Presentation of group equal to trivial group
Problem: Show that the group given by the presentation $$langle x,y,z mid xyx^{-1}y^{-2}, , , yzy^{-1}z^{-2}, , , zxz^{-1}x^{-2} rangle $$ is equivalent to the trivial group.
I have tried all sorts of manners to try to show that the relations given by the presentation above imply that $x=y=z=e$. However, I am stuck and would appreciate any hints as to how I should move forward.
abstract-algebra group-theory algebraic-topology group-presentation
add a comment |
Problem: Show that the group given by the presentation $$langle x,y,z mid xyx^{-1}y^{-2}, , , yzy^{-1}z^{-2}, , , zxz^{-1}x^{-2} rangle $$ is equivalent to the trivial group.
I have tried all sorts of manners to try to show that the relations given by the presentation above imply that $x=y=z=e$. However, I am stuck and would appreciate any hints as to how I should move forward.
abstract-algebra group-theory algebraic-topology group-presentation
Do you know any reference where you found this question. I saw it in previous year question paper of my institution.
– Bhaskar Vashishth
Feb 5 '15 at 21:53
add a comment |
Problem: Show that the group given by the presentation $$langle x,y,z mid xyx^{-1}y^{-2}, , , yzy^{-1}z^{-2}, , , zxz^{-1}x^{-2} rangle $$ is equivalent to the trivial group.
I have tried all sorts of manners to try to show that the relations given by the presentation above imply that $x=y=z=e$. However, I am stuck and would appreciate any hints as to how I should move forward.
abstract-algebra group-theory algebraic-topology group-presentation
Problem: Show that the group given by the presentation $$langle x,y,z mid xyx^{-1}y^{-2}, , , yzy^{-1}z^{-2}, , , zxz^{-1}x^{-2} rangle $$ is equivalent to the trivial group.
I have tried all sorts of manners to try to show that the relations given by the presentation above imply that $x=y=z=e$. However, I am stuck and would appreciate any hints as to how I should move forward.
abstract-algebra group-theory algebraic-topology group-presentation
abstract-algebra group-theory algebraic-topology group-presentation
edited Feb 5 '15 at 21:44
Paul Plummer
5,20721950
5,20721950
asked Nov 15 '14 at 20:39
JessJess
476616
476616
Do you know any reference where you found this question. I saw it in previous year question paper of my institution.
– Bhaskar Vashishth
Feb 5 '15 at 21:53
add a comment |
Do you know any reference where you found this question. I saw it in previous year question paper of my institution.
– Bhaskar Vashishth
Feb 5 '15 at 21:53
Do you know any reference where you found this question. I saw it in previous year question paper of my institution.
– Bhaskar Vashishth
Feb 5 '15 at 21:53
Do you know any reference where you found this question. I saw it in previous year question paper of my institution.
– Bhaskar Vashishth
Feb 5 '15 at 21:53
add a comment |
2 Answers
2
active
oldest
votes
This is a very well-known presentation of the trivial group, to be compared with the presentation of Higman's infinite group with no finite quotient. I do not know of any easy proof.
The proof I'm going to give is due to Bernhard Neumann in An Essay on Free Products of Groups with Amalgamations (Philosophical Transactions of the Royal Society of London, Series A, 246, 919 (1954), pp. 503-554.)
So we have to prove that in a group satisfying
$$ xyx^{-1} = y^2 qquad (R_1)$$
$$ yzy^{-1} = z^2 qquad (R_2)$$
$$ zxz^{-1} = x^2 qquad (R_3)$$
the elements $x$, $y$ and $z$ are trivial.
By inverting $(R_1)$, multiplying on the left by $y$ and on the right by $x$, we get $$yxy^{-1} = y^{-1}x.$$ This easily gives $$y^i x y^{-i} = y^{-i} x qquad(R_1^{[i]}),$$ for every integer $i$, by induction. The same argument on the second relation gives $$z^i y z^{-i} = z^{-i} y. qquad (R_2^{[i]})$$
If we now conjugate $(R_3)$ by $y$, the left-hand side becomes
$$begin{align}yzxz^{-1}y^{-1} &= z^2ycdot x cdot y^{-1}z^{-2}\
& = z^2y^{-1}xz^{-2}\
& = z^2 y^{-1}z^{-2}cdot z^2xz^{-2}\
& = y^{-1}z^2cdot x^4
end{align}$$
(the first equality is a double use of the relation $yz = z^2y$, a reformulation of $(R_2)$ ; the second uses $(R_1^{[1]})$ and the last uses the inverse of $(R_2^{[2]})$ and $R_3$ twice).
