Mixing
Example of three-generator abélien by cyclic
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Can you give me example of three-generator group abelien by cyclic(i.e there exist normal subgroup $N$ in $G$ abelien and $G/N$ is cyclic) which is not finite by nilpotent (i.e there isn't finite subgroup$ M$ in $G$ such that $G/M$ is nilpotent) but has each of its two-generator subnormal subgroups abelien.
for example let $A$ be free abelien group of rank $2$ generated by $a$,$b$ and let $x$ be the automorphism of order 2 which invert evry element of $A$ ,let $G$ be split extension of A by$<x>$ .
subnormal subgroups if There is a finite ascending chain of subgroups starting from the subgroup and going till the whole group.
group-theory finite-groups
asked Mar 20 '15 at 17:08
user220373user220373
1177
$endgroup$
|
show 1 more comment
$begingroup$
Can you give me example of three-generator group abelien by cyclic(i.e there exist normal subgroup $N$ in $G$ abelien and $G/N$ is cyclic) which is not finite by nilpotent (i.e there isn't finite subgroup$ M$ in $G$ such that $G/M$ is nilpotent) but has each of its two-generator subnormal subgroups abelien.
for example let $A$ be free abelien group of rank $2$ generated by $a$,$b$ and let $x$ be the automorphism of order 2 which invert evry element of $A$ ,let $G$ be split extension of A by$<x>$ .
subnormal subgroups if There is a finite ascending chain of subgroups starting from the subgroup and going till the whole group.
group-theory finite-groups
asked Mar 20 '15 at 17:08
user220373user220373
1177
$endgroup$
$begingroup$
Please supply more context.
$endgroup$
– Derek Holt
Mar 20 '15 at 19:39
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@DerekHolt iwill supply more context
$endgroup$
– user220373
Mar 21 '15 at 14:06
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I think the example you mention works. What is the problem? Are you having difficulty proving that it satisfies one of the required properties?
$endgroup$
– Derek Holt
Mar 22 '15 at 14:08
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@DerekHolt yes exactly
$endgroup$
– user220373
Mar 22 '15 at 14:11
$begingroup$
So which property are you having difficulty proving? Surely it is clearly abelian by cyclic?
$endgroup$
– Derek Holt
Mar 22 '15 at 16:02
|
show 1 more comment
$begingroup$
Can you give me example of three-generator group abelien by cyclic(i.e there exist normal subgroup $N$ in $G$ abelien and $G/N$ is cyclic) which is not finite by nilpotent (i.e there isn't finite subgroup$ M$ in $G$ such that $G/M$ is nilpotent) but has each of its two-generator subnormal subgroups abelien.
for example let $A$ be free abelien group of rank $2$ generated by $a$,$b$ and let $x$ be the automorphism of order 2 which invert evry element of $A$ ,let $G$ be split extension of A by$<x>$ .
subnormal subgroups if There is a finite ascending chain of subgroups starting from the subgroup and going till the whole group.
group-theory finite-groups
asked Mar 20 '15 at 17:08
user220373user220373
1177
$endgroup$
Can you give me example of three-generator group abelien by cyclic(i.e there exist normal subgroup $N$ in $G$ abelien and $G/N$ is cyclic) which is not finite by nilpotent (i.e there isn't finite subgroup$ M$ in $G$ such that $G/M$ is nilpotent) but has each of its two-generator subnormal subgroups abelien.
for example let $A$ be free abelien group of rank $2$ generated by $a$,$b$ and let $x$ be the automorphism of order 2 which invert evry element of $A$ ,let $G$ be split extension of A by$<x>$ .
subnormal subgroups if There is a finite ascending chain of subgroups starting from the subgroup and going till the whole group.
group-theory finite-groups
group-theory finite-groups
asked Mar 20 '15 at 17:08
user220373user220373
1177
asked Mar 20 '15 at 17:08
user220373user220373
1177
edited Mar 21 '15 at 14:12
user220373
asked Mar 20 '15 at 17:08
user220373user220373
1177
asked Mar 20 '15 at 17:08
user220373user220373
1177
asked Mar 20 '15 at 17:08
user220373user220373
1177
1177
$begingroup$
Please supply more context.
$endgroup$
– Derek Holt
Mar 20 '15 at 19:39
$begingroup$
@DerekHolt iwill supply more context
$endgroup$
– user220373
Mar 21 '15 at 14:06
$begingroup$
I think the example you mention works. What is the problem? Are you having difficulty proving that it satisfies one of the required properties?
