Mixing
Subgroups needing more generators in number
2
$begingroup$
Let G be an odd order group such that [G,G] is central. If G is a 2-generator group, can it have a 3-generator subgroup?
This can't happen if G is abelian.
I tried to go for p-groups but their classification wasn't very helpful.
Next, I looked for the smallest order but I didn't pass 33.
Thanks in advance.
abstract-algebra group-theory finite-groups
asked Jan 9 at 17:55
CharlesACharlesA
111
$endgroup$
|
show 3 more comments
2
$begingroup$
Let G be an odd order group such that [G,G] is central. If G is a 2-generator group, can it have a 3-generator subgroup?
This can't happen if G is abelian.
I tried to go for p-groups but their classification wasn't very helpful.
Next, I looked for the smallest order but I didn't pass 33.
Thanks in advance.
abstract-algebra group-theory finite-groups
asked Jan 9 at 17:55
CharlesACharlesA
111
$endgroup$
1
$begingroup$
Let $G=langle x,y,zmid x^9=y^9=z^9=1, xz=zx, yz=zy, yx=xyzrangle$. Then look at $H=langle x^3,y^3,zrangle$.
$endgroup$
– Arturo Magidin
Jan 9 at 18:10
1
$begingroup$
Smallest group is probably of order $p^4$: $langle x,y,zmid x^{p^2}=y^p=z^p=1, xz=zx, yz=zy, yx=xyzrangle$, with $H=langle x^p,y,zrangle$.
$endgroup$
– Arturo Magidin
Jan 9 at 18:15
$begingroup$
I'm not very good with generators and relations, but I can more or less see that your example is correct. Is the given group a semidirect product or a matrix group? Why that group has order p^4? Why do you think the smallest overall example has order 81? Thanks!
$endgroup$
– CharlesA
Jan 9 at 18:48
$begingroup$
I believe that you can realize the first group as a group of $3times 3$ upper triangular matrices with $1$s in the diagonal and entries in $mathbb{Z}/9mathbb{Z}$, with $x$ corresponding to $left(begin{array}{ccc}1&1&0\0&1&0\0&0&1end{array}right)$, $y$ to $left(begin{array}{ccc}1&0&0\0&1&1\0&0&1end{array}right)$, and $z=left(begin{array}{ccc}1&0&1\0&1&0\0&0&1end{array}right)$. For the second group, take the $(1,2)$ entry in $mathbb{Z}/p^2mathbb{Z}$, and the $(1,3)$ and $(2,3)$ entries in $mathbb{Z}/pmathbb{Z}$.
$endgroup$
– Arturo Magidin
Jan 9 at 19:15
$begingroup$
Groups of order $p^2$ are abelian, so they don't give you the example you want. The nonabelian groups of order $p^3$ don't have "enough room" for an example: proper subgroups are abelian of order at most $p^2$, hence $2$-generated or cyclic. My second example is of order $p^4$, hence that's the smallest. It has order $p^4$ because every element can be written as $x^ay^bz^c$ with $0leq alt p^2$, $0leq blt p$, $0leq clt p$, so you have $p^2pp=p^4$ elements.
$endgroup$
– Arturo Magidin
Jan 9 at 19:17
|
show 3 more comments
2
2
2
$begingroup$
Let G be an odd order group such that [G,G] is central. If G is a 2-generator group, can it have a 3-generator subgroup?
This can't happen if G is abelian.
I tried to go for p-groups but their classification wasn't very helpful.
Next, I looked for the smallest order but I didn't pass 33.
Thanks in advance.
abstract-algebra group-theory finite-groups
asked Jan 9 at 17:55
CharlesACharlesA
111
$endgroup$
Let G be an odd order group such that [G,G] is central. If G is a 2-generator group, can it have a 3-generator subgroup?
This can't happen if G is abelian.
I tried to go for p-groups but their classification wasn't very helpful.
