How do I find $left|langle a,bmid a^2=b^3=erangleright|$?
$begingroup$
Suppose $G$ is a group satisfying $G=langle a,bmid a^2=b^3=erangle$. Find $|G|$.
group-theory finite-groups group-presentation combinatorial-group-theory
edited Dec 31 '18 at 13:59
Shaun
9,310113684
asked Mar 22 '13 at 9:03
Hung nguyenHung nguyen
30028
$endgroup$
add a comment |
$begingroup$
Suppose $G$ is a group satisfying $G=langle a,bmid a^2=b^3=erangle$. Find $|G|$.
group-theory finite-groups group-presentation combinatorial-group-theory
edited Dec 31 '18 at 13:59
Shaun
9,310113684
asked Mar 22 '13 at 9:03
Hung nguyenHung nguyen
30028
$endgroup$
2
$begingroup$
Do you mean for the relations to read $a^2 = b^3 = e$, where $e$ is the identity?
$endgroup$
– Sammy Black
Mar 22 '13 at 9:07
$begingroup$
Also, the trivial group $G={e}$ with $a=b=e$ satisfies your conditions.
$endgroup$
– Stefan
Mar 22 '13 at 9:29
3
$begingroup$
you should post the problems carefully
$endgroup$
– Mathematician
Mar 22 '13 at 9:43
$begingroup$
The author's original question seemed to imply that we are free to assume $G$ is a group. We should clarify as the question only makes sense with such an assumption.
$endgroup$
– user47805
Mar 22 '13 at 9:59
$begingroup$
Why don't use GAP to help you finding the order?
$endgroup$
– mrs
Mar 22 '13 at 20:16
add a comment |
$begingroup$
Suppose $G$ is a group satisfying $G=langle a,bmid a^2=b^3=erangle$. Find $|G|$.
group-theory finite-groups group-presentation combinatorial-group-theory
edited Dec 31 '18 at 13:59
Shaun
9,310113684
asked Mar 22 '13 at 9:03
Hung nguyenHung nguyen
30028
$endgroup$
Suppose $G$ is a group satisfying $G=langle a,bmid a^2=b^3=erangle$. Find $|G|$.
group-theory finite-groups group-presentation combinatorial-group-theory
group-theory finite-groups group-presentation combinatorial-group-theory
edited Dec 31 '18 at 13:59
Shaun
9,310113684
asked Mar 22 '13 at 9:03
Hung nguyenHung nguyen
30028
edited Dec 31 '18 at 13:59
Shaun
9,310113684
asked Mar 22 '13 at 9:03
Hung nguyenHung nguyen
30028
edited Dec 31 '18 at 13:59
Shaun
9,310113684
edited Dec 31 '18 at 13:59
Shaun
9,310113684
edited Dec 31 '18 at 13:59
Shaun
9,310113684
9,310113684
asked Mar 22 '13 at 9:03
Hung nguyenHung nguyen
30028
asked Mar 22 '13 at 9:03
Hung nguyenHung nguyen
30028
asked Mar 22 '13 at 9:03
Hung nguyenHung nguyen
30028
30028
2
$begingroup$
Do you mean for the relations to read $a^2 = b^3 = e$, where $e$ is the identity?
$endgroup$
– Sammy Black
Mar 22 '13 at 9:07
$begingroup$
Also, the trivial group $G={e}$ with $a=b=e$ satisfies your conditions.
$endgroup$
– Stefan
Mar 22 '13 at 9:29
3
$begingroup$
you should post the problems carefully
$endgroup$
– Mathematician
Mar 22 '13 at 9:43
$begingroup$
The author's original question seemed to imply that we are free to assume $G$ is a group. We should clarify as the question only makes sense with such an assumption.
$endgroup$
– user47805
Mar 22 '13 at 9:59
$begingroup$
Why don't use GAP to help you finding the order?
$endgroup$
– mrs
Mar 22 '13 at 20:16
add a comment |
2
$begingroup$
Do you mean for the relations to read $a^2 = b^3 = e$, where $e$ is the identity?
$endgroup$
– Sammy Black
Mar 22 '13 at 9:07
$begingroup$
Also, the trivial group $G={e}$ with $a=b=e$ satisfies your conditions.
