomorphism from a subgroup to a group. I don't know what you mean by 'between two groups without generators'
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 1:52
$begingroup$
In general, I don't think there's a natural homomorphism between a group and a subgroup. A quotient group, yes, but that's not necessarily a subgroup. The image of a homomorphism G->G is a subgroup but not all subgroups can be expressed this way.
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 2:00
$begingroup$
For the groups with no generators I meant non cyclic groups. And what do u mean by the identity giving a homomorphism from a subgroup to a group?
$endgroup$
– learning_mathematics
Dec 31 '16 at 2:26
$begingroup$
I guess the more proper term is 'inclusion', not identity. If H is a subgroup of G, there is a map H->G that just takes h->h. It's a homomorphism.
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 11:13
|
show 3 more comments
$begingroup$
Are you familiar with quotient maps/cosets?
$endgroup$
– πr8
Dec 31 '16 at 1:40
1
$begingroup$
I don't think the subgroup of squares of the elements of Z/15Z is 1,4, did you mean Z/5Z? To get a homomorphism from an abelian group to the subgroup of squares, the map x-> x^2 would work. Also, going the other way, the identity gives a homomorphism from a subgroup to a group. I don't know what you mean by 'between two groups without generators'
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 1:52
$begingroup$
In general, I don't think there's a natural homomorphism between a group and a subgroup. A quotient group, yes, but that's not necessarily a subgroup. The image of a homomorphism G->G is a subgroup but not all subgroups can be expressed this way.
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 2:00
$begingroup$
For the groups with no generators I meant non cyclic groups. And what do u mean by the identity giving a homomorphism from a subgroup to a group?
$endgroup$
– learning_mathematics
Dec 31 '16 at 2:26
$begingroup$
I guess the more proper term is 'inclusion', not identity. If H is a subgroup of G, there is a map H->G that just takes h->h. It's a homomorphism.
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 11:13
$begingroup$
Are you familiar with quotient maps/cosets?
$endgroup$
– πr8
Dec 31 '16 at 1:40
$begingroup$
Are you familiar with quotient maps/cosets?
$endgroup$
– πr8
Dec 31 '16 at 1:40
1
1
$begingroup$
I don't think the subgroup of squares of the elements of Z/15Z is 1,4, did you mean Z/5Z? To get a homomorphism from an abelian group to the subgroup of squares, the map x-> x^2 would work. Also, going the other way, the identity gives a homomorphism from a subgroup to a group. I don't know what you mean by 'between two groups without generators'
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 1:52
$begingroup$
I don't think the subgroup of squares of the elements of Z/15Z is 1,4, did you mean Z/5Z? To get a homomorphism from an abelian group to the subgroup of squares, the map x-> x^2 would work. Also, going the other way, the identity gives a homomorphism from a subgroup to a group. I don't know what you mean by 'between two groups without generators'
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 1:52
$begingroup$
In general, I don't think there's a natural homomorphism between a group and a subgroup. A quotient group, yes, but that's not necessarily a subgroup. The image of a homomorphism G->G is a subgroup but not all subgroups can be expressed this way.
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 2:00
$begingroup$
In general, I don't think there's a natural homomorphism between a group and a subgroup. A quotient group, yes, but that's not necessarily a subgroup. The image of a homomorphism G->G is a subgroup but not all subgroups can be expressed this way.
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 2:00
$begingroup$
For the groups with no generators I meant non cyclic groups. And what do u mean by the identity giving a homomorphism from a subgroup to a group?
$endgroup$
– learning_mathematics
Dec 31 '16 at 2:26
$begingroup$
For the groups with no generators I meant non cyclic groups. And what do u mean by the identity giving a homomorphism from a subgroup to a group?
$endgroup$
– learning_mathematics
Dec 31 '16 at 2:26
$begingroup$
I guess the more proper term is 'inclusion', not identity. If H is a subgroup of G, there is a map H->G that just takes h->h. It's a homomorphism.
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 11:13
$begingroup$
I guess the more proper term is 'inclusion', not identity. If H is a subgroup of G, there is a map H->G that just takes h->h. It's a homomorphism.
