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“Simple problem†to find order of a product in a Group [duplicate]
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Inverse of $ab$ and its order in a group with relations $a^2=e=b^6$ and $ab=b^4a$
2 answers
Let a and b be elements of a group, with $a^2 = e, b^6 = e $ and $ ab =b^4 a $ . Find the order of ab, and
express its inverse in each of the forms $a^mb^n$ and $b^ma^n?$
Though it seems very simple I'm unable to find a power such that $(ab)^x =e$. Please help.
Im all confused with this problem.
PS: to put in context, this question was asked for
maths optional UPSC IAS exam in India
abstract-algebra group-theory
edited Jul 26 at 13:26
rschwieb
99.7k1191226
asked Jul 26 at 9:50
johny
515
marked as duplicate by rschwieb abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
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This question already has an answer here:
Inverse of $ab$ and its order in a group with relations $a^2=e=b^6$ and $ab=b^4a$
2 answers
Let a and b be elements of a group, with $a^2 = e, b^6 = e $ and $ ab =b^4 a $ . Find the order of ab, and
express its inverse in each of the forms $a^mb^n$ and $b^ma^n?$
Though it seems very simple I'm unable to find a power such that $(ab)^x =e$. Please help.
Im all confused with this problem.
PS: to put in context, this question was asked for
maths optional UPSC IAS exam in India
abstract-algebra group-theory
edited Jul 26 at 13:26
rschwieb
99.7k1191226
asked Jul 26 at 9:50
johny
515
marked as duplicate by rschwieb abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
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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
For all we know $ab$ could be equal to $e$, hence of order $1$.
– Arnaud Mortier
Jul 26 at 9:53
$(ab)^2 = abcdot ab = b^4acdot ab = b^4a^2b = b^4eb = b^5$.
– Bill Wallis
Jul 26 at 9:54
2
Why do you think that $ab=(ab)^4$? If so, then the order of $ab$ would be $1$ or $3$.
– Arnaud Mortier
Jul 26 at 9:58
@BillWallis i dont understand the operation is $ xy=y^4x $, how can you simply multiply $b^4a$ and $ab$??
– johny
Jul 26 at 10:34
@johny You are given that $ab = b^4a$. Thus, any time you see $ab$ you can replace it with $b^4a$, and vice versa. This means that the product $(ab)^2 = abab$ is equal to the product $b^4aab$ by replacing (one of) $ab$ with $b^4a$.
– Bill Wallis
Jul 26 at 10:38
 |Â
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up vote
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down vote
favorite
This question already has an answer here:
Inverse of $ab$ and its order in a group with relations $a^2=e=b^6$ and $ab=b^4a$
2 answers
Let a and b be elements of a group, with $a^2 = e, b^6 = e $ and $ ab =b^4 a $ . Find the order of ab, and
express its inverse in each of the forms $a^mb^n$ and $b^ma^n?$
Though it seems very simple I'm unable to find a power such that $(ab)^x =e$. Please help.
Im all confused with this problem.
PS: to put in context, this question was asked for
maths optional UPSC IAS exam in India
abstract-algebra group-theory
edited Jul 26 at 13:26
rschwieb
99.7k1191226
asked Jul 26 at 9:50
johny
515
This question already has an answer here:
Inverse of $ab$ and its order in a group with relations $a^2=e=b^6$ and $ab=b^4a$
2 answers
Let a and b be elements of a group, with $a^2 = e, b^6 = e $ and $ ab =b^4 a $ . Find the order of ab, and
express its inverse in each of the forms $a^mb^n$ and $b^ma^n?$
Though it seems very simple I'm unable to find a power such that $(ab)^x =e$. Please help.
Im all confused with this problem.
PS: to put in context, this question was asked for
maths optional UPSC IAS exam in India
This question already has an answer here:
Inverse of $ab$ and its order in a group with relations $a^2=e=b^6$ and $ab=b^4a$
2 answers
abstract-algebra group-theory
edited Jul 26 at 13:26
rschwieb
99.7k1191226
asked Jul 26 at 9:50
johny
515
edited Jul 26 at 13:26
rschwieb
99.7k1191226
edited Jul 26 at 13:26
rschwieb
99.7k1191226
edited Jul 26 at 13:26
rschwieb
99.7k1191226
99.7k1191226
asked Jul 26 at 9:50
johny
515
asked Jul 26 at 9:50
johny
515
asked Jul 26 at 9:50
johny
515
515
marked as duplicate by rschwieb abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
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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
For all we know $ab$ could be equal to $e$, hence of order $1$.
– Arnaud Mortier
Jul 26 at 9:53
$(ab)^2 = abcdot ab = b^4acdot ab = b^4a^2b = b^4eb = b^5$.
– Bill Wallis
Jul 26 at 9:54
2
Why do you think that $ab=(ab)^4$? If so, then the order of $ab$ would be $1$ or $3$.