On the other hand, the left side becomes
$$begin{align} yx^2y^{-1} &= (y^{-1}x)^2 \
&= y^{-1}xy^{-1}x \
&= y^{-3}x^2
end{align}$$
(the first equality uses the $(R_1^{[1]})$ twice, the last uses the inverse of $(R_1)$).
Put together, we have proven $y^{-1}z^2x^4 = y^{-3}x^2$, which gives
$$z^2 = y^{-2}x^{-2}.qquad (R^*)$$
If we conjugate $y$ by $z^{-2}$, we now get on the one hand
$$begin{align}z^{-2}y z^2 &= x^2 y^2 cdot y cdot y^{-2} x^{-2} \
&= x^2 y x^{-2} \
&= y^4
end{align}$$
(the first equality uses $(R^*)$ twice, the last uses $(R_1)$ twice.) But, on the other hand, $z^{-2}yz^2 = z^2 y$ because of $(R_2^{[-2]}$). So we finally get $z^2 y = y^4$, which translates to
$$z^2 = y^3.$$
This proves that $y$ and $z^2$ commute. The relation $(R_2)$ then boils down to $z = z^2$, which gives $z = 1$. Because of the symmetries in the presentation, this proves that the group is trivial.
Not very enlightening, but the fact that the corresponding group with 4 generators is highly nontrivial somehow reduces my hopes of ever finding a "good reason" for this group to be trivial.
add a comment |
Here is a proof-
$G=langle x,y,z mid xyx^{-1}=y^{2}, , , yzy^{-1}=z^{2}, , , zxz^{-1}=x^{2} rangle$. See that $xyx^{-1}y^{-1}=y$ tells you that $y$ belongs in the Commutator subgroup generated by $x$ and $y$ and similarly $x$ belongs in Commutator subgroup generated by $x$ and $z$ and $z$ belongs in Commutator subgroup generated by $z$ and $y$ and Now this gives you that $G$ is perfect i.e. $G=G'$. Now If you can prove $G$ is solvable , you are done as only perfect solvable group is trivial group.
So consider the subgroup generated by $H=langle x,y rangle$ and show that $H$ is solvable. For that consider $H_1=langle y rangle < H$ and it is easily seen that $H_1 unlhd H$ so what is factor group $H/H_1$? Yeah correct, $H/H_1 cong langle x rangle$ which is abelian and hence $H$ is solvable.
Now only thing remains is to check that $H=G$, i.e. $z in langle x,y rangle$. Now is the tedious calculation work you will have to do, in order to prove this, use the relators given and express $z$ in terms of $x$ and $y$.
I hope this helps!
add a comment |
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2 Answers
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2 Answers
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active
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This is a very well-known presentation of the trivial group, to be compared with the presentation of Higman's infinite group with no finite quotient. I do not know of any easy proof.
The proof I'm going to give is due to Bernhard Neumann in An Essay on Free Products of Groups with Amalgamations (Philosophical Transactions of the Royal Society of London, Series A, 246, 919 (1954), pp. 503-554.)
So we have to prove that in a group satisfying
$$ xyx^{-1} = y^2 qquad (R_1)$$
$$ yzy^{-1} = z^2 qquad (R_2)$$
$$ zxz^{-1} = x^2 qquad (R_3)$$
the elements $x$, $y$ and $z$ are trivial.
