Mixing
Is it true that the order of $ab$ is always equal to the order of $ba$?
$begingroup$
How do I prove that if $a$, $b$ are elements of group, then $o(ab) = o(ba)$?
For some reason I end up doing the proof for abelian(ness?), i.e., I assume that the order of $ab$ is $2$ and do the steps that lead me to conclude that $ab=ba$, so the orders must be the same. Is that the right way to do it?
abstract-algebra group-theory
asked Nov 15 '12 at 20:45
SiyandaSiyanda
1,08052236
$endgroup$
add a comment |
$begingroup$
How do I prove that if $a$, $b$ are elements of group, then $o(ab) = o(ba)$?
For some reason I end up doing the proof for abelian(ness?), i.e., I assume that the order of $ab$ is $2$ and do the steps that lead me to conclude that $ab=ba$, so the orders must be the same. Is that the right way to do it?
abstract-algebra group-theory
asked Nov 15 '12 at 20:45
SiyandaSiyanda
1,08052236
$endgroup$
4
$begingroup$
Why on earth are you assuming that the order of $ab$ is $2$?
$endgroup$
– Chris Eagle
Dec 8 '12 at 11:14
$begingroup$
This question is related to math.stackexchange.com/questions/225942/…
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– user26857
Apr 22 '13 at 21:33
$begingroup$
See this question for answers that go more to the essence of the matter (conjugation).
$endgroup$
– Bill Dubuque
Dec 20 '16 at 15:36
add a comment |
$begingroup$
How do I prove that if $a$, $b$ are elements of group, then $o(ab) = o(ba)$?
For some reason I end up doing the proof for abelian(ness?), i.e., I assume that the order of $ab$ is $2$ and do the steps that lead me to conclude that $ab=ba$, so the orders must be the same. Is that the right way to do it?
abstract-algebra group-theory
asked Nov 15 '12 at 20:45
SiyandaSiyanda
1,08052236
$endgroup$
How do I prove that if $a$, $b$ are elements of group, then $o(ab) = o(ba)$?
For some reason I end up doing the proof for abelian(ness?), i.e., I assume that the order of $ab$ is $2$ and do the steps that lead me to conclude that $ab=ba$, so the orders must be the same. Is that the right way to do it?
abstract-algebra group-theory
abstract-algebra group-theory
asked Nov 15 '12 at 20:45
SiyandaSiyanda
1,08052236
asked Nov 15 '12 at 20:45
SiyandaSiyanda
1,08052236
edited Apr 22 '13 at 20:26
user26857
asked Nov 15 '12 at 20:45
SiyandaSiyanda
1,08052236
asked Nov 15 '12 at 20:45
SiyandaSiyanda
1,08052236
asked Nov 15 '12 at 20:45
SiyandaSiyanda
1,08052236
1,08052236
4
$begingroup$
Why on earth are you assuming that the order of $ab$ is $2$?
$endgroup$
– Chris Eagle
Dec 8 '12 at 11:14
$begingroup$
This question is related to math.stackexchange.com/questions/225942/…
$endgroup$
– user26857
Apr 22 '13 at 21:33
$begingroup$
See this question for answers that go more to the essence of the matter (conjugation).
$endgroup$
– Bill Dubuque
Dec 20 '16 at 15:36
add a comment |
4
$begingroup$
Why on earth are you assuming that the order of $ab$ is $2$?
$endgroup$
– Chris Eagle
Dec 8 '12 at 11:14
$begingroup$
This question is related to math.stackexchange.com/questions/225942/…
$endgroup$
– user26857
Apr 22 '13 at 21:33
$begingroup$
See this question for answers that go more to the essence of the matter (conjugation).
$endgroup$
– Bill Dubuque
Dec 20 '16 at 15:36
4
4
$begingroup$
Why on earth are you assuming that the order of $ab$ is $2$?
$endgroup$
– Chris Eagle
Dec 8 '12 at 11:14
$begingroup$
Why on earth are you assuming that the order of $ab$ is $2$?
