Isomorphic groups with 4 elements [duplicate]
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This question already has an answer here:
Any group of order four is either cyclic or isomorphic to $V$
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I have written that all the groups with 4 element are isomorphic with Z4 or the Klein group. Can somebody explain me why,please?
group-theory
asked Jan 13 at 17:26
GaboruGaboru
4428
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marked as duplicate by Parcly Taxel, José Carlos Santos, John Douma, Derek Holt, Shaun Jan 13 at 18:44
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
Any group of order four is either cyclic or isomorphic to $V$
2 answers
I have written that all the groups with 4 element are isomorphic with Z4 or the Klein group. Can somebody explain me why,please?
group-theory
asked Jan 13 at 17:26
GaboruGaboru
4428
$endgroup$
marked as duplicate by Parcly Taxel, José Carlos Santos, John Douma, Derek Holt, Shaun Jan 13 at 18:44
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Any group of order four is either cyclic or isomorphic to $V$
2 answers
I have written that all the groups with 4 element are isomorphic with Z4 or the Klein group. Can somebody explain me why,please?
group-theory
asked Jan 13 at 17:26
GaboruGaboru
4428
$endgroup$
This question already has an answer here:
Any group of order four is either cyclic or isomorphic to $V$
2 answers
I have written that all the groups with 4 element are isomorphic with Z4 or the Klein group. Can somebody explain me why,please?
This question already has an answer here:
Any group of order four is either cyclic or isomorphic to $V$
2 answers
group-theory
group-theory
asked Jan 13 at 17:26
GaboruGaboru
4428
asked Jan 13 at 17:26
GaboruGaboru
4428
asked Jan 13 at 17:26
GaboruGaboru
4428
asked Jan 13 at 17:26
GaboruGaboru
4428
asked Jan 13 at 17:26
GaboruGaboru
4428
4428
marked as duplicate by Parcly Taxel, José Carlos Santos, John Douma, Derek Holt, Shaun Jan 13 at 18:44
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Parcly Taxel, José Carlos Santos, John Douma, Derek Holt, Shaun Jan 13 at 18:44
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
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2 Answers
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If it's not cyclic, then all of its non-identity elements must have order 2 by Lagrange, and the subgroups that they separately generate must all be normal, as they have index 2, so it must be abelian, hence it must be the Klein group.
answered Jan 13 at 17:36
user3482749user3482749
4,3111019
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You have 4 elements in the group, one element being the identity. So you're left with 3 elements to 'fill in' the rest of the group, by Lagrange's Theorem the orders must divide 4 i.e be 2 or 4 (can't be 1 since it would be equal to identity). So the remaining 3 elements could potentially have orders 2,2,2 or 2,2,4 or 2,4,4 or 4,4,4. If you do some simple maths or construct a composition table for the group you'll very quickly find it's impossible to have the 3 elements having orders 2,2,4 or 4,4,4, so the only possibility is to have orders
1,2,2,2 or 1,2,4,4 in a group of 4 elements (these correspond to Klein group and Z/4Z, respectively.)
answered Jan 13 at 17:43
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2 Answers
2
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oldest
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
If it's not cyclic, then all of its non-identity elements must have order 2 by Lagrange, and the subgroups that they separately generate must all be normal, as they have index 2, so it must be abelian, hence it must be the Klein group.
answered Jan 13 at 17:36
user3482749user3482749
4,3111019
$endgroup$
add a comment |
$begingroup$
If it's not cyclic, then all of its non-identity elements must have order 2 by Lagrange, and the subgroups that they separately generate must all be normal, as they have index 2, so it must be abelian, hence it must be the Klein group.
answered Jan 13 at 17:36
user3482749user3482749
4,3111019
$endgroup$
add a comment |
$begingroup$
If it's not cyclic, then all of its non-identity elements must have order 2 by Lagrange, and the subgroups that they separately generate must all be normal, as they have index 2, so it must be abelian, hence it must be the Klein group.