$endgroup$
– Derek Holt
Mar 22 '15 at 14:08
$begingroup$
@DerekHolt yes exactly
$endgroup$
– user220373
Mar 22 '15 at 14:11
$begingroup$
So which property are you having difficulty proving? Surely it is clearly abelian by cyclic?
$endgroup$
– Derek Holt
Mar 22 '15 at 16:02
|
show 1 more comment
$begingroup$
Please supply more context.
$endgroup$
– Derek Holt
Mar 20 '15 at 19:39
$begingroup$
@DerekHolt iwill supply more context
$endgroup$
– user220373
Mar 21 '15 at 14:06
$begingroup$
I think the example you mention works. What is the problem? Are you having difficulty proving that it satisfies one of the required properties?
$endgroup$
– Derek Holt
Mar 22 '15 at 14:08
$begingroup$
@DerekHolt yes exactly
$endgroup$
– user220373
Mar 22 '15 at 14:11
$begingroup$
So which property are you having difficulty proving? Surely it is clearly abelian by cyclic?
$endgroup$
– Derek Holt
Mar 22 '15 at 16:02
$begingroup$
Please supply more context.
$endgroup$
– Derek Holt
Mar 20 '15 at 19:39
$begingroup$
Please supply more context.
$endgroup$
– Derek Holt
Mar 20 '15 at 19:39
$begingroup$
@DerekHolt iwill supply more context
$endgroup$
– user220373
Mar 21 '15 at 14:06
$begingroup$
@DerekHolt iwill supply more context
$endgroup$
– user220373
Mar 21 '15 at 14:06
$begingroup$
I think the example you mention works. What is the problem? Are you having difficulty proving that it satisfies one of the required properties?
$endgroup$
– Derek Holt
Mar 22 '15 at 14:08
$begingroup$
I think the example you mention works. What is the problem? Are you having difficulty proving that it satisfies one of the required properties?
$endgroup$
– Derek Holt
Mar 22 '15 at 14:08
$begingroup$
@DerekHolt yes exactly
$endgroup$
– user220373
Mar 22 '15 at 14:11
$begingroup$
@DerekHolt yes exactly
$endgroup$
– user220373
Mar 22 '15 at 14:11
$begingroup$
So which property are you having difficulty proving? Surely it is clearly abelian by cyclic?
$endgroup$
– Derek Holt
Mar 22 '15 at 16:02
$begingroup$
So which property are you having difficulty proving? Surely it is clearly abelian by cyclic?
$endgroup$
– Derek Holt
Mar 22 '15 at 16:02
|
show 1 more comment
1 Answer
1
active
oldest
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$begingroup$
We want to prove that every subnormal subgroup of the group $G=langle a,b,t mid ab=ba, t^2=1, a^t=a^{-1},b^t=b^{-1} rangle$ is abelian.
Let $H$ be a nonabelian $2$-generator subgroup of $G$. Then $H$ contains some element outside of $langle a,b rangle$, and since all such elements have order $2$ and are equivalent under an automorphism of $G$, we can assume that $t in H$. We can assume also that the other generator lies in $langle a,b rangle$, so $H = langle t,a^ib^j rangle$ for some $i,j in {mathbb Z}$.
We need to show that $H$ is not subnormal in $G$. Let $N = N_G(H)$. If $a^kb^l in N$, then $a^{-k}b^{-l}ta^kb^l = t a^{2k}b^{2l} in H$, so $a^{2k}b^{2l} in H$. Hence $|N:H| le 2$ and $N cap langle a,b rangle$ is still cyclic. So, if $|N:H| = 2$ we can replace $H$ by $N$ and calculate its normalizer. Repeating this process, the ascending chain of normalizers must eventually stabilize with a $2$-generator subgroup $N'$ with $N' cap langle a,b rangle$ cyclic. So $H$ is not subnormal in $G$.
answered Mar 22 '15 at 19:48
Derek HoltDerek Holt
53.7k53571
$endgroup$
$begingroup$
I don't understand something in the last paragraph: $N$ is the normalizer of $H$, so $N$ contains $H$, and $Ncap H = H$. Then it is not cyclic. What is going on?
$endgroup$
– Ben Blum-Smith
Jan 13 at 22:26
2
$begingroup$
I meant $N cap langle a,b rangle$. I have corrected this, as well as correcting the problem that $x$ unaccountably changed into $t$ halfway through the answer.
$endgroup$
– Derek Holt
Jan 14 at 8:52
add a comment |
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$begingroup$
We want to prove that every subnormal subgroup of the group $G=langle a,b,t mid ab=ba, t^2=1, a^t=a^{-1},b^t=b^{-1} rangle$ is abelian.