– Arnaud Mortier
Jul 26 at 9:58
@BillWallis i dont understand the operation is $ xy=y^4x $, how can you simply multiply $b^4a$ and $ab$??
– johny
Jul 26 at 10:34
@johny You are given that $ab = b^4a$. Thus, any time you see $ab$ you can replace it with $b^4a$, and vice versa. This means that the product $(ab)^2 = abab$ is equal to the product $b^4aab$ by replacing (one of) $ab$ with $b^4a$.
– Bill Wallis
Jul 26 at 10:38
 |Â
show 2 more comments
1
For all we know $ab$ could be equal to $e$, hence of order $1$.
– Arnaud Mortier
Jul 26 at 9:53
$(ab)^2 = abcdot ab = b^4acdot ab = b^4a^2b = b^4eb = b^5$.
– Bill Wallis
Jul 26 at 9:54
2
Why do you think that $ab=(ab)^4$? If so, then the order of $ab$ would be $1$ or $3$.
– Arnaud Mortier
Jul 26 at 9:58
@BillWallis i dont understand the operation is $ xy=y^4x $, how can you simply multiply $b^4a$ and $ab$??
– johny
Jul 26 at 10:34
@johny You are given that $ab = b^4a$. Thus, any time you see $ab$ you can replace it with $b^4a$, and vice versa. This means that the product $(ab)^2 = abab$ is equal to the product $b^4aab$ by replacing (one of) $ab$ with $b^4a$.
– Bill Wallis
Jul 26 at 10:38
1
1
For all we know $ab$ could be equal to $e$, hence of order $1$.
– Arnaud Mortier
Jul 26 at 9:53
For all we know $ab$ could be equal to $e$, hence of order $1$.
– Arnaud Mortier
Jul 26 at 9:53
$(ab)^2 = abcdot ab = b^4acdot ab = b^4a^2b = b^4eb = b^5$.
– Bill Wallis
Jul 26 at 9:54
$(ab)^2 = abcdot ab = b^4acdot ab = b^4a^2b = b^4eb = b^5$.
– Bill Wallis
Jul 26 at 9:54
2
2
Why do you think that $ab=(ab)^4$? If so, then the order of $ab$ would be $1$ or $3$.
– Arnaud Mortier
Jul 26 at 9:58
Why do you think that $ab=(ab)^4$? If so, then the order of $ab$ would be $1$ or $3$.
– Arnaud Mortier
Jul 26 at 9:58
@BillWallis i dont understand the operation is $ xy=y^4x $, how can you simply multiply $b^4a$ and $ab$??
– johny
Jul 26 at 10:34
@BillWallis i dont understand the operation is $ xy=y^4x $, how can you simply multiply $b^4a$ and $ab$??
– johny
Jul 26 at 10:34
@johny You are given that $ab = b^4a$. Thus, any time you see $ab$ you can replace it with $b^4a$, and vice versa. This means that the product $(ab)^2 = abab$ is equal to the product $b^4aab$ by replacing (one of) $ab$ with $b^4a$.
– Bill Wallis
Jul 26 at 10:38
@johny You are given that $ab = b^4a$. Thus, any time you see $ab$ you can replace it with $b^4a$, and vice versa. This means that the product $(ab)^2 = abab$ is equal to the product $b^4aab$ by replacing (one of) $ab$ with $b^4a$.
– Bill Wallis
Jul 26 at 10:38
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
From $b^6=e$ and $ab = b^4a$ we can conclude $b^3 = e$: we have
$$
b^3 = a^-1(aba^-1)^3a = a^-1(b^4)^3a = a^-1b^12a = a^-1ea = e.
$$
Hence, $ab = b^4a = ba$, i.e. $a$ and $b$ commute. Now, it is easy to see that $(ab)^6 = e$. From this, we can only conclude $lvert abrvert in 1,2,3,6$.
For the second part of the question: Notice that $(ab)^6=e$ implies $(ab)^-1 = (ab)^5 = a^5b^5 = ab^2 = b^2a$. (Here we have used $a^2 = b^3 = e$ and $ab = ba$.)
answered Jul 26 at 10:10
Claudius
3,7111515
The conclusion that $b^3=e$ is correct. The conclusion that the original group $G$ is isomorphic to $H=langle a,b | a^2=b^3=e, ab=baranglesimeqmathbbZ_6$ is wrong. Note that the trivial group satisfies all requirements (the OP never stated that $aneq b$ or that $bneq e$). The correct conclusion is that there is a group epimorphism $Hto G$ meaning $G$ is a quotient of $H$ making the order $|ab|$ a bit more complicated.
– freakish
Jul 26 at 10:23
@Claudius how did you made $b^3$ complicated like that, I didnt get that step. Why you people are taking $ab=b^4a $ whenever you like and at other times simply multiplying it as it is ab. I mean inside the bracket, why didnt you expand like $ba^-1 = a^-4 b$? Thats my confusion sometimes it's taken as though there is a binary operation a*b , other times simply as ab. Its confusing me.