$endgroup$
– Chris Eagle
Dec 8 '12 at 11:14
$begingroup$
This question is related to math.stackexchange.com/questions/225942/…
$endgroup$
– user26857
Apr 22 '13 at 21:33
$begingroup$
This question is related to math.stackexchange.com/questions/225942/…
$endgroup$
– user26857
Apr 22 '13 at 21:33
$begingroup$
See this question for answers that go more to the essence of the matter (conjugation).
$endgroup$
– Bill Dubuque
Dec 20 '16 at 15:36
$begingroup$
See this question for answers that go more to the essence of the matter (conjugation).
$endgroup$
– Bill Dubuque
Dec 20 '16 at 15:36
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Here's an approach that allows you to do some hand-waving and not do any calculations at all. $ab$ and $ba$ are conjugate: indeed, $ba=a^{-1}(ab)a$. It is obvious (and probably already known at this point) that conjugation is an automorphism of the group, and it is obvious that automorphisms preserve orders of elements.
answered Nov 15 '12 at 21:12
Dan ShvedDan Shved
12.7k2045
$endgroup$
11
$begingroup$
Calling this «hand-waving» is quite misguided!
$endgroup$
– Mariano Suárez-Álvarez
Apr 22 '16 at 7:33
add a comment |
$begingroup$
Hint: Suppose $ab$ has order $n$, and consider $(ba)^{n+1}$.
Another hint is greyed out below (hover over with a mouse to display it):
Notice that $(ba)^{n+1} = b(ab)^na$.
answered Nov 15 '12 at 20:49
Clive NewsteadClive Newstead
52k474136
$endgroup$
$begingroup$
simple proof +1
$endgroup$
– viru
Apr 29 '18 at 5:19
add a comment |
$begingroup$
If $(ab)^n=e$ then $(ab)^na=a$. Since $(ab)^na=a(ba)^n$, $(ba)^n=e$. This proves that the order of $ba$ divides the order of $ab$. By symmetry, the order of $ab$ divides the order of $ba$. Hence the order of $ab$ and the order of $ba$ coincide.
answered Nov 15 '12 at 20:50
DidDid
249k23226466
$endgroup$
$begingroup$
I think the OP should note that the orders of $a$ and $b$ are both finite.
$endgroup$
– mrs
Nov 15 '12 at 20:58
8
$begingroup$
@BabakSorouh Why? The order of ab may be finite while those of a and b are infinite.
$endgroup$
– Did
Nov 15 '12 at 21:11
2
$begingroup$
@GeoffreyCritzer Try $b=a^{-1}$.
$endgroup$
– Did
Jun 5 '15 at 10:33
1
$begingroup$
@GeoffreyCritzer In the plane, try $a$ the translation by $(1,0)$ and $ba$ the symmetry $(x,y)mapsto(y,x)$.
$endgroup$
– Did
Jun 5 '15 at 10:44
1
$begingroup$
@GeoffreyCritzer On this, allow me to send you back to my post, the answer is there. By the way, if you have a new question, adding comments to some answer more than 30 months old is not the way to go: do not be shy, ask your own question.
$endgroup$
– Did
Jun 5 '15 at 11:37
|
show 3 more comments
$begingroup$
By associativity, $(ab)^p=a(ba)^{p-1}b$ for $pgeqslant 1$. If $(ab)^p=e$ then $a(ba)^{p-1}b=e$, so $a(ba)^p=a$ and $(ba)^p=e$. We conclude that for $pgeqslant 1$,
$$(ab)^p=eLeftrightarrow (ba)^p=e.$$
answered Nov 15 '12 at 20:49
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
$begingroup$
Another elementary way
On contrary suppose $|ab|,|ba|$ are different
With out loss of generality assume $|ab|=n>|ba|=k$
$(ab)^n=
abababab........ab=e$
$a(ba)^{n-1}b=e$ as form assumption k
$a(ba)^{n-1-k}b=e=(ab)^{n-k}$ that implies order of ab is n-k which contradition to assumption.
n-k
answered Sep 12 '18 at 7:03
SRJSRJ
1,9101620
$endgroup$
add a comment |
$begingroup$
(1)
$(ab)^n = e$
$Rightarrow$
$(ba)^n = (ba)^nbb^{-1} = b(ab)^nb^{-1} = beb^{-1} = e$.
