finding the order of an element in a group presentation

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Let G be a group given by



$G = langle x, y | x^4 = y^4=1, yx = x^2 y^2 rangle$
I found $ G/G'$ to be isomorphic to C4
What can we say about the order of $ x$ in G?
The final answer is 4
But $ x^4 = 1$ in G means that x belong to the normal closure of R where R is the relations in G.
I can't see the connection. Can any one help




group-theory combinatorial-group-theory




share|cite|improve this question





















  • I'm just curious: how did you find G/G' isomorphic to C4?
    – Yip Jung Hon
    Jul 29 at 7:53











  • @YipJungHon: If $G$ is presented by $langle X;|;Rrangle$, then $G/G'$ can be presented by $langle X;|;Rcup Crangle$, where $C = [x_i,x_j];$. From this it is not hard to see that, for the group $G$ of this problem, $G/G'$ contains only the elements $G', xG', (xG')^2, (xG')^3=yG'$. To see that these four elements must be distinct in the presented group, it suffices to observe that there exists a group, namely $C_4=1,g,g^2,g^3$, which has an element $x:=g$ of order $4$ and another element $y:=g^-1$ which satisfies all relations in the presentation.
    – Keith Kearnes
    Jul 29 at 10:01











  • Just curious, if I define the group slightly differently, with it being: $G=<x,y | x^4=y^4=5>,$ it will still be $cong$ to $C_4$ right?
    – Yip Jung Hon
    Jul 29 at 11:06










  • @Yip Jung Hon by adding the relation xy=yx the group become abelian and hence from the last relation you have x=y and simply remove the generator y you left with x^4=1
    – Rosa1
    Jul 29 at 14:08










  • In G/G' you only make the generators abelian
    – Rosa1
    Jul 29 at 14:09














up vote
4
down vote

favorite
1












Let G be a group given by



$G = langle x, y | x^4 = y^4=1, yx = x^2 y^2 rangle$
I found $ G/G'$ to be isomorphic to C4
What can we say about the order of $ x$ in G?
The final answer is 4
But $ x^4 = 1$ in G means that x belong to the normal closure of R where R is the relations in G.
I can't see the connection. Can any one help




group-theory combinatorial-group-theory




share|cite|improve this question





















  • I'm just curious: how did you find G/G' isomorphic to C4?
    – Yip Jung Hon
    Jul 29 at 7:53











  • @YipJungHon: If $G$ is presented by $langle X;|;Rrangle$, then $G/G'$ can be presented by $langle X;|;Rcup Crangle$, where $C = [x_i,x_j];$. From this it is not hard to see that, for the group $G$ of this problem, $G/G'$ contains only the elements $G', xG', (xG')^2, (xG')^3=yG'$. To see that these four elements must be distinct in the presented group, it suffices to observe that there exists a group, namely $C_4=1,g,g^2,g^3$, which has an element $x:=g$ of order $4$ and another element $y:=g^-1$ which satisfies all relations in the presentation.
    – Keith Kearnes
    Jul 29 at 10:01











  • Just curious, if I define the group slightly differently, with it being: $G=<x,y | x^4=y^4=5>,$ it will still be $cong$ to $C_4$ right?
    – Yip Jung Hon
    Jul 29 at 11:06










  • @Yip Jung Hon by adding the relation xy=yx the group become abelian and hence from the last relation you have x=y and simply remove the generator y you left with x^4=1
    – Rosa1
    Jul 29 at 14:08










  • In G/G' you only make the generators abelian
    – Rosa1
    Jul 29 at 14:09












up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





Let G be a group given by



$G = langle x, y | x^4 = y^4=1, yx = x^2 y^2 rangle$
I found $ G/G'$ to be isomorphic to C4
What can we say about the order of $ x$ in G?
The final answer is 4
But $ x^4 = 1$ in G means that x belong to the normal closure of R where R is the relations in G.
I can't see the connection. Can any one help




group-theory combinatorial-group-theory




share|cite|improve this question













Let G be a group given by



$G = langle x, y | x^4 = y^4=1, yx = x^2 y^2 rangle$
I found $ G/G'$ to be isomorphic to C4
What can we say about the order of $ x$ in G?
The final answer is 4
But $ x^4 = 1$ in G means that x belong to the normal closure of R where R is the relations in G.
I can't see the connection. Can any one help





group-theory combinatorial-group-theory





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 29 at 7:29
























asked Jul 29 at 7:22









Rosa1

405




405











  • I'm just curious: how did you find G/G' isomorphic to C4?
    – Yip Jung Hon
    Jul 29 at 7:53