$endgroup$
– Stefan
Mar 22 '13 at 9:29
3
$begingroup$
you should post the problems carefully
$endgroup$
– Mathematician
Mar 22 '13 at 9:43
$begingroup$
The author's original question seemed to imply that we are free to assume $G$ is a group. We should clarify as the question only makes sense with such an assumption.
$endgroup$
– user47805
Mar 22 '13 at 9:59
$begingroup$
Why don't use GAP to help you finding the order?
$endgroup$
– mrs
Mar 22 '13 at 20:16
2
2
$begingroup$
Do you mean for the relations to read $a^2 = b^3 = e$, where $e$ is the identity?
$endgroup$
– Sammy Black
Mar 22 '13 at 9:07
$begingroup$
Do you mean for the relations to read $a^2 = b^3 = e$, where $e$ is the identity?
$endgroup$
– Sammy Black
Mar 22 '13 at 9:07
$begingroup$
Also, the trivial group $G={e}$ with $a=b=e$ satisfies your conditions.
$endgroup$
– Stefan
Mar 22 '13 at 9:29
$begingroup$
Also, the trivial group $G={e}$ with $a=b=e$ satisfies your conditions.
$endgroup$
– Stefan
Mar 22 '13 at 9:29
3
3
$begingroup$
you should post the problems carefully
$endgroup$
– Mathematician
Mar 22 '13 at 9:43
$begingroup$
you should post the problems carefully
$endgroup$
– Mathematician
Mar 22 '13 at 9:43
$begingroup$
The author's original question seemed to imply that we are free to assume $G$ is a group. We should clarify as the question only makes sense with such an assumption.
$endgroup$
– user47805
Mar 22 '13 at 9:59
$begingroup$
The author's original question seemed to imply that we are free to assume $G$ is a group. We should clarify as the question only makes sense with such an assumption.
$endgroup$
– user47805
Mar 22 '13 at 9:59
$begingroup$
Why don't use GAP to help you finding the order?
$endgroup$
– mrs
Mar 22 '13 at 20:16
$begingroup$
Why don't use GAP to help you finding the order?
$endgroup$
– mrs
Mar 22 '13 at 20:16
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Note: I am assuming that you intended $G = langle a, b ,|, a^2 = b^3 = e rangle$.
This is not a finite group: it is the free product $mathbb{Z}_2 * mathbb{Z}_3$. It has distinct elements that are words of any length that alternate between $a$ and either $b$ or $b^2$, such as $abab^2ab^2abababa$.
answered Mar 22 '13 at 9:22
Sammy BlackSammy Black
12.3k21736
$endgroup$
2
$begingroup$
That free product is known to be isomorphic to the quotient of $SL_2(mathbb{Z})$ by its center $langle -I_2rangle$. Clearly an infinite group.
$endgroup$
– Jyrki Lahtonen
Mar 22 '13 at 9:50
add a comment |
$begingroup$
To form a group using a presentation you must write $langle S | R rangle$ where $S$ is a set of generators and $R$ is a set of relations among those generators. $3$ is not in the set of generators in what you have written, so this is not a valid group presentation.
If instead of $3$ you meant to write $1$ or $e$ or $text{id}$, the group formed does not have finite order, because $a^{-1}b^{-1}ab$ is not present in the relators in any form. Thus we can form an infinite number of elements that look like $abababababldots$ as we have no way of reducing these words.
answered Mar 22 '13 at 9:41
Alexander Gruber♦Alexander Gruber
20.1k25102172
$endgroup$
$begingroup$
I think we have two answer for this question . 1) $|G|=6$ when G is a finite group . 2) G is infinite when it's free group .
$endgroup$
– Hung nguyen
Mar 22 '13 at 9:59
3
$begingroup$
@Hungnguyen There are many finite groups of order greater than $6$ generated by an element of order $2$ and an element of order $3$. For example $A_4$ is generated by $(12)(34)$ and $(123)$. The only way the question makes sense is if it's a group presentation, and in that case, the group is not finite.
$endgroup$
– Alexander Gruber♦
Mar 22 '13 at 20:35
add a comment |
$begingroup$
With the edit to the problem, this answer is
=====
Assuming we mean $G=langle a,b|a^2=b^3=erangle$.
Consider $S_3$ where $(1,2)^2=(1,2,3)^3=e$ and $|S_3|=6$.