(2)
$(ba)^n = e$
$Rightarrow$
$(ab)^n = (ab)^naa^{-1} = a(ba)^na^{-1} = aea^{-1} = e$.
answered Jan 29 at 7:19
tchappy hatchappy ha
788412
$endgroup$
add a comment |
protected by Zev Chonoles Apr 22 '16 at 6:54
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's an approach that allows you to do some hand-waving and not do any calculations at all. $ab$ and $ba$ are conjugate: indeed, $ba=a^{-1}(ab)a$. It is obvious (and probably already known at this point) that conjugation is an automorphism of the group, and it is obvious that automorphisms preserve orders of elements.
answered Nov 15 '12 at 21:12
Dan ShvedDan Shved
12.7k2045
$endgroup$
11
$begingroup$
Calling this «hand-waving» is quite misguided!
$endgroup$
– Mariano Suárez-Álvarez
Apr 22 '16 at 7:33
add a comment |
$begingroup$
Here's an approach that allows you to do some hand-waving and not do any calculations at all. $ab$ and $ba$ are conjugate: indeed, $ba=a^{-1}(ab)a$. It is obvious (and probably already known at this point) that conjugation is an automorphism of the group, and it is obvious that automorphisms preserve orders of elements.
answered Nov 15 '12 at 21:12
Dan ShvedDan Shved
12.7k2045
$endgroup$
11
$begingroup$
Calling this «hand-waving» is quite misguided!
$endgroup$
– Mariano Suárez-Álvarez
Apr 22 '16 at 7:33
add a comment |
$begingroup$
Here's an approach that allows you to do some hand-waving and not do any calculations at all. $ab$ and $ba$ are conjugate: indeed, $ba=a^{-1}(ab)a$. It is obvious (and probably already known at this point) that conjugation is an automorphism of the group, and it is obvious that automorphisms preserve orders of elements.
answered Nov 15 '12 at 21:12
Dan ShvedDan Shved
12.7k2045
$endgroup$
Here's an approach that allows you to do some hand-waving and not do any calculations at all. $ab$ and $ba$ are conjugate: indeed, $ba=a^{-1}(ab)a$. It is obvious (and probably already known at this point) that conjugation is an automorphism of the group, and it is obvious that automorphisms preserve orders of elements.
answered Nov 15 '12 at 21:12
Dan ShvedDan Shved
12.7k2045
answered Nov 15 '12 at 21:12
Dan ShvedDan Shved
12.7k2045
answered Nov 15 '12 at 21:12
Dan ShvedDan Shved
12.7k2045
answered Nov 15 '12 at 21:12
Dan ShvedDan Shved
12.7k2045
12.7k2045
11
$begingroup$
Calling this «hand-waving» is quite misguided!
$endgroup$
– Mariano Suárez-Álvarez
Apr 22 '16 at 7:33
add a comment |
11
$begingroup$
Calling this «hand-waving» is quite misguided!
$endgroup$
– Mariano Suárez-Álvarez
Apr 22 '16 at 7:33
11
11
$begingroup$
Calling this «hand-waving» is quite misguided!
$endgroup$
– Mariano Suárez-Álvarez
Apr 22 '16 at 7:33
$begingroup$
Calling this «hand-waving» is quite misguided!
$endgroup$
– Mariano Suárez-Álvarez
Apr 22 '16 at 7:33
add a comment |
$begingroup$
Hint: Suppose $ab$ has order $n$, and consider $(ba)^{n+1}$.
Another hint is greyed out below (hover over with a mouse to display it):
Notice that $(ba)^{n+1} = b(ab)^na$.
answered Nov 15 '12 at 20:49
Clive NewsteadClive Newstead
52k474136
$endgroup$
$begingroup$
simple proof +1
$endgroup$
– viru
Apr 29 '18 at 5:19
add a comment |
$begingroup$
Hint: Suppose $ab$ has order $n$, and consider $(ba)^{n+1}$.
