How do I find $left|langle a,bmid a^2=b^3=erangleright|$?












3












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















3












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














3












3








3


2



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 13:59









Shaun

9,310113684




9,310113684










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














  • 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










3 Answers
3






active

oldest

votes


















8












$begingroup$

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 $abab^2ab^2abababa$.






share|cite|improve this answer









$endgroup$









  • 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





















3












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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





















1












$begingroup$

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)



$(1,2)^2=e$



$(1,2)=(1,2)$



$(1,2,3)=(1,2,3)$



$(1,2)(1,2,3)=(2,3)$



$(1,2,3)(1,2)=(1,3)$



$(1,2,3)^2=(1,3,2)$






share|cite|improve this answer











$endgroup$









  • 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













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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

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 $abab^2ab^2abababa$.






share|cite|improve this answer









$endgroup$









  • 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


















8












$begingroup$

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 $abab^2ab^2abababa$.






share|cite|improve this answer









$endgroup$









  • 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
















8












8








8





$begingroup$

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 $abab^2ab^2abababa$.






share|cite|improve this answer









$endgroup$



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 $abab^2ab^2abababa$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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
















  • 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













3












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















3












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















3












3








3





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 22 '13 at 9:47

























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




















  • $begingroup$
    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













1












$begingroup$

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)



$(1,2)^2=e$



$(1,2)=(1,2)$



$(1,2,3)=(1,2,3)$



$(1,2)(1,2,3)=(2,3)$



$(1,2,3)(1,2)=(1,3)$



$(1,2,3)^2=(1,3,2)$






share|cite|improve this answer











$endgroup$









  • 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


















1












$begingroup$

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)



$(1,2)^2=e$



$(1,2)=(1,2)$



$(1,2,3)=(1,2,3)$



$(1,2)(1,2,3)=(2,3)$



$(1,2,3)(1,2)=(1,3)$



$(1,2,3)^2=(1,3,2)$






share|cite|improve this answer











$endgroup$









  • 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
















1












1








1





$begingroup$

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)



$(1,2)^2=e$



$(1,2)=(1,2)$



$(1,2,3)=(1,2,3)$



$(1,2)(1,2,3)=(2,3)$



$(1,2,3)(1,2)=(1,3)$



$(1,2,3)^2=(1,3,2)$






share|cite|improve this answer











$endgroup$



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)



$(1,2)^2=e$



$(1,2)=(1,2)$



$(1,2,3)=(1,2,3)$



$(1,2)(1,2,3)=(2,3)$



$(1,2,3)(1,2)=(1,3)$



$(1,2,3)^2=(1,3,2)$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Apr 1 '13 at 15:38

























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
















  • 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




















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