Mixing
Example that in general $|abc|neq |cba|$ [duplicate]
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This question already has an answer here:
how to prove that $abc$ and $cba$ do not necessarily have the same order?
2 answers
I tried to provide an example of group elements $a,b,c$ that shows that in general $|abc|neq |cba|$.
Please could you tell me if my example is correct?
Let $G$ be the dihedral group $D_6$ (the symmetries of a hexagon). Let us label the axes of symmetry with numbers $1$ to $6$ and call the corresponding reflections $F_1$ to $F_6$.
Now consider two "adjacent" reflections $F_1$ and $F_2$. Then
$$ F_1 F_2$$
is rotation clockwise by $120$ degrees and
$$ F_2 F_1$$
is roation by $120$ degrees counterclockwise.
Let $R$ be rotation by $120$ degrees counterclockwise. Then
$$ R F_1 F_2 = e$$
has order $1$ but
$$ F_2 F_1 R$$
is rotation by $240$ counterclockwise which has order not equal to $1$.
group-theory proof-verification examples-counterexamples
edited Jan 22 at 13:44
Martin Sleziak
44.8k10119273
asked Nov 27 '15 at 5:55
a studenta student
1,958726
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marked as duplicate by user1729 group-theory
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Jan 22 at 16:42
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:
how to prove that $abc$ and $cba$ do not necessarily have the same order?
2 answers
I tried to provide an example of group elements $a,b,c$ that shows that in general $|abc|neq |cba|$.
Please could you tell me if my example is correct?
Let $G$ be the dihedral group $D_6$ (the symmetries of a hexagon). Let us label the axes of symmetry with numbers $1$ to $6$ and call the corresponding reflections $F_1$ to $F_6$.
Now consider two "adjacent" reflections $F_1$ and $F_2$. Then
$$ F_1 F_2$$
is rotation clockwise by $120$ degrees and
$$ F_2 F_1$$
is roation by $120$ degrees counterclockwise.
Let $R$ be rotation by $120$ degrees counterclockwise. Then
$$ R F_1 F_2 = e$$
has order $1$ but
$$ F_2 F_1 R$$
is rotation by $240$ counterclockwise which has order not equal to $1$.
group-theory proof-verification examples-counterexamples
edited Jan 22 at 13:44
Martin Sleziak
44.8k10119273
asked Nov 27 '15 at 5:55
a studenta student
1,958726
$endgroup$
marked as duplicate by user1729 group-theory
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Jan 22 at 16:42
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.
1
$begingroup$
Looks good to me.
$endgroup$
– cr001
Nov 27 '15 at 6:38
$begingroup$
Yes, this is correct.
$endgroup$
– Crostul
Nov 27 '15 at 8:30
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Thank you very much for your help, cr001 and Crostful!
$endgroup$
– a student
Nov 29 '15 at 2:49
add a comment |
$begingroup$
This question already has an answer here:
how to prove that $abc$ and $cba$ do not necessarily have the same order?
2 answers
I tried to provide an example of group elements $a,b,c$ that shows that in general $|abc|neq |cba|$.
Please could you tell me if my example is correct?
Let $G$ be the dihedral group $D_6$ (the symmetries of a hexagon). Let us label the axes of symmetry with numbers $1$ to $6$ and call the corresponding reflections $F_1$ to $F_6$.
Now consider two "adjacent" reflections $F_1$ and $F_2$. Then
$$ F_1 F_2$$
is rotation clockwise by $120$ degrees and
$$ F_2 F_1$$
is roation by $120$ degrees counterclockwise.
Let $R$ be rotation by $120$ degrees counterclockwise. Then
$$ R F_1 F_2 = e$$
has order $1$ but
$$ F_2 F_1 R$$
is rotation by $240$ counterclockwise which has order not equal to $1$.
group-theory proof-verification examples-counterexamples
edited Jan 22 at 13:44
Martin Sleziak
44.8k10119273
asked Nov 27 '15 at 5:55
a studenta student
1,958726
$endgroup$
This question already has an answer here:
how to prove that $abc$ and $cba$ do not necessarily have the same order?