Another hint is greyed out below (hover over with a mouse to display it):
Notice that $(ba)^{n+1} = b(ab)^na$.
answered Nov 15 '12 at 20:49
Clive NewsteadClive Newstead
52k474136
$endgroup$
$begingroup$
simple proof +1
$endgroup$
– viru
Apr 29 '18 at 5:19
add a comment |
$begingroup$
Hint: Suppose $ab$ has order $n$, and consider $(ba)^{n+1}$.
Another hint is greyed out below (hover over with a mouse to display it):
Notice that $(ba)^{n+1} = b(ab)^na$.
answered Nov 15 '12 at 20:49
Clive NewsteadClive Newstead
52k474136
$endgroup$
Hint: Suppose $ab$ has order $n$, and consider $(ba)^{n+1}$.
Another hint is greyed out below (hover over with a mouse to display it):
Notice that $(ba)^{n+1} = b(ab)^na$.
answered Nov 15 '12 at 20:49
Clive NewsteadClive Newstead
52k474136
answered Nov 15 '12 at 20:49
Clive NewsteadClive Newstead
52k474136
answered Nov 15 '12 at 20:49
Clive NewsteadClive Newstead
52k474136
answered Nov 15 '12 at 20:49
Clive NewsteadClive Newstead
52k474136
52k474136
$begingroup$
simple proof +1
$endgroup$
– viru
Apr 29 '18 at 5:19
add a comment |
$begingroup$
simple proof +1
$endgroup$
– viru
Apr 29 '18 at 5:19
$begingroup$
simple proof +1
$endgroup$
– viru
Apr 29 '18 at 5:19
$begingroup$
simple proof +1
$endgroup$
– viru
Apr 29 '18 at 5:19
add a comment |
$begingroup$
If $(ab)^n=e$ then $(ab)^na=a$. Since $(ab)^na=a(ba)^n$, $(ba)^n=e$. This proves that the order of $ba$ divides the order of $ab$. By symmetry, the order of $ab$ divides the order of $ba$. Hence the order of $ab$ and the order of $ba$ coincide.
answered Nov 15 '12 at 20:50
DidDid
249k23226466
$endgroup$
$begingroup$
I think the OP should note that the orders of $a$ and $b$ are both finite.
$endgroup$
– mrs
Nov 15 '12 at 20:58
8
$begingroup$
@BabakSorouh Why? The order of ab may be finite while those of a and b are infinite.
$endgroup$
– Did
Nov 15 '12 at 21:11
2
$begingroup$
@GeoffreyCritzer Try $b=a^{-1}$.
$endgroup$
– Did
Jun 5 '15 at 10:33
1
$begingroup$
@GeoffreyCritzer In the plane, try $a$ the translation by $(1,0)$ and $ba$ the symmetry $(x,y)mapsto(y,x)$.
$endgroup$
– Did
Jun 5 '15 at 10:44
1
$begingroup$
@GeoffreyCritzer On this, allow me to send you back to my post, the answer is there. By the way, if you have a new question, adding comments to some answer more than 30 months old is not the way to go: do not be shy, ask your own question.
$endgroup$
– Did
Jun 5 '15 at 11:37
|
show 3 more comments
$begingroup$
If $(ab)^n=e$ then $(ab)^na=a$. Since $(ab)^na=a(ba)^n$, $(ba)^n=e$. This proves that the order of $ba$ divides the order of $ab$. By symmetry, the order of $ab$ divides the order of $ba$. Hence the order of $ab$ and the order of $ba$ coincide.
answered Nov 15 '12 at 20:50
DidDid
249k23226466
$endgroup$
$begingroup$
I think the OP should note that the orders of $a$ and $b$ are both finite.
$endgroup$
– mrs
Nov 15 '12 at 20:58
8
$begingroup$
@BabakSorouh Why? The order of ab may be finite while those of a and b are infinite.
$endgroup$
– Did
Nov 15 '12 at 21:11
2
$begingroup$
@GeoffreyCritzer Try $b=a^{-1}$.
$endgroup$
– Did
Jun 5 '15 at 10:33
1
$begingroup$
@GeoffreyCritzer In the plane, try $a$ the translation by $(1,0)$ and $ba$ the symmetry $(x,y)mapsto(y,x)$.