  • @YipJungHon: If $G$ is presented by $langle X;|;Rrangle$, then $G/G'$ can be presented by $langle X;|;Rcup Crangle$, where $C = [x_i,x_j];$. From this it is not hard to see that, for the group $G$ of this problem, $G/G'$ contains only the elements $G', xG', (xG')^2, (xG')^3=yG'$. To see that these four elements must be distinct in the presented group, it suffices to observe that there exists a group, namely $C_4=1,g,g^2,g^3$, which has an element $x:=g$ of order $4$ and another element $y:=g^-1$ which satisfies all relations in the presentation.
    – Keith Kearnes
    Jul 29 at 10:01











  • Just curious, if I define the group slightly differently, with it being: $G=<x,y | x^4=y^4=5>,$ it will still be $cong$ to $C_4$ right?
    – Yip Jung Hon
    Jul 29 at 11:06










  • @Yip Jung Hon by adding the relation xy=yx the group become abelian and hence from the last relation you have x=y and simply remove the generator y you left with x^4=1
    – Rosa1
    Jul 29 at 14:08










  • In G/G' you only make the generators abelian
    – Rosa1
    Jul 29 at 14:09
















  • I'm just curious: how did you find G/G' isomorphic to C4?
    – Yip Jung Hon
    Jul 29 at 7:53











  • @YipJungHon: If $G$ is presented by $langle X;|;Rrangle$, then $G/G'$ can be presented by $langle X;|;Rcup Crangle$, where $C = [x_i,x_j];$. From this it is not hard to see that, for the group $G$ of this problem, $G/G'$ contains only the elements $G', xG', (xG')^2, (xG')^3=yG'$. To see that these four elements must be distinct in the presented group, it suffices to observe that there exists a group, namely $C_4=1,g,g^2,g^3$, which has an element $x:=g$ of order $4$ and another element $y:=g^-1$ which satisfies all relations in the presentation.
    – Keith Kearnes
    Jul 29 at 10:01











  • Just curious, if I define the group slightly differently, with it being: $G=<x,y | x^4=y^4=5>,$ it will still be $cong$ to $C_4$ right?
    – Yip Jung Hon
    Jul 29 at 11:06










  • @Yip Jung Hon by adding the relation xy=yx the group become abelian and hence from the last relation you have x=y and simply remove the generator y you left with x^4=1
    – Rosa1
    Jul 29 at 14:08










  • In G/G' you only make the generators abelian
    – Rosa1
    Jul 29 at 14:09















I'm just curious: how did you find G/G' isomorphic to C4?
– Yip Jung Hon
Jul 29 at 7:53





I'm just curious: how did you find G/G' isomorphic to C4?
– Yip Jung Hon
Jul 29 at 7:53













@YipJungHon: If $G$ is presented by $langle X;|;Rrangle$, then $G/G'$ can be presented by $langle X;|;Rcup Crangle$, where $C = [x_i,x_j];$. From this it is not hard to see that, for the group $G$ of this problem, $G/G'$ contains only the elements $G', xG', (xG')^2, (xG')^3=yG'$. To see that these four elements must be distinct in the presented group, it suffices to observe that there exists a group, namely $C_4=1,g,g^2,g^3$, which has an element $x:=g$ of order $4$ and another element $y:=g^-1$ which satisfies all relations in the presentation.
– Keith Kearnes
Jul 29 at 10:01





