Relation between the order of the elements of 2 groups and isomorphism Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is every bijection that preserves element order an isomorphism?Isomorphism of Direct Product of GroupsIsomorphisms of direct products of finite abelian groupsProving isomorphism between two sets of ordered pairs.Isomorphism between quotient groups with normal subgroupsWhen are two direct products of groups isomorphic?Extend isomorphism of subgroups to homomorphism of groupsExtension of isomorphismWhen are free products of pairs of groups isomorphic?Isomorphism and RelationsLet $G,H$ be groups and $varphi:G times Hto G$ and $H'=ker(varphi)$. Show that $(Gtimes H)/H'cong G$
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Relation between the order of the elements of 2 groups and isomorphism
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is every bijection that preserves element order an isomorphism?Isomorphism of Direct Product of GroupsIsomorphisms of direct products of finite abelian groupsProving isomorphism between two sets of ordered pairs.Isomorphism between quotient groups with normal subgroupsWhen are two direct products of groups isomorphic?Extend isomorphism of subgroups to homomorphism of groupsExtension of isomorphismWhen are free products of pairs of groups isomorphic?Isomorphism and RelationsLet $G,H$ be groups and $varphi:G times Hto G$ and $H'=ker(varphi)$. Show that $(Gtimes H)/H'cong G$
$begingroup$
Let be $G$ and $H$ two equinumerous groups of order $n$.
We label the elements in such a way that
$$
1< |g_1|le|g_2|le cdots le |g_n|
$$
$$
1< |h_1|le|h_2|le cdots le |h_n|
$$
If $(|g_1|,|g_2|,cdots,|g_n|)ne (|h_1|,|h_2|,cdots,|h_n|)$ then $Gnotcong H$.
Is there a result deducing from $(|g_1|,|g_2|,cdots,|g_n|)= (|h_1|,|h_2|,cdots,|h_n|)$ something about the isomorphism of $G$ and $H$?
abstract-algebra group-isomorphism
asked Apr 1 at 13:39
Aaron LenzAaron Lenz
7110
$endgroup$
add a comment |
$begingroup$
Let be $G$ and $H$ two equinumerous groups of order $n$.
We label the elements in such a way that
$$
1< |g_1|le|g_2|le cdots le |g_n|
$$
$$
1< |h_1|le|h_2|le cdots le |h_n|
$$
If $(|g_1|,|g_2|,cdots,|g_n|)ne (|h_1|,|h_2|,cdots,|h_n|)$ then $Gnotcong H$.
Is there a result deducing from $(|g_1|,|g_2|,cdots,|g_n|)= (|h_1|,|h_2|,cdots,|h_n|)$ something about the isomorphism of $G$ and $H$?
abstract-algebra group-isomorphism
asked Apr 1 at 13:39
Aaron LenzAaron Lenz
7110
$endgroup$
1
$begingroup$
Well one obvious property is that the isomorphism will send equal order elements to equal order elements and thus if $(|g_1|,...,|g_n|)$ is strictly increasing then the isomorphism is unique
$endgroup$
– Μάρκος Καραμέρης
Apr 1 at 13:47
add a comment |
$begingroup$
Let be $G$ and $H$ two equinumerous groups of order $n$.
We label the elements in such a way that
$$
1< |g_1|le|g_2|le cdots le |g_n|
$$
$$
1< |h_1|le|h_2|le cdots le |h_n|
$$
If $(|g_1|,|g_2|,cdots,|g_n|)ne (|h_1|,|h_2|,cdots,|h_n|)$ then $Gnotcong H$.
Is there a result deducing from $(|g_1|,|g_2|,cdots,|g_n|)= (|h_1|,|h_2|,cdots,|h_n|)$ something about the isomorphism of $G$ and $H$?
abstract-algebra group-isomorphism
asked Apr 1 at 13:39
Aaron LenzAaron Lenz
7110
$endgroup$
Let be $G$ and $H$ two equinumerous groups of order $n$.
We label the elements in such a way that
$$
1< |g_1|le|g_2|le cdots le |g_n|
$$
$$
1< |h_1|le|h_2|le cdots le |h_n|
$$
If $(|g_1|,|g_2|,cdots,|g_n|)ne (|h_1|,|h_2|,cdots,|h_n|)$ then $Gnotcong H$.
Is there a result deducing from $(|g_1|,|g_2|,cdots,|g_n|)= (|h_1|,|h_2|,cdots,|h_n|)$ something about the isomorphism of $G$ and $H$?
abstract-algebra group-isomorphism
abstract-algebra group-isomorphism
asked Apr 1 at 13:39
Aaron LenzAaron Lenz
7110
asked Apr 1 at 13:39
Aaron LenzAaron Lenz
7110
asked Apr 1 at 13:39
Aaron LenzAaron Lenz
7110
asked Apr 1 at 13:39
Aaron LenzAaron Lenz
7110
asked Apr 1 at 13:39
Aaron LenzAaron Lenz
7110
7110
1
$begingroup$
Well one obvious property is that the isomorphism will send equal order elements to equal order elements and thus if $(|g_1|,...,|g_n|)$ is strictly increasing then the isomorphism is unique
$endgroup$
– Μάρκος Καραμέρης
Apr 1 at 13:47
add a comment |
1
$begingroup$
Well one obvious property is that the isomorphism will send equal order elements to equal order elements and thus if $(|g_1|,...,|g_n|)$ is strictly increasing then the isomorphism is unique
$endgroup$
– Μάρκος Καραμέρης
Apr 1 at 13:47
1
1
$begingroup$
Well one obvious property is that the isomorphism will send equal order elements to equal order elements and thus if $(|g_1|,...,|g_n|)$ is strictly increasing then the isomorphism is unique
$endgroup$
– Μάρκος Καραμέρης
Apr 1 at 13:47
$begingroup$
Well one obvious property is that the isomorphism will send equal order elements to equal order elements and thus if $(|g_1|,...,|g_n|)$ is strictly increasing then the isomorphism is unique
$endgroup$
– Μάρκος Καραμέρης
Apr 1 at 13:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It does not follow that $G$ and $H$ are isomorphic from the given hypothesis. As indicated in this answer to a similar question (https://math.stackexchange.com/a/1478993/81163), $G = mathbbZ/4times mathbbZ/4$ and $H = mathbbZ/2 times Q_8$. where $Q_8$ is the quaternions, are not isomorphic, but they have the same number of elements of each order.