$endgroup$
– Did
Jun 5 '15 at 10:44
1
$begingroup$
@GeoffreyCritzer On this, allow me to send you back to my post, the answer is there. By the way, if you have a new question, adding comments to some answer more than 30 months old is not the way to go: do not be shy, ask your own question.
$endgroup$
– Did
Jun 5 '15 at 11:37
|
show 3 more comments
$begingroup$
If $(ab)^n=e$ then $(ab)^na=a$. Since $(ab)^na=a(ba)^n$, $(ba)^n=e$. This proves that the order of $ba$ divides the order of $ab$. By symmetry, the order of $ab$ divides the order of $ba$. Hence the order of $ab$ and the order of $ba$ coincide.
answered Nov 15 '12 at 20:50
DidDid
249k23226466
$endgroup$
If $(ab)^n=e$ then $(ab)^na=a$. Since $(ab)^na=a(ba)^n$, $(ba)^n=e$. This proves that the order of $ba$ divides the order of $ab$. By symmetry, the order of $ab$ divides the order of $ba$. Hence the order of $ab$ and the order of $ba$ coincide.
answered Nov 15 '12 at 20:50
DidDid
249k23226466
edited Oct 18 '16 at 5:29
answered Nov 15 '12 at 20:50
DidDid
249k23226466
answered Nov 15 '12 at 20:50
DidDid
249k23226466
answered Nov 15 '12 at 20:50
DidDid
249k23226466
249k23226466
$begingroup$
I think the OP should note that the orders of $a$ and $b$ are both finite.
$endgroup$
– mrs
Nov 15 '12 at 20:58
8
$begingroup$
@BabakSorouh Why? The order of ab may be finite while those of a and b are infinite.
$endgroup$
– Did
Nov 15 '12 at 21:11
2
$begingroup$
@GeoffreyCritzer Try $b=a^{-1}$.
$endgroup$
– Did
Jun 5 '15 at 10:33
1
$begingroup$
@GeoffreyCritzer In the plane, try $a$ the translation by $(1,0)$ and $ba$ the symmetry $(x,y)mapsto(y,x)$.
$endgroup$
– Did
Jun 5 '15 at 10:44
1
$begingroup$
@GeoffreyCritzer On this, allow me to send you back to my post, the answer is there. By the way, if you have a new question, adding comments to some answer more than 30 months old is not the way to go: do not be shy, ask your own question.
$endgroup$
– Did
Jun 5 '15 at 11:37
|
show 3 more comments
$begingroup$
I think the OP should note that the orders of $a$ and $b$ are both finite.
$endgroup$
– mrs
Nov 15 '12 at 20:58
8
$begingroup$
@BabakSorouh Why? The order of ab may be finite while those of a and b are infinite.
$endgroup$
– Did
Nov 15 '12 at 21:11
2
$begingroup$
@GeoffreyCritzer Try $b=a^{-1}$.
$endgroup$
– Did
Jun 5 '15 at 10:33
1
$begingroup$
@GeoffreyCritzer In the plane, try $a$ the translation by $(1,0)$ and $ba$ the symmetry $(x,y)mapsto(y,x)$.
$endgroup$
– Did
Jun 5 '15 at 10:44
1
$begingroup$
@GeoffreyCritzer On this, allow me to send you back to my post, the answer is there. By the way, if you have a new question, adding comments to some answer more than 30 months old is not the way to go: do not be shy, ask your own question.
$endgroup$
– Did
Jun 5 '15 at 11:37
$begingroup$
I think the OP should note that the orders of $a$ and $b$ are both finite.
$endgroup$
– mrs
Nov 15 '12 at 20:58
$begingroup$
I think the OP should note that the orders of $a$ and $b$ are both finite.
$endgroup$
– mrs
Nov 15 '12 at 20:58
8
8
$begingroup$
@BabakSorouh Why? The order of ab may be finite while those of a and b are infinite.
$endgroup$
– Did
Nov 15 '12 at 21:11
$begingroup$
@BabakSorouh Why? The order of ab may be finite while those of a and b are infinite.
$endgroup$
– Did
Nov 15 '12 at 21:11
2
2
$begingroup$
@GeoffreyCritzer Try $b=a^{-1}$.