By inverting $(R_1)$, multiplying on the left by $y$ and on the right by $x$, we get $$yxy^{-1} = y^{-1}x.$$ This easily gives $$y^i x y^{-i} = y^{-i} x qquad(R_1^{[i]}),$$ for every integer $i$, by induction. The same argument on the second relation gives $$z^i y z^{-i} = z^{-i} y. qquad (R_2^{[i]})$$
If we now conjugate $(R_3)$ by $y$, the left-hand side becomes
$$begin{align}yzxz^{-1}y^{-1} &= z^2ycdot x cdot y^{-1}z^{-2}\
& = z^2y^{-1}xz^{-2}\
& = z^2 y^{-1}z^{-2}cdot z^2xz^{-2}\
& = y^{-1}z^2cdot x^4
end{align}$$
(the first equality is a double use of the relation $yz = z^2y$, a reformulation of $(R_2)$ ; the second uses $(R_1^{[1]})$ and the last uses the inverse of $(R_2^{[2]})$ and $R_3$ twice).
On the other hand, the left side becomes
$$begin{align} yx^2y^{-1} &= (y^{-1}x)^2 \
&= y^{-1}xy^{-1}x \
&= y^{-3}x^2
end{align}$$
(the first equality uses the $(R_1^{[1]})$ twice, the last uses the inverse of $(R_1)$).
Put together, we have proven $y^{-1}z^2x^4 = y^{-3}x^2$, which gives
$$z^2 = y^{-2}x^{-2}.qquad (R^*)$$
If we conjugate $y$ by $z^{-2}$, we now get on the one hand
$$begin{align}z^{-2}y z^2 &= x^2 y^2 cdot y cdot y^{-2} x^{-2} \
&= x^2 y x^{-2} \
&= y^4
end{align}$$
(the first equality uses $(R^*)$ twice, the last uses $(R_1)$ twice.) But, on the other hand, $z^{-2}yz^2 = z^2 y$ because of $(R_2^{[-2]}$). So we finally get $z^2 y = y^4$, which translates to
$$z^2 = y^3.$$
This proves that $y$ and $z^2$ commute. The relation $(R_2)$ then boils down to $z = z^2$, which gives $z = 1$. Because of the symmetries in the presentation, this proves that the group is trivial.
Not very enlightening, but the fact that the corresponding group with 4 generators is highly nontrivial somehow reduces my hopes of ever finding a "good reason" for this group to be trivial.
add a comment |
This is a very well-known presentation of the trivial group, to be compared with the presentation of Higman's infinite group with no finite quotient. I do not know of any easy proof.
The proof I'm going to give is due to Bernhard Neumann in An Essay on Free Products of Groups with Amalgamations (Philosophical Transactions of the Royal Society of London, Series A, 246, 919 (1954), pp. 503-554.)
So we have to prove that in a group satisfying
$$ xyx^{-1} = y^2 qquad (R_1)$$
$$ yzy^{-1} = z^2 qquad (R_2)$$
$$ zxz^{-1} = x^2 qquad (R_3)$$
the elements $x$, $y$ and $z$ are trivial.
By inverting $(R_1)$, multiplying on the left by $y$ and on the right by $x$, we get $$yxy^{-1} = y^{-1}x.$$ This easily gives $$y^i x y^{-i} = y^{-i} x qquad(R_1^{[i]}),$$ for every integer $i$, by induction. The same argument on the second relation gives $$z^i y z^{-i} = z^{-i} y. qquad (R_2^{[i]})$$
If we now conjugate $(R_3)$ by $y$, the left-hand side becomes
$$begin{align}yzxz^{-1}y^{-1} &= z^2ycdot x cdot y^{-1}z^{-2}\
& = z^2y^{-1}xz^{-2}\
& = z^2 y^{-1}z^{-2}cdot z^2xz^{-2}\
& = y^{-1}z^2cdot x^4
end{align}$$
(the first equality is a double use of the relation $yz = z^2y$, a reformulation of $(R_2)$ ; the second uses $(R_1^{[1]})$ and the last uses the inverse of $(R_2^{[2]})$ and $R_3$ twice).
On the other hand, the left side becomes
$$begin{align} yx^2y^{-1} &= (y^{-1}x)^2 \
&= y^{-1}xy^{-1}x \
&= y^{-3}x^2
end{align}$$
(the first equality uses the $(R_1^{[1]})$ twice, the last uses the inverse of $(R_1)$).