Let $H$ be a nonabelian $2$-generator subgroup of $G$. Then $H$ contains some element outside of $langle a,b rangle$, and since all such elements have order $2$ and are equivalent under an automorphism of $G$, we can assume that $t in H$. We can assume also that the other generator lies in $langle a,b rangle$, so $H = langle t,a^ib^j rangle$ for some $i,j in {mathbb Z}$.
We need to show that $H$ is not subnormal in $G$. Let $N = N_G(H)$. If $a^kb^l in N$, then $a^{-k}b^{-l}ta^kb^l = t a^{2k}b^{2l} in H$, so $a^{2k}b^{2l} in H$. Hence $|N:H| le 2$ and $N cap langle a,b rangle$ is still cyclic. So, if $|N:H| = 2$ we can replace $H$ by $N$ and calculate its normalizer. Repeating this process, the ascending chain of normalizers must eventually stabilize with a $2$-generator subgroup $N'$ with $N' cap langle a,b rangle$ cyclic. So $H$ is not subnormal in $G$.
answered Mar 22 '15 at 19:48
Derek HoltDerek Holt
53.7k53571
$endgroup$
$begingroup$
I don't understand something in the last paragraph: $N$ is the normalizer of $H$, so $N$ contains $H$, and $Ncap H = H$. Then it is not cyclic. What is going on?
$endgroup$
– Ben Blum-Smith
Jan 13 at 22:26
2
$begingroup$
I meant $N cap langle a,b rangle$. I have corrected this, as well as correcting the problem that $x$ unaccountably changed into $t$ halfway through the answer.
$endgroup$
– Derek Holt
Jan 14 at 8:52
add a comment |
$begingroup$
We want to prove that every subnormal subgroup of the group $G=langle a,b,t mid ab=ba, t^2=1, a^t=a^{-1},b^t=b^{-1} rangle$ is abelian.
Let $H$ be a nonabelian $2$-generator subgroup of $G$. Then $H$ contains some element outside of $langle a,b rangle$, and since all such elements have order $2$ and are equivalent under an automorphism of $G$, we can assume that $t in H$. We can assume also that the other generator lies in $langle a,b rangle$, so $H = langle t,a^ib^j rangle$ for some $i,j in {mathbb Z}$.
We need to show that $H$ is not subnormal in $G$. Let $N = N_G(H)$. If $a^kb^l in N$, then $a^{-k}b^{-l}ta^kb^l = t a^{2k}b^{2l} in H$, so $a^{2k}b^{2l} in H$. Hence $|N:H| le 2$ and $N cap langle a,b rangle$ is still cyclic. So, if $|N:H| = 2$ we can replace $H$ by $N$ and calculate its normalizer. Repeating this process, the ascending chain of normalizers must eventually stabilize with a $2$-generator subgroup $N'$ with $N' cap langle a,b rangle$ cyclic. So $H$ is not subnormal in $G$.
answered Mar 22 '15 at 19:48
Derek HoltDerek Holt
53.7k53571
$endgroup$
$begingroup$
I don't understand something in the last paragraph: $N$ is the normalizer of $H$, so $N$ contains $H$, and $Ncap H = H$. Then it is not cyclic. What is going on?
$endgroup$
– Ben Blum-Smith
Jan 13 at 22:26
2
$begingroup$
I meant $N cap langle a,b rangle$. I have corrected this, as well as correcting the problem that $x$ unaccountably changed into $t$ halfway through the answer.
$endgroup$
– Derek Holt
Jan 14 at 8:52
add a comment |
$begingroup$
We want to prove that every subnormal subgroup of the group $G=langle a,b,t mid ab=ba, t^2=1, a^t=a^{-1},b^t=b^{-1} rangle$ is abelian.
Let $H$ be a nonabelian $2$-generator subgroup of $G$. Then $H$ contains some element outside of $langle a,b rangle$, and since all such elements have order $2$ and are equivalent under an automorphism of $G$, we can assume that $t in H$. We can assume also that the other generator lies in $langle a,b rangle$, so $H = langle t,a^ib^j rangle$ for some $i,j in {mathbb Z}$.
We need to show that $H$ is not subnormal in $G$. Let $N = N_G(H)$. If $a^kb^l in N$, then $a^{-k}b^{-l}ta^kb^l = t a^{2k}b^{2l} in H$, so $a^{2k}b^{2l} in H$. Hence $|N:H| le 2$ and $N cap langle a,b rangle$ is still cyclic. So, if $|N:H| = 2$ we can replace $H$ by $N$ and calculate its normalizer. Repeating this process, the ascending chain of normalizers must eventually stabilize with a $2$-generator subgroup $N'$ with $N' cap langle a,b rangle$ cyclic. So $H$ is not subnormal in $G$.
answered Mar 22 '15 at 19:48
Derek HoltDerek Holt
53.7k53571
$endgroup$
We want to prove that every subnormal subgroup of the group $G=langle a,b,t mid ab=ba, t^2=1, a^t=a^{-1},b^t=b^{-1} rangle$ is abelian.