$endgroup$
– Did
Jun 5 '15 at 10:33
$begingroup$
@GeoffreyCritzer Try $b=a^{-1}$.
$endgroup$
– Did
Jun 5 '15 at 10:33
1
1
$begingroup$
@GeoffreyCritzer In the plane, try $a$ the translation by $(1,0)$ and $ba$ the symmetry $(x,y)mapsto(y,x)$.
$endgroup$
– Did
Jun 5 '15 at 10:44
$begingroup$
@GeoffreyCritzer In the plane, try $a$ the translation by $(1,0)$ and $ba$ the symmetry $(x,y)mapsto(y,x)$.
$endgroup$
– Did
Jun 5 '15 at 10:44
1
1
$begingroup$
@GeoffreyCritzer On this, allow me to send you back to my post, the answer is there. By the way, if you have a new question, adding comments to some answer more than 30 months old is not the way to go: do not be shy, ask your own question.
$endgroup$
– Did
Jun 5 '15 at 11:37
$begingroup$
@GeoffreyCritzer On this, allow me to send you back to my post, the answer is there. By the way, if you have a new question, adding comments to some answer more than 30 months old is not the way to go: do not be shy, ask your own question.
$endgroup$
– Did
Jun 5 '15 at 11:37
|
show 3 more comments
$begingroup$
By associativity, $(ab)^p=a(ba)^{p-1}b$ for $pgeqslant 1$. If $(ab)^p=e$ then $a(ba)^{p-1}b=e$, so $a(ba)^p=a$ and $(ba)^p=e$. We conclude that for $pgeqslant 1$,
$$(ab)^p=eLeftrightarrow (ba)^p=e.$$
answered Nov 15 '12 at 20:49
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
$begingroup$
By associativity, $(ab)^p=a(ba)^{p-1}b$ for $pgeqslant 1$. If $(ab)^p=e$ then $a(ba)^{p-1}b=e$, so $a(ba)^p=a$ and $(ba)^p=e$. We conclude that for $pgeqslant 1$,
$$(ab)^p=eLeftrightarrow (ba)^p=e.$$
answered Nov 15 '12 at 20:49
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
$begingroup$
By associativity, $(ab)^p=a(ba)^{p-1}b$ for $pgeqslant 1$. If $(ab)^p=e$ then $a(ba)^{p-1}b=e$, so $a(ba)^p=a$ and $(ba)^p=e$. We conclude that for $pgeqslant 1$,
$$(ab)^p=eLeftrightarrow (ba)^p=e.$$
answered Nov 15 '12 at 20:49
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
By associativity, $(ab)^p=a(ba)^{p-1}b$ for $pgeqslant 1$. If $(ab)^p=e$ then $a(ba)^{p-1}b=e$, so $a(ba)^p=a$ and $(ba)^p=e$. We conclude that for $pgeqslant 1$,
$$(ab)^p=eLeftrightarrow (ba)^p=e.$$
answered Nov 15 '12 at 20:49
Davide GiraudoDavide Giraudo
128k17156268
answered Nov 15 '12 at 20:49
Davide GiraudoDavide Giraudo
128k17156268
answered Nov 15 '12 at 20:49
Davide GiraudoDavide Giraudo
128k17156268
answered Nov 15 '12 at 20:49
Davide GiraudoDavide Giraudo
128k17156268
128k17156268
add a comment |
add a comment |
$begingroup$
Another elementary way
On contrary suppose $|ab|,|ba|$ are different
With out loss of generality assume $|ab|=n>|ba|=k$
$(ab)^n=
abababab........ab=e$
$a(ba)^{n-1}b=e$ as form assumption k
$a(ba)^{n-1-k}b=e=(ab)^{n-k}$ that implies order of ab is n-k which contradition to assumption.
n-k
answered Sep 12 '18 at 7:03
SRJSRJ
1,9101620
$endgroup$
add a comment |
$begingroup$
Another elementary way
On contrary suppose $|ab|,|ba|$ are different
With out loss of generality assume $|ab|=n>|ba|=k$
$(ab)^n=
abababab........ab=e$
$a(ba)^{n-1}b=e$ as form assumption k
$a(ba)^{n-1-k}b=e=(ab)^{n-k}$ that implies order of ab is n-k which contradition to assumption.