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 11:13
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If your group $G$ is a finite abelian group then every subgroup of $G$ is a homomorphic image of $G$. This follows from the classification of these groups (the fundamental theorem for finite abelian groups). The multiplicative group of $mathbbZ/15mathbbZ$ is abelian so there exists a homomophism to the subgroup consisting of the squares of elements: it is simply the map $xmapsto x^2$, as was pointed out in the comments.
However, in general if $H$ is a subgroup of $G$ there does not exist a surjective homomorphism $Gtwoheadrightarrow H$. For example, consider the alternating group $G=A_5$. This is a subgroup of the permutation group $S_5$ (if you are going through a text book, these groups will almost certainly be covered!). Now $A_5$ is simple. This means that every normal subgroup is either trivial or the whole group; in terms of homomorphic images, this means that every homomorphism $phi: A_5rightarrow H$ has trivial image or its image is isomorphic to $A_5$ (by the first isomorphism theorem). In particular, if we have $phi:A_5rightarrow H$ where $H$ is a subgroup of $A_5$ then the image of $phi$ is trivial. (Note that the group $A_5$ contains proper, non-trivial subgroups; abelian simple groups, such as the cyclic group of order $3$, do not contain any proper, non-trivial subgroups.)
answered Jan 18 '17 at 15:00
user1729user1729
17.7k64294
$endgroup$
add a comment |
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$begingroup$
If your group $G$ is a finite abelian group then every subgroup of $G$ is a homomorphic image of $G$. This follows from the classification of these groups (the fundamental theorem for finite abelian groups). The multiplicative group of $mathbbZ/15mathbbZ$ is abelian so there exists a homomophism to the subgroup consisting of the squares of elements: it is simply the map $xmapsto x^2$, as was pointed out in the comments.
However, in general if $H$ is a subgroup of $G$ there does not exist a surjective homomorphism $Gtwoheadrightarrow H$. For example, consider the alternating group $G=A_5$. This is a subgroup of the permutation group $S_5$ (if you are going through a text book, these groups will almost certainly be covered!). Now $A_5$ is simple. This means that every normal subgroup is either trivial or the whole group; in terms of homomorphic images, this means that every homomorphism $phi: A_5rightarrow H$ has trivial image or its image is isomorphic to $A_5$ (by the first isomorphism theorem). In particular, if we have $phi:A_5rightarrow H$ where $H$ is a subgroup of $A_5$ then the image of $phi$ is trivial. (Note that the group $A_5$ contains proper, non-trivial subgroups; abelian simple groups, such as the cyclic group of order $3$, do not contain any proper, non-trivial subgroups.)
answered Jan 18 '17 at 15:00
user1729user1729
17.7k64294
$endgroup$
add a comment |
$begingroup$
If your group $G$ is a finite abelian group then every subgroup of $G$ is a homomorphic image of $G$. This follows from the classification of these groups (the fundamental theorem for finite abelian groups). The multiplicative group of $mathbbZ/15mathbbZ$ is abelian so there exists a homomophism to the subgroup consisting of the squares of elements: it is simply the map $xmapsto x^2$, as was pointed out in the comments.
However, in general if $H$ is a subgroup of $G$ there does not exist a surjective homomorphism $Gtwoheadrightarrow H$. For example, consider the alternating group $G=A_5$. This is a subgroup of the permutation group $S_5$ (if you are going through a text book, these groups will almost certainly be covered!). Now $A_5$ is simple. This means that every normal subgroup is either trivial or the whole group; in terms of homomorphic images, this means that every homomorphism $phi: A_5rightarrow H$ has trivial image or its image is isomorphic to $A_5$ (by the first isomorphism theorem). In particular, if we have $phi:A_5rightarrow H$ where $H$ is a subgroup of $A_5$ then the image of $phi$ is trivial. (Note that the group $A_5$ contains proper, non-trivial subgroups; abelian simple groups, such as the cyclic group of order $3$, do not contain any proper, non-trivial subgroups.)
answered Jan 18 '17 at 15:00
user1729user1729
17.7k64294
$endgroup$
add a comment |
$begingroup$
If your group $G$ is a finite abelian group then every subgroup of $G$ is a homomorphic image of $G$. This follows from the classification of these groups (the fundamental theorem for finite abelian groups). The multiplicative group of $mathbbZ/15mathbbZ$ is abelian so there exists a homomophism to the subgroup consisting of the squares of elements: it is simply the map $xmapsto x^2$, as was pointed out in the comments.