– johny
Jul 26 at 10:42
@johny There's an implicit, standard assumption that $xy$ is the same as $x*y$. Also you cannot expand $ba^-1$ because there's no formula for that. Your formulas are precisely about two fixed elements $a,b$ and their product $ab$, nothing else. Unless there's an explicit statement "for all $a,b$" and there's no such thing.
– freakish
Jul 26 at 10:46
Also after getting $b^3 = e$ why cant we just take $(ab)^6 = $$(a^2)^3$$(b^3)^2$ =e. What is the need to prove ab = ba?
– johny
Jul 26 at 10:49
@johny The same thing: $b^3=e$ is a formula about $b$ only. You cannot substitute $ab$ for $b$ because it doesn't hold for $ab$. To make that substitution valid you would have to start your problem with "for all $a,b$ in a group the following holds: ..." but you've started with "let $a,b$ be two elements of a group" which implicitly means that they are fixed.
– freakish
Jul 26 at 10:52
 |Â
show 15 more comments
up vote
0
down vote
Let be $|ab|=m$. Then $(ab)^m=e$.
Observe that $(ab)^3=(ab)(ab)(ab)=ab(b^4a)(ab)=ab^5a^2b=a$. So
$((ab)^3)^2=a^2=e$. So $|ab|=mmid 6$. Hence $|ab|=min 1,2,3,6$.
We conclude easy that $ab=ba$ and $a^2=b^3=e$ and inverse of $ab$ of the form $a^mb^n=ab^2$ and of the form $b^ma^n=b^2a$.
This means that general form of group $G=1,a,b,b^2, ab, ab^2$.
If $|ab|=1$ then $G=e, a$ i.e., $Gcong Bbb Z_2$,
IF $|ab|=2$ then $G=e,a,b,ab$ i.e., $Gcong Bbb Z_ztimes Z_2$ ( (can not be isomorph $Bbb Z_4$ It hasn't any element order $4$),
If $|ab|=3$ then $G=1,a,b,b^2, ab, ab^2$ i.e., then $Gcong Bbb Z_6$,
If $|ab|=6$ then $langle abrangle= G=1,a,b,b^2, ab, ab^2$ $Gcong Bbb Z_6$.
answered Jul 26 at 11:01
1ENÄ°GMA1
762315
Is this method correct? this look very straightforward and easy, did you check the answer below, why they are going that way if it can be done like this? Are you sure?
– johny
Jul 26 at 11:10
please look at the below answer. It's saying your method is not correct. can u comment on that?
– johny
Jul 26 at 11:32
It is easy see that m>2Wrong. If $G$ is trivial then $|ab|=1$. If $G$ is $mathbbZ_2$ then $|ab|=2$. Also the only valid conclusion from $(ab)^6=e$ is that $|ab|$ divides $6$.
– freakish
Jul 26 at 11:57
no as pointed above they are saying your proof is wrong. Do you agree with that? Can u give any counter argument on point raised by @freakish?
– johny
Jul 26 at 12:03
@freakish. You are right :). I edited my answer for you. But, We can exclude easy concept there.
– 1ENÄ°GMA1
Jul 26 at 12:09
 |Â
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
From $b^6=e$ and $ab = b^4a$ we can conclude $b^3 = e$: we have
$$
b^3 = a^-1(aba^-1)^3a = a^-1(b^4)^3a = a^-1b^12a = a^-1ea = e.
$$
Hence, $ab = b^4a = ba$, i.e. $a$ and $b$ commute. Now, it is easy to see that $(ab)^6 = e$. From this, we can only conclude $lvert abrvert in 1,2,3,6$.
For the second part of the question: Notice that $(ab)^6=e$ implies $(ab)^-1 = (ab)^5 = a^5b^5 = ab^2 = b^2a$. (Here we have used $a^2 = b^3 = e$ and $ab = ba$.)
answered Jul 26 at 10:10
Claudius
3,7111515
The conclusion that $b^3=e$ is correct. The conclusion that the original group $G$ is isomorphic to $H=langle a,b | a^2=b^3=e, ab=baranglesimeqmathbbZ_6$ is wrong. Note that the trivial group satisfies all requirements (the OP never stated that $aneq b$ or that $bneq e$). The correct conclusion is that there is a group epimorphism $Hto G$ meaning $G$ is a quotient of $H$ making the order $|ab|$ a bit more complicated.
– freakish
Jul 26 at 10:23
@Claudius how did you made $b^3$ complicated like that, I didnt get that step. Why you people are taking $ab=b^4a $ whenever you like and at other times simply multiplying it as it is ab. I mean inside the bracket, why didnt you expand like $ba^-1 = a^-4 b$? Thats my confusion sometimes it's taken as though there is a binary operation a*b , other times simply as ab. Its confusing me.