Let $H$ be a nonabelian $2$-generator subgroup of $G$. Then $H$ contains some element outside of $langle a,b rangle$, and since all such elements have order $2$ and are equivalent under an automorphism of $G$, we can assume that $t in H$. We can assume also that the other generator lies in $langle a,b rangle$, so $H = langle t,a^ib^j rangle$ for some $i,j in {mathbb Z}$.
We need to show that $H$ is not subnormal in $G$. Let $N = N_G(H)$. If $a^kb^l in N$, then $a^{-k}b^{-l}ta^kb^l = t a^{2k}b^{2l} in H$, so $a^{2k}b^{2l} in H$. Hence $|N:H| le 2$ and $N cap langle a,b rangle$ is still cyclic. So, if $|N:H| = 2$ we can replace $H$ by $N$ and calculate its normalizer. Repeating this process, the ascending chain of normalizers must eventually stabilize with a $2$-generator subgroup $N'$ with $N' cap langle a,b rangle$ cyclic. So $H$ is not subnormal in $G$.
answered Mar 22 '15 at 19:48
Derek HoltDerek Holt
53.7k53571
edited Jan 14 at 8:51
answered Mar 22 '15 at 19:48
Derek HoltDerek Holt
53.7k53571
answered Mar 22 '15 at 19:48
Derek HoltDerek Holt
53.7k53571
answered Mar 22 '15 at 19:48
Derek HoltDerek Holt
53.7k53571
53.7k53571
$begingroup$
I don't understand something in the last paragraph: $N$ is the normalizer of $H$, so $N$ contains $H$, and $Ncap H = H$. Then it is not cyclic. What is going on?
$endgroup$
– Ben Blum-Smith
Jan 13 at 22:26
2
$begingroup$
I meant $N cap langle a,b rangle$. I have corrected this, as well as correcting the problem that $x$ unaccountably changed into $t$ halfway through the answer.
$endgroup$
– Derek Holt
Jan 14 at 8:52
add a comment |
$begingroup$
I don't understand something in the last paragraph: $N$ is the normalizer of $H$, so $N$ contains $H$, and $Ncap H = H$. Then it is not cyclic. What is going on?
$endgroup$
– Ben Blum-Smith
Jan 13 at 22:26
2
$begingroup$
I meant $N cap langle a,b rangle$. I have corrected this, as well as correcting the problem that $x$ unaccountably changed into $t$ halfway through the answer.
$endgroup$
– Derek Holt
Jan 14 at 8:52
$begingroup$
I don't understand something in the last paragraph: $N$ is the normalizer of $H$, so $N$ contains $H$, and $Ncap H = H$. Then it is not cyclic. What is going on?
$endgroup$
– Ben Blum-Smith
Jan 13 at 22:26
$begingroup$
I don't understand something in the last paragraph: $N$ is the normalizer of $H$, so $N$ contains $H$, and $Ncap H = H$. Then it is not cyclic. What is going on?
$endgroup$
– Ben Blum-Smith
Jan 13 at 22:26
2
2
$begingroup$
I meant $N cap langle a,b rangle$. I have corrected this, as well as correcting the problem that $x$ unaccountably changed into $t$ halfway through the answer.
$endgroup$
– Derek Holt
Jan 14 at 8:52
$begingroup$
I meant $N cap langle a,b rangle$. I have corrected this, as well as correcting the problem that $x$ unaccountably changed into $t$ halfway through the answer.
$endgroup$
– Derek Holt
Jan 14 at 8:52
add a comment |
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$begingroup$
Please supply more context.
$endgroup$
– Derek Holt
Mar 20 '15 at 19:39
$begingroup$
@DerekHolt iwill supply more context
$endgroup$
– user220373
Mar 21 '15 at 14:06
$begingroup$
I think the example you mention works. What is the problem? Are you having difficulty proving that it satisfies one of the required properties?
$endgroup$
– Derek Holt
Mar 22 '15 at 14:08
$begingroup$
@DerekHolt yes exactly
$endgroup$
– user220373
Mar 22 '15 at 14:11
$begingroup$
So which property are you having difficulty proving? Surely it is clearly abelian by cyclic?
$endgroup$
– Derek Holt
Mar 22 '15 at 16:02