We have (I'm sure there is an elegant way to do this)
$(1,2)^2=e$
$(1,2)=(1,2)$
$(1,2,3)=(1,2,3)$
$(1,2)(1,2,3)=(2,3)$
$(1,2,3)(1,2)=(1,3)$
$(1,2,3)^2=(1,3,2)$
answered Mar 22 '13 at 9:27
user47805user47805
366316
$endgroup$
2
$begingroup$
The symmetric group needs an addition relation, such as $ba = ab^2$.
$endgroup$
– Sammy Black
Mar 22 '13 at 9:38
1
$begingroup$
Yes, but the group satisfies all given conditions. With the edit you are correct.
$endgroup$
– user47805
Mar 22 '13 at 9:40
4
$begingroup$
The notation involving the angle brackets usually suggests a group presentation, which is a quotient of the free group on the given set of symbols (the generators) by the normal subgroup generated by expressions on the right (the relations).
$endgroup$
– Sammy Black
Mar 22 '13 at 9:43
1
$begingroup$
I believe the original question was something along the lines of "If G is a group that satisfies..." in which case the above answer holds. With the edit to the question, you are correct and we require additional assumptions about G.
$endgroup$
– user47805
Mar 22 '13 at 9:48
$begingroup$
$G=infty$ because $forall xin G$ we has $x=(ab)^nquad (fofall n in mathbb{Z}$ . With $S_3$ we need add a relation is $(ab)^2=id$
$endgroup$
– Hung nguyen
Mar 24 '13 at 2:00
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note: I am assuming that you intended $G = langle a, b ,|, a^2 = b^3 = e rangle$.
This is not a finite group: it is the free product $mathbb{Z}_2 * mathbb{Z}_3$. It has distinct elements that are words of any length that alternate between $a$ and either $b$ or $b^2$, such as $abab^2ab^2abababa$.
answered Mar 22 '13 at 9:22
Sammy BlackSammy Black
12.3k21736
$endgroup$
2
$begingroup$
That free product is known to be isomorphic to the quotient of $SL_2(mathbb{Z})$ by its center $langle -I_2rangle$. Clearly an infinite group.
$endgroup$
– Jyrki Lahtonen
Mar 22 '13 at 9:50
add a comment |
$begingroup$
Note: I am assuming that you intended $G = langle a, b ,|, a^2 = b^3 = e rangle$.
This is not a finite group: it is the free product $mathbb{Z}_2 * mathbb{Z}_3$. It has distinct elements that are words of any length that alternate between $a$ and either $b$ or $b^2$, such as $abab^2ab^2abababa$.
answered Mar 22 '13 at 9:22
Sammy BlackSammy Black
12.3k21736
$endgroup$
2
$begingroup$
That free product is known to be isomorphic to the quotient of $SL_2(mathbb{Z})$ by its center $langle -I_2rangle$. Clearly an infinite group.
$endgroup$
– Jyrki Lahtonen
Mar 22 '13 at 9:50
add a comment |
$begingroup$
Note: I am assuming that you intended $G = langle a, b ,|, a^2 = b^3 = e rangle$.
This is not a finite group: it is the free product $mathbb{Z}_2 * mathbb{Z}_3$. It has distinct elements that are words of any length that alternate between $a$ and either $b$ or $b^2$, such as $abab^2ab^2abababa$.
answered Mar 22 '13 at 9:22
Sammy BlackSammy Black
12.3k21736
$endgroup$
Note: I am assuming that you intended $G = langle a, b ,|, a^2 = b^3 = e rangle$.
This is not a finite group: it is the free product $mathbb{Z}_2 * mathbb{Z}_3$. It has distinct elements that are words of any length that alternate between $a$ and either $b$ or $b^2$, such as $abab^2ab^2abababa$.
answered Mar 22 '13 at 9:22
Sammy BlackSammy Black
12.3k21736
answered Mar 22 '13 at 9:22
Sammy BlackSammy Black
12.3k21736
answered Mar 22 '13 at 9:22
Sammy BlackSammy Black
12.3k21736
answered Mar 22 '13 at 9:22
Sammy BlackSammy Black
12.3k21736
12.3k21736
2
$begingroup$
That free product is known to be isomorphic to the quotient of $SL_2(mathbb{Z})$ by its center $langle -I_2rangle$. Clearly an infinite group.
$endgroup$
– Jyrki Lahtonen
Mar 22 '13 at 9:50
add a comment |
2
$begingroup$
That free product is known to be isomorphic to the quotient of $SL_2(mathbb{Z})$ by its center $langle -I_2rangle$. Clearly an infinite group.