Next, I looked for the smallest order but I didn't pass 33.
Thanks in advance.
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
asked Jan 9 at 17:55
CharlesACharlesA
111
asked Jan 9 at 17:55
CharlesACharlesA
111
asked Jan 9 at 17:55
CharlesACharlesA
111
asked Jan 9 at 17:55
CharlesACharlesA
111
asked Jan 9 at 17:55
CharlesACharlesA
111
111
1
$begingroup$
Let $G=langle x,y,zmid x^9=y^9=z^9=1, xz=zx, yz=zy, yx=xyzrangle$. Then look at $H=langle x^3,y^3,zrangle$.
$endgroup$
– Arturo Magidin
Jan 9 at 18:10
1
$begingroup$
Smallest group is probably of order $p^4$: $langle x,y,zmid x^{p^2}=y^p=z^p=1, xz=zx, yz=zy, yx=xyzrangle$, with $H=langle x^p,y,zrangle$.
$endgroup$
– Arturo Magidin
Jan 9 at 18:15
$begingroup$
I'm not very good with generators and relations, but I can more or less see that your example is correct. Is the given group a semidirect product or a matrix group? Why that group has order p^4? Why do you think the smallest overall example has order 81? Thanks!
$endgroup$
– CharlesA
Jan 9 at 18:48
$begingroup$
I believe that you can realize the first group as a group of $3times 3$ upper triangular matrices with $1$s in the diagonal and entries in $mathbb{Z}/9mathbb{Z}$, with $x$ corresponding to $left(begin{array}{ccc}1&1&0\0&1&0\0&0&1end{array}right)$, $y$ to $left(begin{array}{ccc}1&0&0\0&1&1\0&0&1end{array}right)$, and $z=left(begin{array}{ccc}1&0&1\0&1&0\0&0&1end{array}right)$. For the second group, take the $(1,2)$ entry in $mathbb{Z}/p^2mathbb{Z}$, and the $(1,3)$ and $(2,3)$ entries in $mathbb{Z}/pmathbb{Z}$.
$endgroup$
– Arturo Magidin
Jan 9 at 19:15
$begingroup$
Groups of order $p^2$ are abelian, so they don't give you the example you want. The nonabelian groups of order $p^3$ don't have "enough room" for an example: proper subgroups are abelian of order at most $p^2$, hence $2$-generated or cyclic. My second example is of order $p^4$, hence that's the smallest. It has order $p^4$ because every element can be written as $x^ay^bz^c$ with $0leq alt p^2$, $0leq blt p$, $0leq clt p$, so you have $p^2pp=p^4$ elements.
$endgroup$
– Arturo Magidin
Jan 9 at 19:17
|
show 3 more comments
1
$begingroup$
Let $G=langle x,y,zmid x^9=y^9=z^9=1, xz=zx, yz=zy, yx=xyzrangle$. Then look at $H=langle x^3,y^3,zrangle$.
$endgroup$
– Arturo Magidin
Jan 9 at 18:10
1
$begingroup$
Smallest group is probably of order $p^4$: $langle x,y,zmid x^{p^2}=y^p=z^p=1, xz=zx, yz=zy, yx=xyzrangle$, with $H=langle x^p,y,zrangle$.
$endgroup$
– Arturo Magidin
Jan 9 at 18:15
$begingroup$
I'm not very good with generators and relations, but I can more or less see that your example is correct. Is the given group a semidirect product or a matrix group? Why that group has order p^4? Why do you think the smallest overall example has order 81? Thanks!
$endgroup$
– CharlesA
Jan 9 at 18:48
$begingroup$
I believe that you can realize the first group as a group of $3times 3$ upper triangular matrices with $1$s in the diagonal and entries in $mathbb{Z}/9mathbb{Z}$, with $x$ corresponding to $left(begin{array}{ccc}1&1&0\0&1&0\0&0&1end{array}right)$, $y$ to $left(begin{array}{ccc}1&0&0\0&1&1\0&0&1end{array}right)$, and $z=left(begin{array}{ccc}1&0&1\0&1&0\0&0&1end{array}right)$. For the second group, take the $(1,2)$ entry in $mathbb{Z}/p^2mathbb{Z}$, and the $(1,3)$ and $(2,3)$ entries in $mathbb{Z}/pmathbb{Z}$.