– johny
Jul 26 at 10:42
@johny There's an implicit, standard assumption that $xy$ is the same as $x*y$. Also you cannot expand $ba^-1$ because there's no formula for that. Your formulas are precisely about two fixed elements $a,b$ and their product $ab$, nothing else. Unless there's an explicit statement "for all $a,b$" and there's no such thing.
– freakish
Jul 26 at 10:46
Also after getting $b^3 = e$ why cant we just take $(ab)^6 = $$(a^2)^3$$(b^3)^2$ =e. What is the need to prove ab = ba?
– johny
Jul 26 at 10:49
@johny The same thing: $b^3=e$ is a formula about $b$ only. You cannot substitute $ab$ for $b$ because it doesn't hold for $ab$. To make that substitution valid you would have to start your problem with "for all $a,b$ in a group the following holds: ..." but you've started with "let $a,b$ be two elements of a group" which implicitly means that they are fixed.
– freakish
Jul 26 at 10:52
 |Â
show 15 more comments
up vote
1
down vote
accepted
From $b^6=e$ and $ab = b^4a$ we can conclude $b^3 = e$: we have
$$
b^3 = a^-1(aba^-1)^3a = a^-1(b^4)^3a = a^-1b^12a = a^-1ea = e.
$$
Hence, $ab = b^4a = ba$, i.e. $a$ and $b$ commute. Now, it is easy to see that $(ab)^6 = e$. From this, we can only conclude $lvert abrvert in 1,2,3,6$.
For the second part of the question: Notice that $(ab)^6=e$ implies $(ab)^-1 = (ab)^5 = a^5b^5 = ab^2 = b^2a$. (Here we have used $a^2 = b^3 = e$ and $ab = ba$.)
answered Jul 26 at 10:10
Claudius
3,7111515
The conclusion that $b^3=e$ is correct. The conclusion that the original group $G$ is isomorphic to $H=langle a,b | a^2=b^3=e, ab=baranglesimeqmathbbZ_6$ is wrong. Note that the trivial group satisfies all requirements (the OP never stated that $aneq b$ or that $bneq e$). The correct conclusion is that there is a group epimorphism $Hto G$ meaning $G$ is a quotient of $H$ making the order $|ab|$ a bit more complicated.
– freakish
Jul 26 at 10:23
@Claudius how did you made $b^3$ complicated like that, I didnt get that step. Why you people are taking $ab=b^4a $ whenever you like and at other times simply multiplying it as it is ab. I mean inside the bracket, why didnt you expand like $ba^-1 = a^-4 b$? Thats my confusion sometimes it's taken as though there is a binary operation a*b , other times simply as ab. Its confusing me.
– johny
Jul 26 at 10:42
@johny There's an implicit, standard assumption that $xy$ is the same as $x*y$. Also you cannot expand $ba^-1$ because there's no formula for that. Your formulas are precisely about two fixed elements $a,b$ and their product $ab$, nothing else. Unless there's an explicit statement "for all $a,b$" and there's no such thing.
– freakish
Jul 26 at 10:46
Also after getting $b^3 = e$ why cant we just take $(ab)^6 = $$(a^2)^3$$(b^3)^2$ =e. What is the need to prove ab = ba?
– johny
Jul 26 at 10:49
@johny The same thing: $b^3=e$ is a formula about $b$ only. You cannot substitute $ab$ for $b$ because it doesn't hold for $ab$. To make that substitution valid you would have to start your problem with "for all $a,b$ in a group the following holds: ..." but you've started with "let $a,b$ be two elements of a group" which implicitly means that they are fixed.
– freakish
Jul 26 at 10:52
 |Â
show 15 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
From $b^6=e$ and $ab = b^4a$ we can conclude $b^3 = e$: we have
$$
b^3 = a^-1(aba^-1)^3a = a^-1(b^4)^3a = a^-1b^12a = a^-1ea = e.
$$
Hence, $ab = b^4a = ba$, i.e. $a$ and $b$ commute. Now, it is easy to see that $(ab)^6 = e$. From this, we can only conclude $lvert abrvert in 1,2,3,6$.
For the second part of the question: Notice that $(ab)^6=e$ implies $(ab)^-1 = (ab)^5 = a^5b^5 = ab^2 = b^2a$. (Here we have used $a^2 = b^3 = e$ and $ab = ba$.)
answered Jul 26 at 10:10
Claudius
3,7111515
From $b^6=e$ and $ab = b^4a$ we can conclude $b^3 = e$: we have
$$
b^3 = a^-1(aba^-1)^3a = a^-1(b^4)^3a = a^-1b^12a = a^-1ea = e.
$$
Hence, $ab = b^4a = ba$, i.e. $a$ and $b$ commute. Now, it is easy to see that $(ab)^6 = e$. From this, we can only conclude $lvert abrvert in 1,2,3,6$.