Put together, we have proven $y^{-1}z^2x^4 = y^{-3}x^2$, which gives
$$z^2 = y^{-2}x^{-2}.qquad (R^*)$$
If we conjugate $y$ by $z^{-2}$, we now get on the one hand
$$begin{align}z^{-2}y z^2 &= x^2 y^2 cdot y cdot y^{-2} x^{-2} \
&= x^2 y x^{-2} \
&= y^4
end{align}$$
(the first equality uses $(R^*)$ twice, the last uses $(R_1)$ twice.) But, on the other hand, $z^{-2}yz^2 = z^2 y$ because of $(R_2^{[-2]}$). So we finally get $z^2 y = y^4$, which translates to
$$z^2 = y^3.$$
This proves that $y$ and $z^2$ commute. The relation $(R_2)$ then boils down to $z = z^2$, which gives $z = 1$. Because of the symmetries in the presentation, this proves that the group is trivial.
Not very enlightening, but the fact that the corresponding group with 4 generators is highly nontrivial somehow reduces my hopes of ever finding a "good reason" for this group to be trivial.
add a comment |
This is a very well-known presentation of the trivial group, to be compared with the presentation of Higman's infinite group with no finite quotient. I do not know of any easy proof.
The proof I'm going to give is due to Bernhard Neumann in An Essay on Free Products of Groups with Amalgamations (Philosophical Transactions of the Royal Society of London, Series A, 246, 919 (1954), pp. 503-554.)
So we have to prove that in a group satisfying
$$ xyx^{-1} = y^2 qquad (R_1)$$
$$ yzy^{-1} = z^2 qquad (R_2)$$
$$ zxz^{-1} = x^2 qquad (R_3)$$
the elements $x$, $y$ and $z$ are trivial.
By inverting $(R_1)$, multiplying on the left by $y$ and on the right by $x$, we get $$yxy^{-1} = y^{-1}x.$$ This easily gives $$y^i x y^{-i} = y^{-i} x qquad(R_1^{[i]}),$$ for every integer $i$, by induction. The same argument on the second relation gives $$z^i y z^{-i} = z^{-i} y. qquad (R_2^{[i]})$$
If we now conjugate $(R_3)$ by $y$, the left-hand side becomes
$$begin{align}yzxz^{-1}y^{-1} &= z^2ycdot x cdot y^{-1}z^{-2}\
& = z^2y^{-1}xz^{-2}\
& = z^2 y^{-1}z^{-2}cdot z^2xz^{-2}\
& = y^{-1}z^2cdot x^4
end{align}$$
(the first equality is a double use of the relation $yz = z^2y$, a reformulation of $(R_2)$ ; the second uses $(R_1^{[1]})$ and the last uses the inverse of $(R_2^{[2]})$ and $R_3$ twice).
On the other hand, the left side becomes
$$begin{align} yx^2y^{-1} &= (y^{-1}x)^2 \
&= y^{-1}xy^{-1}x \
&= y^{-3}x^2
end{align}$$
(the first equality uses the $(R_1^{[1]})$ twice, the last uses the inverse of $(R_1)$).
Put together, we have proven $y^{-1}z^2x^4 = y^{-3}x^2$, which gives
$$z^2 = y^{-2}x^{-2}.qquad (R^*)$$
If we conjugate $y$ by $z^{-2}$, we now get on the one hand
$$begin{align}z^{-2}y z^2 &= x^2 y^2 cdot y cdot y^{-2} x^{-2} \
&= x^2 y x^{-2} \
&= y^4
end{align}$$
(the first equality uses $(R^*)$ twice, the last uses $(R_1)$ twice.) But, on the other hand, $z^{-2}yz^2 = z^2 y$ because of $(R_2^{[-2]}$). So we finally get $z^2 y = y^4$, which translates to
$$z^2 = y^3.$$
This proves that $y$ and $z^2$ commute. The relation $(R_2)$ then boils down to $z = z^2$, which gives $z = 1$. Because of the symmetries in the presentation, this proves that the group is trivial.
Not very enlightening, but the fact that the corresponding group with 4 generators is highly nontrivial somehow reduces my hopes of ever finding a "good reason" for this group to be trivial.
This is a very well-known presentation of the trivial group, to be compared with the presentation of Higman's infinite group with no finite quotient. I do not know of any easy proof.