@YipJungHon: If $G$ is presented by $langle X;|;Rrangle$, then $G/G'$ can be presented by $langle X;|;Rcup Crangle$, where $C = [x_i,x_j];$. From this it is not hard to see that, for the group $G$ of this problem, $G/G'$ contains only the elements $G', xG', (xG')^2, (xG')^3=yG'$. To see that these four elements must be distinct in the presented group, it suffices to observe that there exists a group, namely $C_4=1,g,g^2,g^3$, which has an element $x:=g$ of order $4$ and another element $y:=g^-1$ which satisfies all relations in the presentation.
– Keith Kearnes
Jul 29 at 10:01













Just curious, if I define the group slightly differently, with it being: $G=<x,y | x^4=y^4=5>,$ it will still be $cong$ to $C_4$ right?
– Yip Jung Hon
Jul 29 at 11:06




Just curious, if I define the group slightly differently, with it being: $G=<x,y | x^4=y^4=5>,$ it will still be $cong$ to $C_4$ right?
– Yip Jung Hon
Jul 29 at 11:06












@Yip Jung Hon by adding the relation xy=yx the group become abelian and hence from the last relation you have x=y and simply remove the generator y you left with x^4=1
– Rosa1
Jul 29 at 14:08




@Yip Jung Hon by adding the relation xy=yx the group become abelian and hence from the last relation you have x=y and simply remove the generator y you left with x^4=1
– Rosa1
Jul 29 at 14:08












In G/G' you only make the generators abelian
– Rosa1
Jul 29 at 14:09




In G/G' you only make the generators abelian
– Rosa1
Jul 29 at 14:09










2 Answers
2






active

oldest

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up vote
1
down vote



accepted










The relation $x^4=1$ guarantees that the order of $x$ in $G$ is at most $4$. The fact that $xG'$ generates $G/G'cong C_4$ guarantees that the order of $x$ in $G$ is at least $4$.






share|cite|improve this answer





















  • Thank you . @Keith Kearnes The relation you mentioned is saying that $ x^4 $ belong to the normal closer of R so $ x^4R = R$ Where can we say that the order must be at least 4
    – Rosa1
    Jul 29 at 10:05







  • 1




    The answer explains exactly why the order is at least $4$.
    – Derek Holt
    Jul 29 at 12:27










  • @Derek Holt but he talked about the order of x in G/G'. How this is connected to the order of x in G
    – Rosa1
    Jul 29 at 14:27







  • 1




    The order of $xG'$ in $G/G'$ is $4$, so the order of $x$ in $G$ is at least $4$.
    – Derek Holt
    Jul 29 at 15:15

















up vote
1
down vote













Well, the normal closure is even bigger than the subgroup generated by the relation. The only thing that has to be shown is that the order is, in fact, 4 and not less. Since the order divides 4, this boils down to showing that $x^2$ is not in the normal closure. Let $N$ be this normal closure, and let $R le N$ be the subgroup generated by the relations. By considering the powers of $x$ and $y$ occuring in any element of $R$, we find that $x^2 notin R$. But the normal closure is precisely the union of $fRf^-1$ ($f in Flangle x, y rangle$, the free group over $x$ and $y$), and we see that $x^2$ is not in any of these sets. Thus, $x^2$ is nonzero in your group $G$.






share|cite|improve this answer





















  • Thank you . But why you are saying that $ x^2$ is not in R can you clarify
    – Rosa1
    Jul 29 at 9:57











  • can you clarify
    – Rosa1
    Jul 29 at 10:09










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2 Answers
2






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2 Answers
2






active

oldest

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active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










The relation $x^4=1$ guarantees that the order of $x$ in $G$ is at most $4$. The fact that $xG'$ generates $G/G'cong C_4$ guarantees that the order of $x$ in $G$ is at least $4$.






share|cite|improve this answer





















  • Thank you . @Keith Kearnes The relation you mentioned is saying that $ x^4 $ belong to the normal closer of R so $ x^4R = R$ Where can we say that the order must be at least 4
    – Rosa1
    Jul 29 at 10:05







  • 1




    The answer explains exactly why the order is at least $4$.
    – Derek Holt
    Jul 29 at 12:27










  • @Derek Holt but he talked about the order of x in G/G'. How this is connected to the order of x in G
    – Rosa1
    Jul 29 at 14:27