$endgroup$
– Jyrki Lahtonen
Mar 22 '13 at 9:50
2
2
$begingroup$
That free product is known to be isomorphic to the quotient of $SL_2(mathbb{Z})$ by its center $langle -I_2rangle$. Clearly an infinite group.
$endgroup$
– Jyrki Lahtonen
Mar 22 '13 at 9:50
$begingroup$
That free product is known to be isomorphic to the quotient of $SL_2(mathbb{Z})$ by its center $langle -I_2rangle$. Clearly an infinite group.
$endgroup$
– Jyrki Lahtonen
Mar 22 '13 at 9:50
add a comment |
$begingroup$
To form a group using a presentation you must write $langle S | R rangle$ where $S$ is a set of generators and $R$ is a set of relations among those generators. $3$ is not in the set of generators in what you have written, so this is not a valid group presentation.
If instead of $3$ you meant to write $1$ or $e$ or $text{id}$, the group formed does not have finite order, because $a^{-1}b^{-1}ab$ is not present in the relators in any form. Thus we can form an infinite number of elements that look like $abababababldots$ as we have no way of reducing these words.
answered Mar 22 '13 at 9:41
Alexander Gruber♦Alexander Gruber
20.1k25102172
$endgroup$
$begingroup$
I think we have two answer for this question . 1) $|G|=6$ when G is a finite group . 2) G is infinite when it's free group .
$endgroup$
– Hung nguyen
Mar 22 '13 at 9:59
3
$begingroup$
@Hungnguyen There are many finite groups of order greater than $6$ generated by an element of order $2$ and an element of order $3$. For example $A_4$ is generated by $(12)(34)$ and $(123)$. The only way the question makes sense is if it's a group presentation, and in that case, the group is not finite.
$endgroup$
– Alexander Gruber♦
Mar 22 '13 at 20:35
add a comment |
$begingroup$
To form a group using a presentation you must write $langle S | R rangle$ where $S$ is a set of generators and $R$ is a set of relations among those generators. $3$ is not in the set of generators in what you have written, so this is not a valid group presentation.
If instead of $3$ you meant to write $1$ or $e$ or $text{id}$, the group formed does not have finite order, because $a^{-1}b^{-1}ab$ is not present in the relators in any form. Thus we can form an infinite number of elements that look like $abababababldots$ as we have no way of reducing these words.
answered Mar 22 '13 at 9:41
Alexander Gruber♦Alexander Gruber
20.1k25102172
$endgroup$
$begingroup$
I think we have two answer for this question . 1) $|G|=6$ when G is a finite group . 2) G is infinite when it's free group .
$endgroup$
– Hung nguyen
Mar 22 '13 at 9:59
3
$begingroup$
@Hungnguyen There are many finite groups of order greater than $6$ generated by an element of order $2$ and an element of order $3$. For example $A_4$ is generated by $(12)(34)$ and $(123)$. The only way the question makes sense is if it's a group presentation, and in that case, the group is not finite.
$endgroup$
– Alexander Gruber♦
Mar 22 '13 at 20:35
add a comment |
$begingroup$
To form a group using a presentation you must write $langle S | R rangle$ where $S$ is a set of generators and $R$ is a set of relations among those generators. $3$ is not in the set of generators in what you have written, so this is not a valid group presentation.
If instead of $3$ you meant to write $1$ or $e$ or $text{id}$, the group formed does not have finite order, because $a^{-1}b^{-1}ab$ is not present in the relators in any form. Thus we can form an infinite number of elements that look like $abababababldots$ as we have no way of reducing these words.
answered Mar 22 '13 at 9:41
Alexander Gruber♦Alexander Gruber
20.1k25102172
$endgroup$
To form a group using a presentation you must write $langle S | R rangle$ where $S$ is a set of generators and $R$ is a set of relations among those generators. $3$ is not in the set of generators in what you have written, so this is not a valid group presentation.