The proof I'm going to give is due to Bernhard Neumann in An Essay on Free Products of Groups with Amalgamations (Philosophical Transactions of the Royal Society of London, Series A, 246, 919 (1954), pp. 503-554.)
So we have to prove that in a group satisfying
$$ xyx^{-1} = y^2 qquad (R_1)$$
$$ yzy^{-1} = z^2 qquad (R_2)$$
$$ zxz^{-1} = x^2 qquad (R_3)$$
the elements $x$, $y$ and $z$ are trivial.
By inverting $(R_1)$, multiplying on the left by $y$ and on the right by $x$, we get $$yxy^{-1} = y^{-1}x.$$ This easily gives $$y^i x y^{-i} = y^{-i} x qquad(R_1^{[i]}),$$ for every integer $i$, by induction. The same argument on the second relation gives $$z^i y z^{-i} = z^{-i} y. qquad (R_2^{[i]})$$
If we now conjugate $(R_3)$ by $y$, the left-hand side becomes
$$begin{align}yzxz^{-1}y^{-1} &= z^2ycdot x cdot y^{-1}z^{-2}\
& = z^2y^{-1}xz^{-2}\
& = z^2 y^{-1}z^{-2}cdot z^2xz^{-2}\
& = y^{-1}z^2cdot x^4
end{align}$$
(the first equality is a double use of the relation $yz = z^2y$, a reformulation of $(R_2)$ ; the second uses $(R_1^{[1]})$ and the last uses the inverse of $(R_2^{[2]})$ and $R_3$ twice).
On the other hand, the left side becomes
$$begin{align} yx^2y^{-1} &= (y^{-1}x)^2 \
&= y^{-1}xy^{-1}x \
&= y^{-3}x^2
end{align}$$
(the first equality uses the $(R_1^{[1]})$ twice, the last uses the inverse of $(R_1)$).
Put together, we have proven $y^{-1}z^2x^4 = y^{-3}x^2$, which gives
$$z^2 = y^{-2}x^{-2}.qquad (R^*)$$
If we conjugate $y$ by $z^{-2}$, we now get on the one hand
$$begin{align}z^{-2}y z^2 &= x^2 y^2 cdot y cdot y^{-2} x^{-2} \
&= x^2 y x^{-2} \
&= y^4
end{align}$$
(the first equality uses $(R^*)$ twice, the last uses $(R_1)$ twice.) But, on the other hand, $z^{-2}yz^2 = z^2 y$ because of $(R_2^{[-2]}$). So we finally get $z^2 y = y^4$, which translates to
$$z^2 = y^3.$$
This proves that $y$ and $z^2$ commute. The relation $(R_2)$ then boils down to $z = z^2$, which gives $z = 1$. Because of the symmetries in the presentation, this proves that the group is trivial.
Not very enlightening, but the fact that the corresponding group with 4 generators is highly nontrivial somehow reduces my hopes of ever finding a "good reason" for this group to be trivial.
edited Jan 4 at 12:16
Shaun
8,820113681
8,820113681
answered Feb 6 '15 at 8:27
PseudoNeoPseudoNeo
6,4441952
6,4441952
add a comment |
add a comment |
Here is a proof-
$G=langle x,y,z mid xyx^{-1}=y^{2}, , , yzy^{-1}=z^{2}, , , zxz^{-1}=x^{2} rangle$. See that $xyx^{-1}y^{-1}=y$ tells you that $y$ belongs in the Commutator subgroup generated by $x$ and $y$ and similarly $x$ belongs in Commutator subgroup generated by $x$ and $z$ and $z$ belongs in Commutator subgroup generated by $z$ and $y$ and Now this gives you that $G$ is perfect i.e. $G=G'$. Now If you can prove $G$ is solvable , you are done as only perfect solvable group is trivial group.
So consider the subgroup generated by $H=langle x,y rangle$ and show that $H$ is solvable. For that consider $H_1=langle y rangle < H$ and it is easily seen that $H_1 unlhd H$ so what is factor group $H/H_1$? Yeah correct, $H/H_1 cong langle x rangle$ which is abelian and hence $H$ is solvable.