  • 1




    The order of $xG'$ in $G/G'$ is $4$, so the order of $x$ in $G$ is at least $4$.
    – Derek Holt
    Jul 29 at 15:15














up vote
1
down vote



accepted










The relation $x^4=1$ guarantees that the order of $x$ in $G$ is at most $4$. The fact that $xG'$ generates $G/G'cong C_4$ guarantees that the order of $x$ in $G$ is at least $4$.






share|cite|improve this answer





















  • Thank you . @Keith Kearnes The relation you mentioned is saying that $ x^4 $ belong to the normal closer of R so $ x^4R = R$ Where can we say that the order must be at least 4
    – Rosa1
    Jul 29 at 10:05







  • 1




    The answer explains exactly why the order is at least $4$.
    – Derek Holt
    Jul 29 at 12:27










  • @Derek Holt but he talked about the order of x in G/G'. How this is connected to the order of x in G
    – Rosa1
    Jul 29 at 14:27







  • 1




    The order of $xG'$ in $G/G'$ is $4$, so the order of $x$ in $G$ is at least $4$.
    – Derek Holt
    Jul 29 at 15:15












up vote
1
down vote



accepted







up vote
1
down vote



accepted






The relation $x^4=1$ guarantees that the order of $x$ in $G$ is at most $4$. The fact that $xG'$ generates $G/G'cong C_4$ guarantees that the order of $x$ in $G$ is at least $4$.






share|cite|improve this answer













The relation $x^4=1$ guarantees that the order of $x$ in $G$ is at most $4$. The fact that $xG'$ generates $G/G'cong C_4$ guarantees that the order of $x$ in $G$ is at least $4$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 29 at 7:46









Keith Kearnes

5,1691626




5,1691626











  • Thank you . @Keith Kearnes The relation you mentioned is saying that $ x^4 $ belong to the normal closer of R so $ x^4R = R$ Where can we say that the order must be at least 4
    – Rosa1
    Jul 29 at 10:05







  • 1




    The answer explains exactly why the order is at least $4$.
    – Derek Holt
    Jul 29 at 12:27










  • @Derek Holt but he talked about the order of x in G/G'. How this is connected to the order of x in G
    – Rosa1
    Jul 29 at 14:27







  • 1




    The order of $xG'$ in $G/G'$ is $4$, so the order of $x$ in $G$ is at least $4$.
    – Derek Holt
    Jul 29 at 15:15
















  • Thank you . @Keith Kearnes The relation you mentioned is saying that $ x^4 $ belong to the normal closer of R so $ x^4R = R$ Where can we say that the order must be at least 4
    – Rosa1
    Jul 29 at 10:05







  • 1




    The answer explains exactly why the order is at least $4$.
    – Derek Holt
    Jul 29 at 12:27










  • @Derek Holt but he talked about the order of x in G/G'. How this is connected to the order of x in G
    – Rosa1
    Jul 29 at 14:27







  • 1




    The order of $xG'$ in $G/G'$ is $4$, so the order of $x$ in $G$ is at least $4$.
    – Derek Holt
    Jul 29 at 15:15















Thank you . @Keith Kearnes The relation you mentioned is saying that $ x^4 $ belong to the normal closer of R so $ x^4R = R$ Where can we say that the order must be at least 4
– Rosa1
Jul 29 at 10:05





Thank you . @Keith Kearnes The relation you mentioned is saying that $ x^4 $ belong to the normal closer of R so $ x^4R = R$ Where can we say that the order must be at least 4
– Rosa1
Jul 29 at 10:05





1




1




The answer explains exactly why the order is at least $4$.
– Derek Holt
Jul 29 at 12:27




The answer explains exactly why the order is at least $4$.
– Derek Holt
Jul 29 at 12:27












@Derek Holt but he talked about the order of x in G/G'. How this is connected to the order of x in G
– Rosa1
Jul 29 at 14:27





@Derek Holt but he talked about the order of x in G/G'. How this is connected to the order of x in G
– Rosa1
Jul 29 at 14:27





1




1




The order of $xG'$ in $G/G'$ is $4$, so the order of $x$ in $G$ is at least $4$.
– Derek Holt
Jul 29 at 15:15