n-k
answered Sep 12 '18 at 7:03
SRJSRJ
1,9101620
$endgroup$
add a comment |
$begingroup$
Another elementary way
On contrary suppose $|ab|,|ba|$ are different
With out loss of generality assume $|ab|=n>|ba|=k$
$(ab)^n=
abababab........ab=e$
$a(ba)^{n-1}b=e$ as form assumption k
$a(ba)^{n-1-k}b=e=(ab)^{n-k}$ that implies order of ab is n-k which contradition to assumption.
n-k
answered Sep 12 '18 at 7:03
SRJSRJ
1,9101620
$endgroup$
Another elementary way
On contrary suppose $|ab|,|ba|$ are different
With out loss of generality assume $|ab|=n>|ba|=k$
$(ab)^n=
abababab........ab=e$
$a(ba)^{n-1}b=e$ as form assumption k
$a(ba)^{n-1-k}b=e=(ab)^{n-k}$ that implies order of ab is n-k which contradition to assumption.
n-k
answered Sep 12 '18 at 7:03
SRJSRJ
1,9101620
answered Sep 12 '18 at 7:03
SRJSRJ
1,9101620
answered Sep 12 '18 at 7:03
SRJSRJ
1,9101620
answered Sep 12 '18 at 7:03
SRJSRJ
1,9101620
1,9101620
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$begingroup$
(1)
$(ab)^n = e$
$Rightarrow$
$(ba)^n = (ba)^nbb^{-1} = b(ab)^nb^{-1} = beb^{-1} = e$.
(2)
$(ba)^n = e$
$Rightarrow$
$(ab)^n = (ab)^naa^{-1} = a(ba)^na^{-1} = aea^{-1} = e$.
answered Jan 29 at 7:19
tchappy hatchappy ha
788412
$endgroup$
add a comment |
$begingroup$
(1)
$(ab)^n = e$
$Rightarrow$
$(ba)^n = (ba)^nbb^{-1} = b(ab)^nb^{-1} = beb^{-1} = e$.
(2)
$(ba)^n = e$
$Rightarrow$
$(ab)^n = (ab)^naa^{-1} = a(ba)^na^{-1} = aea^{-1} = e$.
answered Jan 29 at 7:19
tchappy hatchappy ha
788412
$endgroup$
add a comment |
$begingroup$
(1)
$(ab)^n = e$
$Rightarrow$
$(ba)^n = (ba)^nbb^{-1} = b(ab)^nb^{-1} = beb^{-1} = e$.
(2)
$(ba)^n = e$
$Rightarrow$
$(ab)^n = (ab)^naa^{-1} = a(ba)^na^{-1} = aea^{-1} = e$.
answered Jan 29 at 7:19
tchappy hatchappy ha
788412
$endgroup$
(1)
$(ab)^n = e$
$Rightarrow$
$(ba)^n = (ba)^nbb^{-1} = b(ab)^nb^{-1} = beb^{-1} = e$.
(2)
$(ba)^n = e$
$Rightarrow$
$(ab)^n = (ab)^naa^{-1} = a(ba)^na^{-1} = aea^{-1} = e$.
answered Jan 29 at 7:19
tchappy hatchappy ha
788412
answered Jan 29 at 7:19
tchappy hatchappy ha
788412
answered Jan 29 at 7:19
tchappy hatchappy ha
788412
answered Jan 29 at 7:19
tchappy hatchappy ha
788412
788412
add a comment |
add a comment |
protected by Zev Chonoles Apr 22 '16 at 6:54
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4
$begingroup$
Why on earth are you assuming that the order of $ab$ is $2$?
$endgroup$
– Chris Eagle
Dec 8 '12 at 11:14
$begingroup$
This question is related to math.stackexchange.com/questions/225942/…
$endgroup$
– user26857
Apr 22 '13 at 21:33
$begingroup$
See this question for answers that go more to the essence of the matter (conjugation).
$endgroup$
– Bill Dubuque
Dec 20 '16 at 15:36