2 answers
I tried to provide an example of group elements $a,b,c$ that shows that in general $|abc|neq |cba|$.
Please could you tell me if my example is correct?
Let $G$ be the dihedral group $D_6$ (the symmetries of a hexagon). Let us label the axes of symmetry with numbers $1$ to $6$ and call the corresponding reflections $F_1$ to $F_6$.
Now consider two "adjacent" reflections $F_1$ and $F_2$. Then
$$ F_1 F_2$$
is rotation clockwise by $120$ degrees and
$$ F_2 F_1$$
is roation by $120$ degrees counterclockwise.
Let $R$ be rotation by $120$ degrees counterclockwise. Then
$$ R F_1 F_2 = e$$
has order $1$ but
$$ F_2 F_1 R$$
is rotation by $240$ counterclockwise which has order not equal to $1$.
This question already has an answer here:
how to prove that $abc$ and $cba$ do not necessarily have the same order?
2 answers
group-theory proof-verification examples-counterexamples
group-theory proof-verification examples-counterexamples
edited Jan 22 at 13:44
Martin Sleziak
44.8k10119273
asked Nov 27 '15 at 5:55
a studenta student
1,958726
edited Jan 22 at 13:44
Martin Sleziak
44.8k10119273
asked Nov 27 '15 at 5:55
a studenta student
1,958726
edited Jan 22 at 13:44
Martin Sleziak
44.8k10119273
edited Jan 22 at 13:44
Martin Sleziak
44.8k10119273
edited Jan 22 at 13:44
Martin Sleziak
44.8k10119273
44.8k10119273
asked Nov 27 '15 at 5:55
a studenta student
1,958726
asked Nov 27 '15 at 5:55
a studenta student
1,958726
asked Nov 27 '15 at 5:55
a studenta student
1,958726
1,958726
marked as duplicate by user1729 group-theory
Users with the group-theory badge can single-handedly close group-theory questions as duplicates and reopen them as needed.
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Jan 22 at 16:42
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 user1729 group-theory
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Jan 22 at 16:42
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.
1
$begingroup$
Looks good to me.
$endgroup$
– cr001
Nov 27 '15 at 6:38
$begingroup$
Yes, this is correct.
$endgroup$
– Crostul
Nov 27 '15 at 8:30
$begingroup$
Thank you very much for your help, cr001 and Crostful!
$endgroup$
– a student
Nov 29 '15 at 2:49
add a comment |
1
$begingroup$
Looks good to me.
$endgroup$
– cr001
Nov 27 '15 at 6:38
$begingroup$
Yes, this is correct.
$endgroup$
– Crostul
Nov 27 '15 at 8:30
$begingroup$
Thank you very much for your help, cr001 and Crostful!
$endgroup$
– a student
Nov 29 '15 at 2:49
1
1
$begingroup$
Looks good to me.
$endgroup$
– cr001
Nov 27 '15 at 6:38
$begingroup$
Looks good to me.
$endgroup$
– cr001
Nov 27 '15 at 6:38
$begingroup$
Yes, this is correct.
$endgroup$
– Crostul
Nov 27 '15 at 8:30
$begingroup$
Yes, this is correct.
$endgroup$
– Crostul
Nov 27 '15 at 8:30
$begingroup$
Thank you very much for your help, cr001 and Crostful!
$endgroup$
– a student
Nov 29 '15 at 2:49
$begingroup$
Thank you very much for your help, cr001 and Crostful!
$endgroup$
– a student
Nov 29 '15 at 2:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Take G to be any group, let x,y be elements of G and observe that the commutator
[x,y] = x'y'xy (here x' denotes the inverse of x in G) can be written as a product abc such that cba=1. Indeed, just take a=x', b=y'x and c=y.