The order of $xG'$ in $G/G'$ is $4$, so the order of $x$ in $G$ is at least $4$.
– Derek Holt
Jul 29 at 15:15










up vote
1
down vote













Well, the normal closure is even bigger than the subgroup generated by the relation. The only thing that has to be shown is that the order is, in fact, 4 and not less. Since the order divides 4, this boils down to showing that $x^2$ is not in the normal closure. Let $N$ be this normal closure, and let $R le N$ be the subgroup generated by the relations. By considering the powers of $x$ and $y$ occuring in any element of $R$, we find that $x^2 notin R$. But the normal closure is precisely the union of $fRf^-1$ ($f in Flangle x, y rangle$, the free group over $x$ and $y$), and we see that $x^2$ is not in any of these sets. Thus, $x^2$ is nonzero in your group $G$.






share|cite|improve this answer





















  • Thank you . But why you are saying that $ x^2$ is not in R can you clarify
    – Rosa1
    Jul 29 at 9:57











  • can you clarify
    – Rosa1
    Jul 29 at 10:09














up vote
1
down vote













Well, the normal closure is even bigger than the subgroup generated by the relation. The only thing that has to be shown is that the order is, in fact, 4 and not less. Since the order divides 4, this boils down to showing that $x^2$ is not in the normal closure. Let $N$ be this normal closure, and let $R le N$ be the subgroup generated by the relations. By considering the powers of $x$ and $y$ occuring in any element of $R$, we find that $x^2 notin R$. But the normal closure is precisely the union of $fRf^-1$ ($f in Flangle x, y rangle$, the free group over $x$ and $y$), and we see that $x^2$ is not in any of these sets. Thus, $x^2$ is nonzero in your group $G$.






share|cite|improve this answer





















  • Thank you . But why you are saying that $ x^2$ is not in R can you clarify
    – Rosa1
    Jul 29 at 9:57











  • can you clarify
    – Rosa1
    Jul 29 at 10:09












up vote
1
down vote










up vote
1
down vote









Well, the normal closure is even bigger than the subgroup generated by the relation. The only thing that has to be shown is that the order is, in fact, 4 and not less. Since the order divides 4, this boils down to showing that $x^2$ is not in the normal closure. Let $N$ be this normal closure, and let $R le N$ be the subgroup generated by the relations. By considering the powers of $x$ and $y$ occuring in any element of $R$, we find that $x^2 notin R$. But the normal closure is precisely the union of $fRf^-1$ ($f in Flangle x, y rangle$, the free group over $x$ and $y$), and we see that $x^2$ is not in any of these sets. Thus, $x^2$ is nonzero in your group $G$.






share|cite|improve this answer













Well, the normal closure is even bigger than the subgroup generated by the relation. The only thing that has to be shown is that the order is, in fact, 4 and not less. Since the order divides 4, this boils down to showing that $x^2$ is not in the normal closure. Let $N$ be this normal closure, and let $R le N$ be the subgroup generated by the relations. By considering the powers of $x$ and $y$ occuring in any element of $R$, we find that $x^2 notin R$. But the normal closure is precisely the union of $fRf^-1$ ($f in Flangle x, y rangle$, the free group over $x$ and $y$), and we see that $x^2$ is not in any of these sets. Thus, $x^2$ is nonzero in your group $G$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 29 at 7:39









AlgebraicsAnonymous

66611




66611











  • Thank you . But why you are saying that $ x^2$ is not in R can you clarify
    – Rosa1
    Jul 29 at 9:57











  • can you clarify
    – Rosa1
    Jul 29 at 10:09
















  • Thank you . But why you are saying that $ x^2$ is not in R can you clarify
    – Rosa1
    Jul 29 at 9:57











  • can you clarify
    – Rosa1
    Jul 29 at 10:09















Thank you . But why you are saying that $ x^2$ is not in R can you clarify
– Rosa1
Jul 29 at 9:57





Thank you . But why you are saying that $ x^2$ is not in R can you clarify
– Rosa1
Jul 29 at 9:57













can you clarify
– Rosa1
Jul 29 at 10:09




can you clarify
– Rosa1
Jul 29 at 10:09












 

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Relationship between determinant of matrix and determinant of adjoint?

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Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
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