If instead of $3$ you meant to write $1$ or $e$ or $text{id}$, the group formed does not have finite order, because $a^{-1}b^{-1}ab$ is not present in the relators in any form. Thus we can form an infinite number of elements that look like $abababababldots$ as we have no way of reducing these words.
answered Mar 22 '13 at 9:41
Alexander Gruber♦Alexander Gruber
20.1k25102172
edited Mar 22 '13 at 9:47
answered Mar 22 '13 at 9:41
Alexander Gruber♦Alexander Gruber
20.1k25102172
answered Mar 22 '13 at 9:41
Alexander Gruber♦Alexander Gruber
20.1k25102172
answered Mar 22 '13 at 9:41
Alexander Gruber♦Alexander Gruber
20.1k25102172
20.1k25102172
$begingroup$
I think we have two answer for this question . 1) $|G|=6$ when G is a finite group . 2) G is infinite when it's free group .
$endgroup$
– Hung nguyen
Mar 22 '13 at 9:59
3
$begingroup$
@Hungnguyen There are many finite groups of order greater than $6$ generated by an element of order $2$ and an element of order $3$. For example $A_4$ is generated by $(12)(34)$ and $(123)$. The only way the question makes sense is if it's a group presentation, and in that case, the group is not finite.
$endgroup$
– Alexander Gruber♦
Mar 22 '13 at 20:35
add a comment |
$begingroup$
I think we have two answer for this question . 1) $|G|=6$ when G is a finite group . 2) G is infinite when it's free group .
$endgroup$
– Hung nguyen
Mar 22 '13 at 9:59
3
$begingroup$
@Hungnguyen There are many finite groups of order greater than $6$ generated by an element of order $2$ and an element of order $3$. For example $A_4$ is generated by $(12)(34)$ and $(123)$. The only way the question makes sense is if it's a group presentation, and in that case, the group is not finite.
$endgroup$
– Alexander Gruber♦
Mar 22 '13 at 20:35
$begingroup$
I think we have two answer for this question . 1) $|G|=6$ when G is a finite group . 2) G is infinite when it's free group .
$endgroup$
– Hung nguyen
Mar 22 '13 at 9:59
$begingroup$
I think we have two answer for this question . 1) $|G|=6$ when G is a finite group . 2) G is infinite when it's free group .
$endgroup$
– Hung nguyen
Mar 22 '13 at 9:59
3
3
$begingroup$
@Hungnguyen There are many finite groups of order greater than $6$ generated by an element of order $2$ and an element of order $3$. For example $A_4$ is generated by $(12)(34)$ and $(123)$. The only way the question makes sense is if it's a group presentation, and in that case, the group is not finite.
$endgroup$
– Alexander Gruber♦
Mar 22 '13 at 20:35
$begingroup$
@Hungnguyen There are many finite groups of order greater than $6$ generated by an element of order $2$ and an element of order $3$. For example $A_4$ is generated by $(12)(34)$ and $(123)$. The only way the question makes sense is if it's a group presentation, and in that case, the group is not finite.
$endgroup$
– Alexander Gruber♦
Mar 22 '13 at 20:35
add a comment |
$begingroup$
With the edit to the problem, this answer is
=====
Assuming we mean $G=langle a,b|a^2=b^3=erangle$.
Consider $S_3$ where $(1,2)^2=(1,2,3)^3=e$ and $|S_3|=6$.
We have (I'm sure there is an elegant way to do this)
$(1,2)^2=e$
$(1,2)=(1,2)$
$(1,2,3)=(1,2,3)$
$(1,2)(1,2,3)=(2,3)$
$(1,2,3)(1,2)=(1,3)$
$(1,2,3)^2=(1,3,2)$
answered Mar 22 '13 at 9:27
user47805user47805
366316
$endgroup$
2
$begingroup$
The symmetric group needs an addition relation, such as $ba = ab^2$.
$endgroup$
– Sammy Black
Mar 22 '13 at 9:38
1
$begingroup$
Yes, but the group satisfies all given conditions. With the edit you are correct.
$endgroup$
– user47805
Mar 22 '13 at 9:40
4
$begingroup$
The notation involving the angle brackets usually suggests a group presentation, which is a quotient of the free group on the given set of symbols (the generators) by the normal subgroup generated by expressions on the right (the relations).
$endgroup$
– Sammy Black
Mar 22 '13 at 9:43
1
$begingroup$
I believe the original question was something along the lines of "If G is a group that satisfies..." in which case the above answer holds. With the edit to the question, you are correct and we require additional assumptions about G.