For the second part of the question: Notice that $(ab)^6=e$ implies $(ab)^-1 = (ab)^5 = a^5b^5 = ab^2 = b^2a$. (Here we have used $a^2 = b^3 = e$ and $ab = ba$.)
answered Jul 26 at 10:10
Claudius
3,7111515
edited Jul 26 at 11:15
answered Jul 26 at 10:10
Claudius
3,7111515
answered Jul 26 at 10:10
Claudius
3,7111515
answered Jul 26 at 10:10
Claudius
3,7111515
3,7111515
The conclusion that $b^3=e$ is correct. The conclusion that the original group $G$ is isomorphic to $H=langle a,b | a^2=b^3=e, ab=baranglesimeqmathbbZ_6$ is wrong. Note that the trivial group satisfies all requirements (the OP never stated that $aneq b$ or that $bneq e$). The correct conclusion is that there is a group epimorphism $Hto G$ meaning $G$ is a quotient of $H$ making the order $|ab|$ a bit more complicated.
– freakish
Jul 26 at 10:23
@Claudius how did you made $b^3$ complicated like that, I didnt get that step. Why you people are taking $ab=b^4a $ whenever you like and at other times simply multiplying it as it is ab. I mean inside the bracket, why didnt you expand like $ba^-1 = a^-4 b$? Thats my confusion sometimes it's taken as though there is a binary operation a*b , other times simply as ab. Its confusing me.
– johny
Jul 26 at 10:42
@johny There's an implicit, standard assumption that $xy$ is the same as $x*y$. Also you cannot expand $ba^-1$ because there's no formula for that. Your formulas are precisely about two fixed elements $a,b$ and their product $ab$, nothing else. Unless there's an explicit statement "for all $a,b$" and there's no such thing.
– freakish
Jul 26 at 10:46
Also after getting $b^3 = e$ why cant we just take $(ab)^6 = $$(a^2)^3$$(b^3)^2$ =e. What is the need to prove ab = ba?
– johny
Jul 26 at 10:49
@johny The same thing: $b^3=e$ is a formula about $b$ only. You cannot substitute $ab$ for $b$ because it doesn't hold for $ab$. To make that substitution valid you would have to start your problem with "for all $a,b$ in a group the following holds: ..." but you've started with "let $a,b$ be two elements of a group" which implicitly means that they are fixed.
– freakish
Jul 26 at 10:52
 |Â
show 15 more comments
The conclusion that $b^3=e$ is correct. The conclusion that the original group $G$ is isomorphic to $H=langle a,b | a^2=b^3=e, ab=baranglesimeqmathbbZ_6$ is wrong. Note that the trivial group satisfies all requirements (the OP never stated that $aneq b$ or that $bneq e$). The correct conclusion is that there is a group epimorphism $Hto G$ meaning $G$ is a quotient of $H$ making the order $|ab|$ a bit more complicated.
– freakish
Jul 26 at 10:23
@Claudius how did you made $b^3$ complicated like that, I didnt get that step. Why you people are taking $ab=b^4a $ whenever you like and at other times simply multiplying it as it is ab. I mean inside the bracket, why didnt you expand like $ba^-1 = a^-4 b$? Thats my confusion sometimes it's taken as though there is a binary operation a*b , other times simply as ab. Its confusing me.
– johny
Jul 26 at 10:42
@johny There's an implicit, standard assumption that $xy$ is the same as $x*y$. Also you cannot expand $ba^-1$ because there's no formula for that. Your formulas are precisely about two fixed elements $a,b$ and their product $ab$, nothing else. Unless there's an explicit statement "for all $a,b$" and there's no such thing.
– freakish
Jul 26 at 10:46
Also after getting $b^3 = e$ why cant we just take $(ab)^6 = $$(a^2)^3$$(b^3)^2$ =e. What is the need to prove ab = ba?
– johny
Jul 26 at 10:49
@johny The same thing: $b^3=e$ is a formula about $b$ only. You cannot substitute $ab$ for $b$ because it doesn't hold for $ab$. To make that substitution valid you would have to start your problem with "for all $a,b$ in a group the following holds: ..." but you've started with "let $a,b$ be two elements of a group" which implicitly means that they are fixed.
– freakish
Jul 26 at 10:52
The conclusion that $b^3=e$ is correct. The conclusion that the original group $G$ is isomorphic to $H=langle a,b | a^2=b^3=e, ab=baranglesimeqmathbbZ_6$ is wrong. Note that the trivial group satisfies all requirements (the OP never stated that $aneq b$ or that $bneq e$). The correct conclusion is that there is a group epimorphism $Hto G$ meaning $G$ is a quotient of $H$ making the order $|ab|$ a bit more complicated.
