Is it generally true that $langle a,b rangle cong langle c,d rangleRightarrow text{ either }|a|=|c|,|b|=|d|...
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Is it generally true that $langle a,b rangle cong langle c,d rangleRightarrow text{ either }|a|=|c|,|b|=|d| text{ or } |a|=|d|,|b|=|c|$?
free groups: $F_Xcong F_YRightarrow|X|=|Y|$Finding what $langle(135)(246),(12)(34)(56)ranglesubset S_{6}$ is isomorphic toHow to prove that $G/langle Mrangle/langle Nranglecong G/langle Mcup Nrangle$?If an element $c$ of order $d$ belongs to both $langle arangle$ and $langle brangle$, then $langle arangle=langle brangle=langle crangle$?Classifying the factor group $(mathbb{Z} times mathbb{Z})/langle (2, 2) rangle$Find the orders of $3 + langle 6 rangle$ and $2 + langle 6 rangle$ in $mathbb{Z}_{15}/langle 6 rangle$When is $langle x,yrangle$ equal to $langle xranglelangle xyrangle$?Tietze Transformations: $langle a, b, cmid b^2, (bc)^2rangle$ and $langle x, y, zmid y^2, z^2rangle$.The index of the normal subgroup generated by $a^3, b^2, aba^{-1}b^{-1}$ in $langle a, brangle$.What is the order of the subgroup $langle 5rangle times langle 3rangle$ in $Z_{30} times Z_{12} ?$
$begingroup$
Background: We are given two groups $G,H$ generated by two elements, say $G=langle a,brangle$ and $H=langle c,drangle$. Further suppose that the orders of $a,b,c,d$ are finite and ${|a|,|b|}neq{|c|,|d|}$. Can we conclude that $Gnotcong H$?
Motivation: I have two groups $G$ and $H$, both being non-Abelian and of order 8 (in fact, $G$ is the quaternion group and $H$ is the dihedral group of degree 4), and I know a set of generators for both of the two groups, say ${a,b}$ and ${c,d}$ correspondingly. I want to show that $G$ is not isomorphic to $H$ since $|a|=|b|=|d|=4$ but $|c|=2$, where $|a|$ is the order of $ain G$.
abstract-algebra group-theory group-presentation
edited Mar 15 at 17:52
Alex Provost
15.6k22351
asked Mar 15 at 17:40
Z WangZ Wang
645
$endgroup$
add a comment |
$begingroup$
Background: We are given two groups $G,H$ generated by two elements, say $G=langle a,brangle$ and $H=langle c,drangle$. Further suppose that the orders of $a,b,c,d$ are finite and ${|a|,|b|}neq{|c|,|d|}$. Can we conclude that $Gnotcong H$?
Motivation: I have two groups $G$ and $H$, both being non-Abelian and of order 8 (in fact, $G$ is the quaternion group and $H$ is the dihedral group of degree 4), and I know a set of generators for both of the two groups, say ${a,b}$ and ${c,d}$ correspondingly. I want to show that $G$ is not isomorphic to $H$ since $|a|=|b|=|d|=4$ but $|c|=2$, where $|a|$ is the order of $ain G$.
abstract-algebra group-theory group-presentation
edited Mar 15 at 17:52
Alex Provost
15.6k22351
asked Mar 15 at 17:40
Z WangZ Wang
645
$endgroup$
1
$begingroup$
Use$langle Xrangle$for $langle Xrangle$.
$endgroup$
– Shaun
Mar 15 at 17:47
$begingroup$
I fixed the $LaTeX$ and made the background section more readable.
$endgroup$
– Alex Provost
Mar 15 at 17:56
$begingroup$
The dihedral group of order $2n$ can always be generated by either two elements of order $2$, or one element of order $2$ and one element of order $n$. So for $ngt 2$, you get a counterexample.
$endgroup$
– Arturo Magidin
Mar 15 at 22:48
add a comment |
$begingroup$
Background: We are given two groups $G,H$ generated by two elements, say $G=langle a,brangle$ and $H=langle c,drangle$. Further suppose that the orders of $a,b,c,d$ are finite and ${|a|,|b|}neq{|c|,|d|}$. Can we conclude that $Gnotcong H$?