$endgroup$
– Arturo Magidin
Jan 9 at 19:15
$begingroup$
Groups of order $p^2$ are abelian, so they don't give you the example you want. The nonabelian groups of order $p^3$ don't have "enough room" for an example: proper subgroups are abelian of order at most $p^2$, hence $2$-generated or cyclic. My second example is of order $p^4$, hence that's the smallest. It has order $p^4$ because every element can be written as $x^ay^bz^c$ with $0leq alt p^2$, $0leq blt p$, $0leq clt p$, so you have $p^2pp=p^4$ elements.
$endgroup$
– Arturo Magidin
Jan 9 at 19:17
1
1
$begingroup$
Let $G=langle x,y,zmid x^9=y^9=z^9=1, xz=zx, yz=zy, yx=xyzrangle$. Then look at $H=langle x^3,y^3,zrangle$.
$endgroup$
– Arturo Magidin
Jan 9 at 18:10
$begingroup$
Let $G=langle x,y,zmid x^9=y^9=z^9=1, xz=zx, yz=zy, yx=xyzrangle$. Then look at $H=langle x^3,y^3,zrangle$.
$endgroup$
– Arturo Magidin
Jan 9 at 18:10
1
1
$begingroup$
Smallest group is probably of order $p^4$: $langle x,y,zmid x^{p^2}=y^p=z^p=1, xz=zx, yz=zy, yx=xyzrangle$, with $H=langle x^p,y,zrangle$.
$endgroup$
– Arturo Magidin
Jan 9 at 18:15
$begingroup$
Smallest group is probably of order $p^4$: $langle x,y,zmid x^{p^2}=y^p=z^p=1, xz=zx, yz=zy, yx=xyzrangle$, with $H=langle x^p,y,zrangle$.
$endgroup$
– Arturo Magidin
Jan 9 at 18:15
$begingroup$
I'm not very good with generators and relations, but I can more or less see that your example is correct. Is the given group a semidirect product or a matrix group? Why that group has order p^4? Why do you think the smallest overall example has order 81? Thanks!
$endgroup$
– CharlesA
Jan 9 at 18:48
$begingroup$
I'm not very good with generators and relations, but I can more or less see that your example is correct. Is the given group a semidirect product or a matrix group? Why that group has order p^4? Why do you think the smallest overall example has order 81? Thanks!
$endgroup$
– CharlesA
Jan 9 at 18:48
$begingroup$
I believe that you can realize the first group as a group of $3times 3$ upper triangular matrices with $1$s in the diagonal and entries in $mathbb{Z}/9mathbb{Z}$, with $x$ corresponding to $left(begin{array}{ccc}1&1&0\0&1&0\0&0&1end{array}right)$, $y$ to $left(begin{array}{ccc}1&0&0\0&1&1\0&0&1end{array}right)$, and $z=left(begin{array}{ccc}1&0&1\0&1&0\0&0&1end{array}right)$. For the second group, take the $(1,2)$ entry in $mathbb{Z}/p^2mathbb{Z}$, and the $(1,3)$ and $(2,3)$ entries in $mathbb{Z}/pmathbb{Z}$.