– freakish
Jul 26 at 10:23
The conclusion that $b^3=e$ is correct. The conclusion that the original group $G$ is isomorphic to $H=langle a,b | a^2=b^3=e, ab=baranglesimeqmathbbZ_6$ is wrong. Note that the trivial group satisfies all requirements (the OP never stated that $aneq b$ or that $bneq e$). The correct conclusion is that there is a group epimorphism $Hto G$ meaning $G$ is a quotient of $H$ making the order $|ab|$ a bit more complicated.
– freakish
Jul 26 at 10:23
@Claudius how did you made $b^3$ complicated like that, I didnt get that step. Why you people are taking $ab=b^4a $ whenever you like and at other times simply multiplying it as it is ab. I mean inside the bracket, why didnt you expand like $ba^-1 = a^-4 b$? Thats my confusion sometimes it's taken as though there is a binary operation a*b , other times simply as ab. Its confusing me.
– johny
Jul 26 at 10:42
@Claudius how did you made $b^3$ complicated like that, I didnt get that step. Why you people are taking $ab=b^4a $ whenever you like and at other times simply multiplying it as it is ab. I mean inside the bracket, why didnt you expand like $ba^-1 = a^-4 b$? Thats my confusion sometimes it's taken as though there is a binary operation a*b , other times simply as ab. Its confusing me.
– johny
Jul 26 at 10:42
@johny There's an implicit, standard assumption that $xy$ is the same as $x*y$. Also you cannot expand $ba^-1$ because there's no formula for that. Your formulas are precisely about two fixed elements $a,b$ and their product $ab$, nothing else. Unless there's an explicit statement "for all $a,b$" and there's no such thing.
– freakish
Jul 26 at 10:46
@johny There's an implicit, standard assumption that $xy$ is the same as $x*y$. Also you cannot expand $ba^-1$ because there's no formula for that. Your formulas are precisely about two fixed elements $a,b$ and their product $ab$, nothing else. Unless there's an explicit statement "for all $a,b$" and there's no such thing.
– freakish
Jul 26 at 10:46
Also after getting $b^3 = e$ why cant we just take $(ab)^6 = $$(a^2)^3$$(b^3)^2$ =e. What is the need to prove ab = ba?
– johny
Jul 26 at 10:49
Also after getting $b^3 = e$ why cant we just take $(ab)^6 = $$(a^2)^3$$(b^3)^2$ =e. What is the need to prove ab = ba?
– johny
Jul 26 at 10:49
@johny The same thing: $b^3=e$ is a formula about $b$ only. You cannot substitute $ab$ for $b$ because it doesn't hold for $ab$. To make that substitution valid you would have to start your problem with "for all $a,b$ in a group the following holds: ..." but you've started with "let $a,b$ be two elements of a group" which implicitly means that they are fixed.
– freakish
Jul 26 at 10:52
@johny The same thing: $b^3=e$ is a formula about $b$ only. You cannot substitute $ab$ for $b$ because it doesn't hold for $ab$. To make that substitution valid you would have to start your problem with "for all $a,b$ in a group the following holds: ..." but you've started with "let $a,b$ be two elements of a group" which implicitly means that they are fixed.
– freakish
Jul 26 at 10:52
 |Â
show 15 more comments
up vote
0
down vote
Let be $|ab|=m$. Then $(ab)^m=e$.
Observe that $(ab)^3=(ab)(ab)(ab)=ab(b^4a)(ab)=ab^5a^2b=a$. So
$((ab)^3)^2=a^2=e$. So $|ab|=mmid 6$. Hence $|ab|=min 1,2,3,6$.
We conclude easy that $ab=ba$ and $a^2=b^3=e$ and inverse of $ab$ of the form $a^mb^n=ab^2$ and of the form $b^ma^n=b^2a$.
This means that general form of group $G=1,a,b,b^2, ab, ab^2$.
If $|ab|=1$ then $G=e, a$ i.e., $Gcong Bbb Z_2$,
IF $|ab|=2$ then $G=e,a,b,ab$ i.e., $Gcong Bbb Z_ztimes Z_2$ ( (can not be isomorph $Bbb Z_4$ It hasn't any element order $4$),
If $|ab|=3$ then $G=1,a,b,b^2, ab, ab^2$ i.e., then $Gcong Bbb Z_6$,
If $|ab|=6$ then $langle abrangle= G=1,a,b,b^2, ab, ab^2$ $Gcong Bbb Z_6$.
answered Jul 26 at 11:01
1ENÄ°GMA1
762315
Is this method correct? this look very straightforward and easy, did you check the answer below, why they are going that way if it can be done like this? Are you sure?
– johny
Jul 26 at 11:10
please look at the below answer. It's saying your method is not correct. can u comment on that?
– johny
Jul 26 at 11:32
It is easy see that m>2Wrong. If $G$ is trivial then $|ab|=1$. If $G$ is $mathbbZ_2$ then $|ab|=2$. Also the only valid conclusion from $(ab)^6=e$ is that $|ab|$ divides $6$.