Motivation: I have two groups $G$ and $H$, both being non-Abelian and of order 8 (in fact, $G$ is the quaternion group and $H$ is the dihedral group of degree 4), and I know a set of generators for both of the two groups, say ${a,b}$ and ${c,d}$ correspondingly. I want to show that $G$ is not isomorphic to $H$ since $|a|=|b|=|d|=4$ but $|c|=2$, where $|a|$ is the order of $ain G$.
abstract-algebra group-theory group-presentation
edited Mar 15 at 17:52
Alex Provost
15.6k22351
asked Mar 15 at 17:40
Z WangZ Wang
645
$endgroup$
Background: We are given two groups $G,H$ generated by two elements, say $G=langle a,brangle$ and $H=langle c,drangle$. Further suppose that the orders of $a,b,c,d$ are finite and ${|a|,|b|}neq{|c|,|d|}$. Can we conclude that $Gnotcong H$?
Motivation: I have two groups $G$ and $H$, both being non-Abelian and of order 8 (in fact, $G$ is the quaternion group and $H$ is the dihedral group of degree 4), and I know a set of generators for both of the two groups, say ${a,b}$ and ${c,d}$ correspondingly. I want to show that $G$ is not isomorphic to $H$ since $|a|=|b|=|d|=4$ but $|c|=2$, where $|a|$ is the order of $ain G$.
abstract-algebra group-theory group-presentation
abstract-algebra group-theory group-presentation
edited Mar 15 at 17:52
Alex Provost
15.6k22351
asked Mar 15 at 17:40
Z WangZ Wang
645
edited Mar 15 at 17:52
Alex Provost
15.6k22351
asked Mar 15 at 17:40
Z WangZ Wang
645
edited Mar 15 at 17:52
Alex Provost
15.6k22351
edited Mar 15 at 17:52
Alex Provost
15.6k22351
edited Mar 15 at 17:52
Alex Provost
15.6k22351
15.6k22351
asked Mar 15 at 17:40
Z WangZ Wang
645
asked Mar 15 at 17:40
Z WangZ Wang
645
asked Mar 15 at 17:40
Z WangZ Wang
645
645
1
$begingroup$
Use$langle Xrangle$for $langle Xrangle$.
$endgroup$
– Shaun
Mar 15 at 17:47
$begingroup$
I fixed the $LaTeX$ and made the background section more readable.
$endgroup$
– Alex Provost
Mar 15 at 17:56
$begingroup$
The dihedral group of order $2n$ can always be generated by either two elements of order $2$, or one element of order $2$ and one element of order $n$. So for $ngt 2$, you get a counterexample.
$endgroup$
– Arturo Magidin
Mar 15 at 22:48
add a comment |
1
$begingroup$
Use$langle Xrangle$for $langle Xrangle$.
$endgroup$
– Shaun
Mar 15 at 17:47
$begingroup$
I fixed the $LaTeX$ and made the background section more readable.
$endgroup$
– Alex Provost
Mar 15 at 17:56
$begingroup$
The dihedral group of order $2n$ can always be generated by either two elements of order $2$, or one element of order $2$ and one element of order $n$. So for $ngt 2$, you get a counterexample.
$endgroup$
– Arturo Magidin
Mar 15 at 22:48
1
1
$begingroup$
Use
$langle Xrangle$ for $langle Xrangle$.$endgroup$
– Shaun
Mar 15 at 17:47
$begingroup$
Use
$langle Xrangle$ for $langle Xrangle$.$endgroup$
– Shaun
Mar 15 at 17:47
$begingroup$
I fixed the $LaTeX$ and made the background section more readable.
$endgroup$
– Alex Provost
Mar 15 at 17:56
$begingroup$
I fixed the $LaTeX$ and made the background section more readable.
$endgroup$
– Alex Provost
Mar 15 at 17:56
$begingroup$
The dihedral group of order $2n$ can always be generated by either two elements of order $2$, or one element of order $2$ and one element of order $n$. So for $ngt 2$, you get a counterexample.
$endgroup$
– Arturo Magidin
Mar 15 at 22:48
$begingroup$
The dihedral group of order $2n$ can always be generated by either two elements of order $2$, or one element of order $2$ and one element of order $n$. So for $ngt 2$, you get a counterexample.
$endgroup$
– Arturo Magidin
Mar 15 at 22:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, take $G=H=mathbb{Z}_4$ which has two sets of generators $langle 1,2rangle$ and $langle 1, 3rangle$. They satisfy the assumption but the order equalities don't follow.
For non-cyclic case note that if $langle a, brangle$ generates a group then so does $langle a, abrangle$. Now for any integers $n,m,r>1$ there is a (finite) group $G$ and elements $a,bin G$ such that $|a|=n$, $|b|=m$ and $|ab|=r$. For details see Theorem 1.64 here. That gives you more counterexamples.
answered Mar 15 at 17:46
freakishfreakish
12.9k1631
$endgroup$
$begingroup$
Did you mean to write "which do satisfy your conditions" ?