answered Apr 1 at 13:48
JamesJames
4,4651822
$endgroup$
$begingroup$
Thank you for your answer. This was clear to me, because of that I I wrote "deduce something". Perhaps together with additional condition. Or one can restrict the collection of isomorphic classes, or deduce the number thereof. I don't know exactly, any result from which one can deduce something.
$endgroup$
– Aaron Lenz
Apr 1 at 14:45
add a comment |
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$begingroup$
It does not follow that $G$ and $H$ are isomorphic from the given hypothesis. As indicated in this answer to a similar question (https://math.stackexchange.com/a/1478993/81163), $G = mathbbZ/4times mathbbZ/4$ and $H = mathbbZ/2 times Q_8$. where $Q_8$ is the quaternions, are not isomorphic, but they have the same number of elements of each order.
answered Apr 1 at 13:48
JamesJames
4,4651822
$endgroup$
$begingroup$
Thank you for your answer. This was clear to me, because of that I I wrote "deduce something". Perhaps together with additional condition. Or one can restrict the collection of isomorphic classes, or deduce the number thereof. I don't know exactly, any result from which one can deduce something.
$endgroup$
– Aaron Lenz
Apr 1 at 14:45
add a comment |
$begingroup$
It does not follow that $G$ and $H$ are isomorphic from the given hypothesis. As indicated in this answer to a similar question (https://math.stackexchange.com/a/1478993/81163), $G = mathbbZ/4times mathbbZ/4$ and $H = mathbbZ/2 times Q_8$. where $Q_8$ is the quaternions, are not isomorphic, but they have the same number of elements of each order.
answered Apr 1 at 13:48
JamesJames
4,4651822
$endgroup$
$begingroup$
Thank you for your answer. This was clear to me, because of that I I wrote "deduce something". Perhaps together with additional condition. Or one can restrict the collection of isomorphic classes, or deduce the number thereof. I don't know exactly, any result from which one can deduce something.
$endgroup$
– Aaron Lenz
Apr 1 at 14:45
add a comment |
$begingroup$
It does not follow that $G$ and $H$ are isomorphic from the given hypothesis. As indicated in this answer to a similar question (https://math.stackexchange.com/a/1478993/81163), $G = mathbbZ/4times mathbbZ/4$ and $H = mathbbZ/2 times Q_8$. where $Q_8$ is the quaternions, are not isomorphic, but they have the same number of elements of each order.
answered Apr 1 at 13:48
JamesJames
4,4651822
$endgroup$
It does not follow that $G$ and $H$ are isomorphic from the given hypothesis. As indicated in this answer to a similar question (https://math.stackexchange.com/a/1478993/81163), $G = mathbbZ/4times mathbbZ/4$ and $H = mathbbZ/2 times Q_8$. where $Q_8$ is the quaternions, are not isomorphic, but they have the same number of elements of each order.
answered Apr 1 at 13:48
JamesJames
4,4651822
answered Apr 1 at 13:48
JamesJames
4,4651822
answered Apr 1 at 13:48
JamesJames
4,4651822
answered Apr 1 at 13:48
JamesJames
4,4651822
4,4651822
$begingroup$
Thank you for your answer. This was clear to me, because of that I I wrote "deduce something". Perhaps together with additional condition. Or one can restrict the collection of isomorphic classes, or deduce the number thereof. I don't know exactly, any result from which one can deduce something.
$endgroup$
– Aaron Lenz
Apr 1 at 14:45
add a comment |
$begingroup$
Thank you for your answer. This was clear to me, because of that I I wrote "deduce something". Perhaps together with additional condition. Or one can restrict the collection of isomorphic classes, or deduce the number thereof. I don't know exactly, any result from which one can deduce something.
$endgroup$
– Aaron Lenz
Apr 1 at 14:45
$begingroup$
Thank you for your answer. This was clear to me, because of that I I wrote "deduce something". Perhaps together with additional condition. Or one can restrict the collection of isomorphic classes, or deduce the number thereof. I don't know exactly, any result from which one can deduce something.
$endgroup$
– Aaron Lenz
Apr 1 at 14:45
$begingroup$
Thank you for your answer. This was clear to me, because of that I I wrote "deduce something". Perhaps together with additional condition. Or one can restrict the collection of isomorphic classes, or deduce the number thereof. I don't know exactly, any result from which one can deduce something.
$endgroup$
– Aaron Lenz
Apr 1 at 14:45
add a comment |
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$begingroup$
Well one obvious property is that the isomorphism will send equal order elements to equal order elements and thus if $(|g_1|,...,|g_n|)$ is strictly increasing then the isomorphism is unique
$endgroup$
– Μάρκος Καραμέρης
Apr 1 at 13:47