$endgroup$
– Arturo Magidin
Jan 9 at 19:15
$begingroup$
I believe that you can realize the first group as a group of $3times 3$ upper triangular matrices with $1$s in the diagonal and entries in $mathbb{Z}/9mathbb{Z}$, with $x$ corresponding to $left(begin{array}{ccc}1&1&0\0&1&0\0&0&1end{array}right)$, $y$ to $left(begin{array}{ccc}1&0&0\0&1&1\0&0&1end{array}right)$, and $z=left(begin{array}{ccc}1&0&1\0&1&0\0&0&1end{array}right)$. For the second group, take the $(1,2)$ entry in $mathbb{Z}/p^2mathbb{Z}$, and the $(1,3)$ and $(2,3)$ entries in $mathbb{Z}/pmathbb{Z}$.
$endgroup$
– Arturo Magidin
Jan 9 at 19:15
$begingroup$
Groups of order $p^2$ are abelian, so they don't give you the example you want. The nonabelian groups of order $p^3$ don't have "enough room" for an example: proper subgroups are abelian of order at most $p^2$, hence $2$-generated or cyclic. My second example is of order $p^4$, hence that's the smallest. It has order $p^4$ because every element can be written as $x^ay^bz^c$ with $0leq alt p^2$, $0leq blt p$, $0leq clt p$, so you have $p^2pp=p^4$ elements.
$endgroup$
– Arturo Magidin
Jan 9 at 19:17
$begingroup$
Groups of order $p^2$ are abelian, so they don't give you the example you want. The nonabelian groups of order $p^3$ don't have "enough room" for an example: proper subgroups are abelian of order at most $p^2$, hence $2$-generated or cyclic. My second example is of order $p^4$, hence that's the smallest. It has order $p^4$ because every element can be written as $x^ay^bz^c$ with $0leq alt p^2$, $0leq blt p$, $0leq clt p$, so you have $p^2pp=p^4$ elements.
$endgroup$
– Arturo Magidin
Jan 9 at 19:17
|
show 3 more comments
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1
$begingroup$
Let $G=langle x,y,zmid x^9=y^9=z^9=1, xz=zx, yz=zy, yx=xyzrangle$. Then look at $H=langle x^3,y^3,zrangle$.
$endgroup$
– Arturo Magidin
Jan 9 at 18:10
1
$begingroup$
Smallest group is probably of order $p^4$: $langle x,y,zmid x^{p^2}=y^p=z^p=1, xz=zx, yz=zy, yx=xyzrangle$, with $H=langle x^p,y,zrangle$.
$endgroup$
– Arturo Magidin
Jan 9 at 18:15
$begingroup$
I'm not very good with generators and relations, but I can more or less see that your example is correct. Is the given group a semidirect product or a matrix group? Why that group has order p^4? Why do you think the smallest overall example has order 81? Thanks!
$endgroup$
– CharlesA
Jan 9 at 18:48
$begingroup$
I believe that you can realize the first group as a group of $3times 3$ upper triangular matrices with $1$s in the diagonal and entries in $mathbb{Z}/9mathbb{Z}$, with $x$ corresponding to $left(begin{array}{ccc}1&1&0\0&1&0\0&0&1end{array}right)$, $y$ to $left(begin{array}{ccc}1&0&0\0&1&1\0&0&1end{array}right)$, and $z=left(begin{array}{ccc}1&0&1\0&1&0\0&0&1end{array}right)$. For the second group, take the $(1,2)$ entry in $mathbb{Z}/p^2mathbb{Z}$, and the $(1,3)$ and $(2,3)$ entries in $mathbb{Z}/pmathbb{Z}$.
$endgroup$
– Arturo Magidin
Jan 9 at 19:15
$begingroup$
Groups of order $p^2$ are abelian, so they don't give you the example you want. The nonabelian groups of order $p^3$ don't have "enough room" for an example: proper subgroups are abelian of order at most $p^2$, hence $2$-generated or cyclic. My second example is of order $p^4$, hence that's the smallest. It has order $p^4$ because every element can be written as $x^ay^bz^c$ with $0leq alt p^2$, $0leq blt p$, $0leq clt p$, so you have $p^2pp=p^4$ elements.
$endgroup$
– Arturo Magidin
Jan 9 at 19:17