$endgroup$
– user47805
Mar 22 '13 at 9:48
$begingroup$
$G=infty$ because $forall xin G$ we has $x=(ab)^nquad (fofall n in mathbb{Z}$ . With $S_3$ we need add a relation is $(ab)^2=id$
$endgroup$
– Hung nguyen
Mar 24 '13 at 2:00
add a comment |
$begingroup$
With the edit to the problem, this answer is
=====
Assuming we mean $G=langle a,b|a^2=b^3=erangle$.
Consider $S_3$ where $(1,2)^2=(1,2,3)^3=e$ and $|S_3|=6$.
We have (I'm sure there is an elegant way to do this)
$(1,2)^2=e$
$(1,2)=(1,2)$
$(1,2,3)=(1,2,3)$
$(1,2)(1,2,3)=(2,3)$
$(1,2,3)(1,2)=(1,3)$
$(1,2,3)^2=(1,3,2)$
answered Mar 22 '13 at 9:27
user47805user47805
366316
$endgroup$
2
$begingroup$
The symmetric group needs an addition relation, such as $ba = ab^2$.
$endgroup$
– Sammy Black
Mar 22 '13 at 9:38
1
$begingroup$
Yes, but the group satisfies all given conditions. With the edit you are correct.
$endgroup$
– user47805
Mar 22 '13 at 9:40
4
$begingroup$
The notation involving the angle brackets usually suggests a group presentation, which is a quotient of the free group on the given set of symbols (the generators) by the normal subgroup generated by expressions on the right (the relations).
$endgroup$
– Sammy Black
Mar 22 '13 at 9:43
1
$begingroup$
I believe the original question was something along the lines of "If G is a group that satisfies..." in which case the above answer holds. With the edit to the question, you are correct and we require additional assumptions about G.
$endgroup$
– user47805
Mar 22 '13 at 9:48
$begingroup$
$G=infty$ because $forall xin G$ we has $x=(ab)^nquad (fofall n in mathbb{Z}$ . With $S_3$ we need add a relation is $(ab)^2=id$
$endgroup$
– Hung nguyen
Mar 24 '13 at 2:00
add a comment |
$begingroup$
With the edit to the problem, this answer is
=====
Assuming we mean $G=langle a,b|a^2=b^3=erangle$.
Consider $S_3$ where $(1,2)^2=(1,2,3)^3=e$ and $|S_3|=6$.
We have (I'm sure there is an elegant way to do this)
$(1,2)^2=e$
$(1,2)=(1,2)$
$(1,2,3)=(1,2,3)$
$(1,2)(1,2,3)=(2,3)$
$(1,2,3)(1,2)=(1,3)$
$(1,2,3)^2=(1,3,2)$
answered Mar 22 '13 at 9:27
user47805user47805
366316
$endgroup$
With the edit to the problem, this answer is
=====
Assuming we mean $G=langle a,b|a^2=b^3=erangle$.
Consider $S_3$ where $(1,2)^2=(1,2,3)^3=e$ and $|S_3|=6$.
We have (I'm sure there is an elegant way to do this)
$(1,2)^2=e$
$(1,2)=(1,2)$
$(1,2,3)=(1,2,3)$
$(1,2)(1,2,3)=(2,3)$
$(1,2,3)(1,2)=(1,3)$
$(1,2,3)^2=(1,3,2)$
answered Mar 22 '13 at 9:27
user47805user47805
366316
edited Apr 1 '13 at 15:38
answered Mar 22 '13 at 9:27
user47805user47805
366316
answered Mar 22 '13 at 9:27
user47805user47805
366316
answered Mar 22 '13 at 9:27
user47805user47805
366316
366316
2
$begingroup$
The symmetric group needs an addition relation, such as $ba = ab^2$.
$endgroup$
– Sammy Black
Mar 22 '13 at 9:38
1
$begingroup$
Yes, but the group satisfies all given conditions. With the edit you are correct.
$endgroup$
– user47805
Mar 22 '13 at 9:40
4
$begingroup$
The notation involving the angle brackets usually suggests a group presentation, which is a quotient of the free group on the given set of symbols (the generators) by the normal subgroup generated by expressions on the right (the relations).
$endgroup$
– Sammy Black
Mar 22 '13 at 9:43
1
$begingroup$
I believe the original question was something along the lines of "If G is a group that satisfies..." in which case the above answer holds. With the edit to the question, you are correct and we require additional assumptions about G.