– freakish
Jul 26 at 11:57
no as pointed above they are saying your proof is wrong. Do you agree with that? Can u give any counter argument on point raised by @freakish?
– johny
Jul 26 at 12:03
@freakish. You are right :). I edited my answer for you. But, We can exclude easy concept there.
– 1ENÄ°GMA1
Jul 26 at 12:09
 |Â
show 12 more comments
up vote
0
down vote
Let be $|ab|=m$. Then $(ab)^m=e$.
Observe that $(ab)^3=(ab)(ab)(ab)=ab(b^4a)(ab)=ab^5a^2b=a$. So
$((ab)^3)^2=a^2=e$. So $|ab|=mmid 6$. Hence $|ab|=min 1,2,3,6$.
We conclude easy that $ab=ba$ and $a^2=b^3=e$ and inverse of $ab$ of the form $a^mb^n=ab^2$ and of the form $b^ma^n=b^2a$.
This means that general form of group $G=1,a,b,b^2, ab, ab^2$.
If $|ab|=1$ then $G=e, a$ i.e., $Gcong Bbb Z_2$,
IF $|ab|=2$ then $G=e,a,b,ab$ i.e., $Gcong Bbb Z_ztimes Z_2$ ( (can not be isomorph $Bbb Z_4$ It hasn't any element order $4$),
If $|ab|=3$ then $G=1,a,b,b^2, ab, ab^2$ i.e., then $Gcong Bbb Z_6$,
If $|ab|=6$ then $langle abrangle= G=1,a,b,b^2, ab, ab^2$ $Gcong Bbb Z_6$.
answered Jul 26 at 11:01
1ENÄ°GMA1
762315
Is this method correct? this look very straightforward and easy, did you check the answer below, why they are going that way if it can be done like this? Are you sure?
– johny
Jul 26 at 11:10
please look at the below answer. It's saying your method is not correct. can u comment on that?
– johny
Jul 26 at 11:32
It is easy see that m>2Wrong. If $G$ is trivial then $|ab|=1$. If $G$ is $mathbbZ_2$ then $|ab|=2$. Also the only valid conclusion from $(ab)^6=e$ is that $|ab|$ divides $6$.
– freakish
Jul 26 at 11:57
no as pointed above they are saying your proof is wrong. Do you agree with that? Can u give any counter argument on point raised by @freakish?
– johny
Jul 26 at 12:03
@freakish. You are right :). I edited my answer for you. But, We can exclude easy concept there.
– 1ENÄ°GMA1
Jul 26 at 12:09
 |Â
show 12 more comments
up vote
0
down vote
up vote
0
down vote
Let be $|ab|=m$. Then $(ab)^m=e$.
Observe that $(ab)^3=(ab)(ab)(ab)=ab(b^4a)(ab)=ab^5a^2b=a$. So
$((ab)^3)^2=a^2=e$. So $|ab|=mmid 6$. Hence $|ab|=min 1,2,3,6$.
We conclude easy that $ab=ba$ and $a^2=b^3=e$ and inverse of $ab$ of the form $a^mb^n=ab^2$ and of the form $b^ma^n=b^2a$.
This means that general form of group $G=1,a,b,b^2, ab, ab^2$.
If $|ab|=1$ then $G=e, a$ i.e., $Gcong Bbb Z_2$,
IF $|ab|=2$ then $G=e,a,b,ab$ i.e., $Gcong Bbb Z_ztimes Z_2$ ( (can not be isomorph $Bbb Z_4$ It hasn't any element order $4$),
If $|ab|=3$ then $G=1,a,b,b^2, ab, ab^2$ i.e., then $Gcong Bbb Z_6$,
If $|ab|=6$ then $langle abrangle= G=1,a,b,b^2, ab, ab^2$ $Gcong Bbb Z_6$.
answered Jul 26 at 11:01
1ENÄ°GMA1
762315
Let be $|ab|=m$. Then $(ab)^m=e$.
Observe that $(ab)^3=(ab)(ab)(ab)=ab(b^4a)(ab)=ab^5a^2b=a$. So
$((ab)^3)^2=a^2=e$. So $|ab|=mmid 6$. Hence $|ab|=min 1,2,3,6$.
We conclude easy that $ab=ba$ and $a^2=b^3=e$ and inverse of $ab$ of the form $a^mb^n=ab^2$ and of the form $b^ma^n=b^2a$.
This means that general form of group $G=1,a,b,b^2, ab, ab^2$.