Now only thing remains is to check that $H=G$, i.e. $z in langle x,y rangle$. Now is the tedious calculation work you will have to do, in order to prove this, use the relators given and express $z$ in terms of $x$ and $y$.
I hope this helps!
add a comment |
Here is a proof-
$G=langle x,y,z mid xyx^{-1}=y^{2}, , , yzy^{-1}=z^{2}, , , zxz^{-1}=x^{2} rangle$. See that $xyx^{-1}y^{-1}=y$ tells you that $y$ belongs in the Commutator subgroup generated by $x$ and $y$ and similarly $x$ belongs in Commutator subgroup generated by $x$ and $z$ and $z$ belongs in Commutator subgroup generated by $z$ and $y$ and Now this gives you that $G$ is perfect i.e. $G=G'$. Now If you can prove $G$ is solvable , you are done as only perfect solvable group is trivial group.
So consider the subgroup generated by $H=langle x,y rangle$ and show that $H$ is solvable. For that consider $H_1=langle y rangle < H$ and it is easily seen that $H_1 unlhd H$ so what is factor group $H/H_1$? Yeah correct, $H/H_1 cong langle x rangle$ which is abelian and hence $H$ is solvable.
Now only thing remains is to check that $H=G$, i.e. $z in langle x,y rangle$. Now is the tedious calculation work you will have to do, in order to prove this, use the relators given and express $z$ in terms of $x$ and $y$.
I hope this helps!
add a comment |
Here is a proof-
$G=langle x,y,z mid xyx^{-1}=y^{2}, , , yzy^{-1}=z^{2}, , , zxz^{-1}=x^{2} rangle$. See that $xyx^{-1}y^{-1}=y$ tells you that $y$ belongs in the Commutator subgroup generated by $x$ and $y$ and similarly $x$ belongs in Commutator subgroup generated by $x$ and $z$ and $z$ belongs in Commutator subgroup generated by $z$ and $y$ and Now this gives you that $G$ is perfect i.e. $G=G'$. Now If you can prove $G$ is solvable , you are done as only perfect solvable group is trivial group.
So consider the subgroup generated by $H=langle x,y rangle$ and show that $H$ is solvable. For that consider $H_1=langle y rangle < H$ and it is easily seen that $H_1 unlhd H$ so what is factor group $H/H_1$? Yeah correct, $H/H_1 cong langle x rangle$ which is abelian and hence $H$ is solvable.
Now only thing remains is to check that $H=G$, i.e. $z in langle x,y rangle$. Now is the tedious calculation work you will have to do, in order to prove this, use the relators given and express $z$ in terms of $x$ and $y$.
I hope this helps!
Here is a proof-
$G=langle x,y,z mid xyx^{-1}=y^{2}, , , yzy^{-1}=z^{2}, , , zxz^{-1}=x^{2} rangle$. See that $xyx^{-1}y^{-1}=y$ tells you that $y$ belongs in the Commutator subgroup generated by $x$ and $y$ and similarly $x$ belongs in Commutator subgroup generated by $x$ and $z$ and $z$ belongs in Commutator subgroup generated by $z$ and $y$ and Now this gives you that $G$ is perfect i.e. $G=G'$. Now If you can prove $G$ is solvable , you are done as only perfect solvable group is trivial group.
So consider the subgroup generated by $H=langle x,y rangle$ and show that $H$ is solvable. For that consider $H_1=langle y rangle < H$ and it is easily seen that $H_1 unlhd H$ so what is factor group $H/H_1$? Yeah correct, $H/H_1 cong langle x rangle$ which is abelian and hence $H$ is solvable.
Now only thing remains is to check that $H=G$, i.e. $z in langle x,y rangle$. Now is the tedious calculation work you will have to do, in order to prove this, use the relators given and express $z$ in terms of $x$ and $y$.
I hope this helps!
answered Feb 26 '15 at 11:57
Bhaskar VashishthBhaskar Vashishth
7,59712053
7,59712053
add a comment |
add a comment |
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Do you know any reference where you found this question. I saw it in previous year question paper of my institution.
– Bhaskar Vashishth
Feb 5 '15 at 21:53