answered Jan 13 at 17:36
user3482749user3482749
4,3111019
$endgroup$
If it's not cyclic, then all of its non-identity elements must have order 2 by Lagrange, and the subgroups that they separately generate must all be normal, as they have index 2, so it must be abelian, hence it must be the Klein group.
answered Jan 13 at 17:36
user3482749user3482749
4,3111019
answered Jan 13 at 17:36
user3482749user3482749
4,3111019
answered Jan 13 at 17:36
user3482749user3482749
4,3111019
answered Jan 13 at 17:36
user3482749user3482749
4,3111019
4,3111019
add a comment |
add a comment |
$begingroup$
You have 4 elements in the group, one element being the identity. So you're left with 3 elements to 'fill in' the rest of the group, by Lagrange's Theorem the orders must divide 4 i.e be 2 or 4 (can't be 1 since it would be equal to identity). So the remaining 3 elements could potentially have orders 2,2,2 or 2,2,4 or 2,4,4 or 4,4,4. If you do some simple maths or construct a composition table for the group you'll very quickly find it's impossible to have the 3 elements having orders 2,2,4 or 4,4,4, so the only possibility is to have orders
1,2,2,2 or 1,2,4,4 in a group of 4 elements (these correspond to Klein group and Z/4Z, respectively.)
answered Jan 13 at 17:43
DisplaynameDisplayname
25011
$endgroup$
add a comment |
$begingroup$
You have 4 elements in the group, one element being the identity. So you're left with 3 elements to 'fill in' the rest of the group, by Lagrange's Theorem the orders must divide 4 i.e be 2 or 4 (can't be 1 since it would be equal to identity). So the remaining 3 elements could potentially have orders 2,2,2 or 2,2,4 or 2,4,4 or 4,4,4. If you do some simple maths or construct a composition table for the group you'll very quickly find it's impossible to have the 3 elements having orders 2,2,4 or 4,4,4, so the only possibility is to have orders
1,2,2,2 or 1,2,4,4 in a group of 4 elements (these correspond to Klein group and Z/4Z, respectively.)
answered Jan 13 at 17:43
DisplaynameDisplayname
25011
$endgroup$
add a comment |
$begingroup$
You have 4 elements in the group, one element being the identity. So you're left with 3 elements to 'fill in' the rest of the group, by Lagrange's Theorem the orders must divide 4 i.e be 2 or 4 (can't be 1 since it would be equal to identity). So the remaining 3 elements could potentially have orders 2,2,2 or 2,2,4 or 2,4,4 or 4,4,4. If you do some simple maths or construct a composition table for the group you'll very quickly find it's impossible to have the 3 elements having orders 2,2,4 or 4,4,4, so the only possibility is to have orders
1,2,2,2 or 1,2,4,4 in a group of 4 elements (these correspond to Klein group and Z/4Z, respectively.)
answered Jan 13 at 17:43
DisplaynameDisplayname
25011
$endgroup$
You have 4 elements in the group, one element being the identity. So you're left with 3 elements to 'fill in' the rest of the group, by Lagrange's Theorem the orders must divide 4 i.e be 2 or 4 (can't be 1 since it would be equal to identity). So the remaining 3 elements could potentially have orders 2,2,2 or 2,2,4 or 2,4,4 or 4,4,4. If you do some simple maths or construct a composition table for the group you'll very quickly find it's impossible to have the 3 elements having orders 2,2,4 or 4,4,4, so the only possibility is to have orders
1,2,2,2 or 1,2,4,4 in a group of 4 elements (these correspond to Klein group and Z/4Z, respectively.)
answered Jan 13 at 17:43
DisplaynameDisplayname
25011
edited Jan 13 at 17:54
answered Jan 13 at 17:43
DisplaynameDisplayname
25011
answered Jan 13 at 17:43
DisplaynameDisplayname
25011
answered Jan 13 at 17:43
DisplaynameDisplayname
25011
25011
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