Now if x,y don't commute you get that abc is nontrivial, while cba is trivial and presto: you have infinitely many easy examples.
N.B. This trick also shows that in any group G, if a finite product of elements is trivial, then the product of the same elements in reverse order belongs to the derived subgroup G'.
Also: the elements of G' are those finite products of elements in G whose corresponding product in reverse order is the identity.
Hope this helps.
answered Nov 29 '15 at 17:14
nea marinnea marin
111
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take G to be any group, let x,y be elements of G and observe that the commutator
[x,y] = x'y'xy (here x' denotes the inverse of x in G) can be written as a product abc such that cba=1. Indeed, just take a=x', b=y'x and c=y.
Now if x,y don't commute you get that abc is nontrivial, while cba is trivial and presto: you have infinitely many easy examples.
N.B. This trick also shows that in any group G, if a finite product of elements is trivial, then the product of the same elements in reverse order belongs to the derived subgroup G'.
Also: the elements of G' are those finite products of elements in G whose corresponding product in reverse order is the identity.
Hope this helps.
answered Nov 29 '15 at 17:14
nea marinnea marin
111
$endgroup$
add a comment |
$begingroup$
Take G to be any group, let x,y be elements of G and observe that the commutator
[x,y] = x'y'xy (here x' denotes the inverse of x in G) can be written as a product abc such that cba=1. Indeed, just take a=x', b=y'x and c=y.
Now if x,y don't commute you get that abc is nontrivial, while cba is trivial and presto: you have infinitely many easy examples.
N.B. This trick also shows that in any group G, if a finite product of elements is trivial, then the product of the same elements in reverse order belongs to the derived subgroup G'.
Also: the elements of G' are those finite products of elements in G whose corresponding product in reverse order is the identity.
Hope this helps.
answered Nov 29 '15 at 17:14
nea marinnea marin
111
$endgroup$
add a comment |
$begingroup$
Take G to be any group, let x,y be elements of G and observe that the commutator
[x,y] = x'y'xy (here x' denotes the inverse of x in G) can be written as a product abc such that cba=1. Indeed, just take a=x', b=y'x and c=y.
Now if x,y don't commute you get that abc is nontrivial, while cba is trivial and presto: you have infinitely many easy examples.
N.B. This trick also shows that in any group G, if a finite product of elements is trivial, then the product of the same elements in reverse order belongs to the derived subgroup G'.
Also: the elements of G' are those finite products of elements in G whose corresponding product in reverse order is the identity.
Hope this helps.
answered Nov 29 '15 at 17:14
nea marinnea marin
111
$endgroup$
Take G to be any group, let x,y be elements of G and observe that the commutator
[x,y] = x'y'xy (here x' denotes the inverse of x in G) can be written as a product abc such that cba=1. Indeed, just take a=x', b=y'x and c=y.
Now if x,y don't commute you get that abc is nontrivial, while cba is trivial and presto: you have infinitely many easy examples.
N.B. This trick also shows that in any group G, if a finite product of elements is trivial, then the product of the same elements in reverse order belongs to the derived subgroup G'.
Also: the elements of G' are those finite products of elements in G whose corresponding product in reverse order is the identity.
Hope this helps.
answered Nov 29 '15 at 17:14
nea marinnea marin
111
answered Nov 29 '15 at 17:14
nea marinnea marin
111
answered Nov 29 '15 at 17:14
nea marinnea marin
111
answered Nov 29 '15 at 17:14
nea marinnea marin
111
111
add a comment |
add a comment |
1
$begingroup$
Looks good to me.
$endgroup$
– cr001
Nov 27 '15 at 6:38
$begingroup$
Yes, this is correct.
$endgroup$
– Crostul
Nov 27 '15 at 8:30
$begingroup$
Thank you very much for your help, cr001 and Crostful!
$endgroup$
– a student
Nov 29 '15 at 2:49