$endgroup$
– Lee Mosher
Mar 15 at 17:47
add a comment |
Your Answer
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$begingroup$
No, take $G=H=mathbb{Z}_4$ which has two sets of generators $langle 1,2rangle$ and $langle 1, 3rangle$. They satisfy the assumption but the order equalities don't follow.
For non-cyclic case note that if $langle a, brangle$ generates a group then so does $langle a, abrangle$. Now for any integers $n,m,r>1$ there is a (finite) group $G$ and elements $a,bin G$ such that $|a|=n$, $|b|=m$ and $|ab|=r$. For details see Theorem 1.64 here. That gives you more counterexamples.
answered Mar 15 at 17:46
freakishfreakish
12.9k1631
$endgroup$
$begingroup$
Did you mean to write "which do satisfy your conditions" ?
$endgroup$
– Lee Mosher
Mar 15 at 17:47
add a comment |
$begingroup$
No, take $G=H=mathbb{Z}_4$ which has two sets of generators $langle 1,2rangle$ and $langle 1, 3rangle$. They satisfy the assumption but the order equalities don't follow.
For non-cyclic case note that if $langle a, brangle$ generates a group then so does $langle a, abrangle$. Now for any integers $n,m,r>1$ there is a (finite) group $G$ and elements $a,bin G$ such that $|a|=n$, $|b|=m$ and $|ab|=r$. For details see Theorem 1.64 here. That gives you more counterexamples.
answered Mar 15 at 17:46
freakishfreakish
12.9k1631
$endgroup$
$begingroup$
Did you mean to write "which do satisfy your conditions" ?
$endgroup$
– Lee Mosher
Mar 15 at 17:47
add a comment |
$begingroup$
No, take $G=H=mathbb{Z}_4$ which has two sets of generators $langle 1,2rangle$ and $langle 1, 3rangle$. They satisfy the assumption but the order equalities don't follow.
For non-cyclic case note that if $langle a, brangle$ generates a group then so does $langle a, abrangle$. Now for any integers $n,m,r>1$ there is a (finite) group $G$ and elements $a,bin G$ such that $|a|=n$, $|b|=m$ and $|ab|=r$. For details see Theorem 1.64 here. That gives you more counterexamples.
answered Mar 15 at 17:46
freakishfreakish
12.9k1631
$endgroup$
No, take $G=H=mathbb{Z}_4$ which has two sets of generators $langle 1,2rangle$ and $langle 1, 3rangle$. They satisfy the assumption but the order equalities don't follow.
For non-cyclic case note that if $langle a, brangle$ generates a group then so does $langle a, abrangle$. Now for any integers $n,m,r>1$ there is a (finite) group $G$ and elements $a,bin G$ such that $|a|=n$, $|b|=m$ and $|ab|=r$. For details see Theorem 1.64 here. That gives you more counterexamples.
answered Mar 15 at 17:46
freakishfreakish
12.9k1631
edited Mar 15 at 17:53
answered Mar 15 at 17:46
freakishfreakish
12.9k1631
answered Mar 15 at 17:46
freakishfreakish
12.9k1631
answered Mar 15 at 17:46
freakishfreakish
12.9k1631
12.9k1631
$begingroup$
Did you mean to write "which do satisfy your conditions" ?
$endgroup$
– Lee Mosher
Mar 15 at 17:47
add a comment |
$begingroup$
Did you mean to write "which do satisfy your conditions" ?
$endgroup$
– Lee Mosher
Mar 15 at 17:47
$begingroup$
Did you mean to write "which do satisfy your conditions" ?
$endgroup$
– Lee Mosher
Mar 15 at 17:47
$begingroup$
Did you mean to write "which do satisfy your conditions" ?
$endgroup$
– Lee Mosher
Mar 15 at 17:47
add a comment |
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1
$begingroup$
Use
$langle Xrangle$for $langle Xrangle$.$endgroup$
– Shaun
Mar 15 at 17:47
$begingroup$
I fixed the $LaTeX$ and made the background section more readable.
$endgroup$
– Alex Provost
Mar 15 at 17:56
$begingroup$
The dihedral group of order $2n$ can always be generated by either two elements of order $2$, or one element of order $2$ and one element of order $n$. So for $ngt 2$, you get a counterexample.
$endgroup$
– Arturo Magidin
Mar 15 at 22:48