If $|ab|=1$ then $G=e, a$ i.e., $Gcong Bbb Z_2$,
IF $|ab|=2$ then $G=e,a,b,ab$ i.e., $Gcong Bbb Z_ztimes Z_2$ ( (can not be isomorph $Bbb Z_4$ It hasn't any element order $4$),
If $|ab|=3$ then $G=1,a,b,b^2, ab, ab^2$ i.e., then $Gcong Bbb Z_6$,
If $|ab|=6$ then $langle abrangle= G=1,a,b,b^2, ab, ab^2$ $Gcong Bbb Z_6$.
answered Jul 26 at 11:01
1ENÄ°GMA1
762315
edited Jul 26 at 13:37
answered Jul 26 at 11:01
1ENÄ°GMA1
762315
answered Jul 26 at 11:01
1ENÄ°GMA1
762315
answered Jul 26 at 11:01
1ENÄ°GMA1
762315
762315
Is this method correct? this look very straightforward and easy, did you check the answer below, why they are going that way if it can be done like this? Are you sure?
– johny
Jul 26 at 11:10
please look at the below answer. It's saying your method is not correct. can u comment on that?
– johny
Jul 26 at 11:32
It is easy see that m>2Wrong. If $G$ is trivial then $|ab|=1$. If $G$ is $mathbbZ_2$ then $|ab|=2$. Also the only valid conclusion from $(ab)^6=e$ is that $|ab|$ divides $6$.
– freakish
Jul 26 at 11:57
no as pointed above they are saying your proof is wrong. Do you agree with that? Can u give any counter argument on point raised by @freakish?
– johny
Jul 26 at 12:03
@freakish. You are right :). I edited my answer for you. But, We can exclude easy concept there.
– 1ENÄ°GMA1
Jul 26 at 12:09
 |Â
show 12 more comments
Is this method correct? this look very straightforward and easy, did you check the answer below, why they are going that way if it can be done like this? Are you sure?
– johny
Jul 26 at 11:10
please look at the below answer. It's saying your method is not correct. can u comment on that?
– johny
Jul 26 at 11:32
It is easy see that m>2Wrong. If $G$ is trivial then $|ab|=1$. If $G$ is $mathbbZ_2$ then $|ab|=2$. Also the only valid conclusion from $(ab)^6=e$ is that $|ab|$ divides $6$.
– freakish
Jul 26 at 11:57
no as pointed above they are saying your proof is wrong. Do you agree with that? Can u give any counter argument on point raised by @freakish?
– johny
Jul 26 at 12:03
@freakish. You are right :). I edited my answer for you. But, We can exclude easy concept there.
– 1ENÄ°GMA1
Jul 26 at 12:09
Is this method correct? this look very straightforward and easy, did you check the answer below, why they are going that way if it can be done like this? Are you sure?
– johny
Jul 26 at 11:10
Is this method correct? this look very straightforward and easy, did you check the answer below, why they are going that way if it can be done like this? Are you sure?
– johny
Jul 26 at 11:10
please look at the below answer. It's saying your method is not correct. can u comment on that?
– johny
Jul 26 at 11:32
please look at the below answer. It's saying your method is not correct. can u comment on that?
– johny
Jul 26 at 11:32
It is easy see that m>2 Wrong. If $G$ is trivial then $|ab|=1$. If $G$ is $mathbbZ_2$ then $|ab|=2$. Also the only valid conclusion from $(ab)^6=e$ is that $|ab|$ divides $6$.– freakish
Jul 26 at 11:57
It is easy see that m>2 Wrong. If $G$ is trivial then $|ab|=1$. If $G$ is $mathbbZ_2$ then $|ab|=2$. Also the only valid conclusion from $(ab)^6=e$ is that $|ab|$ divides $6$.– freakish
Jul 26 at 11:57
no as pointed above they are saying your proof is wrong. Do you agree with that? Can u give any counter argument on point raised by @freakish?
– johny
Jul 26 at 12:03
no as pointed above they are saying your proof is wrong. Do you agree with that? Can u give any counter argument on point raised by @freakish?
– johny
Jul 26 at 12:03
@freakish. You are right :). I edited my answer for you. But, We can exclude easy concept there.
– 1ENÄ°GMA1
Jul 26 at 12:09
@freakish. You are right :). I edited my answer for you. But, We can exclude easy concept there.
– 1ENÄ°GMA1
Jul 26 at 12:09
 |Â
show 12 more comments
1
For all we know $ab$ could be equal to $e$, hence of order $1$.
– Arnaud Mortier
Jul 26 at 9:53
$(ab)^2 = abcdot ab = b^4acdot ab = b^4a^2b = b^4eb = b^5$.
– Bill Wallis
Jul 26 at 9:54
2
Why do you think that $ab=(ab)^4$? If so, then the order of $ab$ would be $1$ or $3$.
– Arnaud Mortier
Jul 26 at 9:58
@BillWallis i dont understand the operation is $ xy=y^4x $, how can you simply multiply $b^4a$ and $ab$??
– johny
Jul 26 at 10:34
@johny You are given that $ab = b^4a$. Thus, any time you see $ab$ you can replace it with $b^4a$, and vice versa. This means that the product $(ab)^2 = abab$ is equal to the product $b^4aab$ by replacing (one of) $ab$ with $b^4a$.
– Bill Wallis
Jul 26 at 10:38