$endgroup$
– user47805
Mar 22 '13 at 9:48
$begingroup$
$G=infty$ because $forall xin G$ we has $x=(ab)^nquad (fofall n in mathbb{Z}$ . With $S_3$ we need add a relation is $(ab)^2=id$
$endgroup$
– Hung nguyen
Mar 24 '13 at 2:00
add a comment |
2
$begingroup$
The symmetric group needs an addition relation, such as $ba = ab^2$.
$endgroup$
– Sammy Black
Mar 22 '13 at 9:38
1
$begingroup$
Yes, but the group satisfies all given conditions. With the edit you are correct.
$endgroup$
– user47805
Mar 22 '13 at 9:40
4
$begingroup$
The notation involving the angle brackets usually suggests a group presentation, which is a quotient of the free group on the given set of symbols (the generators) by the normal subgroup generated by expressions on the right (the relations).
$endgroup$
– Sammy Black
Mar 22 '13 at 9:43
1
$begingroup$
I believe the original question was something along the lines of "If G is a group that satisfies..." in which case the above answer holds. With the edit to the question, you are correct and we require additional assumptions about G.
$endgroup$
– user47805
Mar 22 '13 at 9:48
$begingroup$
$G=infty$ because $forall xin G$ we has $x=(ab)^nquad (fofall n in mathbb{Z}$ . With $S_3$ we need add a relation is $(ab)^2=id$
$endgroup$
– Hung nguyen
Mar 24 '13 at 2:00
2
2
$begingroup$
The symmetric group needs an addition relation, such as $ba = ab^2$.
$endgroup$
– Sammy Black
Mar 22 '13 at 9:38
$begingroup$
The symmetric group needs an addition relation, such as $ba = ab^2$.
$endgroup$
– Sammy Black
Mar 22 '13 at 9:38
1
1
$begingroup$
Yes, but the group satisfies all given conditions. With the edit you are correct.
$endgroup$
– user47805
Mar 22 '13 at 9:40
$begingroup$
Yes, but the group satisfies all given conditions. With the edit you are correct.
$endgroup$
– user47805
Mar 22 '13 at 9:40
4
4
$begingroup$
The notation involving the angle brackets usually suggests a group presentation, which is a quotient of the free group on the given set of symbols (the generators) by the normal subgroup generated by expressions on the right (the relations).
$endgroup$
– Sammy Black
Mar 22 '13 at 9:43
$begingroup$
The notation involving the angle brackets usually suggests a group presentation, which is a quotient of the free group on the given set of symbols (the generators) by the normal subgroup generated by expressions on the right (the relations).
$endgroup$
– Sammy Black
Mar 22 '13 at 9:43
1
1
$begingroup$
I believe the original question was something along the lines of "If G is a group that satisfies..." in which case the above answer holds. With the edit to the question, you are correct and we require additional assumptions about G.
$endgroup$
– user47805
Mar 22 '13 at 9:48
$begingroup$
I believe the original question was something along the lines of "If G is a group that satisfies..." in which case the above answer holds. With the edit to the question, you are correct and we require additional assumptions about G.
$endgroup$
– user47805
Mar 22 '13 at 9:48
$begingroup$
$G=infty$ because $forall xin G$ we has $x=(ab)^nquad (fofall n in mathbb{Z}$ . With $S_3$ we need add a relation is $(ab)^2=id$
$endgroup$
– Hung nguyen
Mar 24 '13 at 2:00
$begingroup$
$G=infty$ because $forall xin G$ we has $x=(ab)^nquad (fofall n in mathbb{Z}$ . With $S_3$ we need add a relation is $(ab)^2=id$
$endgroup$
– Hung nguyen
Mar 24 '13 at 2:00
add a comment |
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2
$begingroup$
Do you mean for the relations to read $a^2 = b^3 = e$, where $e$ is the identity?
$endgroup$
– Sammy Black
Mar 22 '13 at 9:07
$begingroup$
Also, the trivial group $G={e}$ with $a=b=e$ satisfies your conditions.
$endgroup$
– Stefan
Mar 22 '13 at 9:29
3
$begingroup$
you should post the problems carefully
$endgroup$
– Mathematician
Mar 22 '13 at 9:43
$begingroup$
The author's original question seemed to imply that we are free to assume $G$ is a group. We should clarify as the question only makes sense with such an assumption.
$endgroup$
– user47805
Mar 22 '13 at 9:59
$begingroup$
Why don't use GAP to help you finding the order?
$endgroup$
– mrs
Mar 22 '13 at 20:16