However, in general if $H$ is a subgroup of $G$ there does not exist a surjective homomorphism $Gtwoheadrightarrow H$. For example, consider the alternating group $G=A_5$. This is a subgroup of the permutation group $S_5$ (if you are going through a text book, these groups will almost certainly be covered!). Now $A_5$ is simple. This means that every normal subgroup is either trivial or the whole group; in terms of homomorphic images, this means that every homomorphism $phi: A_5rightarrow H$ has trivial image or its image is isomorphic to $A_5$ (by the first isomorphism theorem). In particular, if we have $phi:A_5rightarrow H$ where $H$ is a subgroup of $A_5$ then the image of $phi$ is trivial. (Note that the group $A_5$ contains proper, non-trivial subgroups; abelian simple groups, such as the cyclic group of order $3$, do not contain any proper, non-trivial subgroups.)
answered Jan 18 '17 at 15:00
user1729user1729
17.7k64294
$endgroup$
If your group $G$ is a finite abelian group then every subgroup of $G$ is a homomorphic image of $G$. This follows from the classification of these groups (the fundamental theorem for finite abelian groups). The multiplicative group of $mathbbZ/15mathbbZ$ is abelian so there exists a homomophism to the subgroup consisting of the squares of elements: it is simply the map $xmapsto x^2$, as was pointed out in the comments.
However, in general if $H$ is a subgroup of $G$ there does not exist a surjective homomorphism $Gtwoheadrightarrow H$. For example, consider the alternating group $G=A_5$. This is a subgroup of the permutation group $S_5$ (if you are going through a text book, these groups will almost certainly be covered!). Now $A_5$ is simple. This means that every normal subgroup is either trivial or the whole group; in terms of homomorphic images, this means that every homomorphism $phi: A_5rightarrow H$ has trivial image or its image is isomorphic to $A_5$ (by the first isomorphism theorem). In particular, if we have $phi:A_5rightarrow H$ where $H$ is a subgroup of $A_5$ then the image of $phi$ is trivial. (Note that the group $A_5$ contains proper, non-trivial subgroups; abelian simple groups, such as the cyclic group of order $3$, do not contain any proper, non-trivial subgroups.)
answered Jan 18 '17 at 15:00
user1729user1729
17.7k64294
edited Jan 18 '17 at 15:22
answered Jan 18 '17 at 15:00
user1729user1729
17.7k64294
answered Jan 18 '17 at 15:00
user1729user1729
17.7k64294
answered Jan 18 '17 at 15:00
user1729user1729
17.7k64294
17.7k64294
add a comment |
add a comment |
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$begingroup$
Are you familiar with quotient maps/cosets?
$endgroup$
– πr8
Dec 31 '16 at 1:40
1
$begingroup$
I don't think the subgroup of squares of the elements of Z/15Z is 1,4, did you mean Z/5Z? To get a homomorphism from an abelian group to the subgroup of squares, the map x-> x^2 would work. Also, going the other way, the identity gives a homomorphism from a subgroup to a group. I don't know what you mean by 'between two groups without generators'
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 1:52
$begingroup$
In general, I don't think there's a natural homomorphism between a group and a subgroup. A quotient group, yes, but that's not necessarily a subgroup. The image of a homomorphism G->G is a subgroup but not all subgroups can be expressed this way.
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 2:00
$begingroup$
For the groups with no generators I meant non cyclic groups. And what do u mean by the identity giving a homomorphism from a subgroup to a group?
$endgroup$
– learning_mathematics
Dec 31 '16 at 2:26
$begingroup$
I guess the more proper term is 'inclusion', not identity. If H is a subgroup of G, there is a map H->G that just takes h->h. It's a homomorphism.
$endgroup$
– spaceisdarkgreen
Dec 31 '16 at 11:13