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edited Jun 26 2011 at 8:18 |
I revise earlier edits to give a coherent account of the construction which shows that such subgroups can exist.
The underlying idea of the strategy is as follows: Let $X$ be a non-trivial finite group with trivial center which admits an automorphism $\alpha$ fixing only the identity ((informally, and by a slight abuse, known as fixed-point free automorphism). Note that $\alpha$ can't have prime order, for otherwise $X$ would be nilpotent by Thompson's theorem, and hence would have non-trivial center. Then the direct product $X \times X$ is a product of two self normalizing subgroups of order $|X|$ which intersect trivially. One is $\Delta(X) = \{ (x,x): x \in X \}$. The other is
$\Delta^{\alpha}(X) = \{ (x,x\alpha): x \in X\}$. That $\Delta(X)$ is self-normalizing is clear,
since $Z(X) = 1$. For notice that if $(x,x)^{(a,b)} \in \Delta(X)$ for each $x \in X$, then
$ab^{-1} \in Z(X)$. The argument for the other subgroup is similar. Now we seek such a group $X$ with trivial center admitting a fixed point free automorphism $\alpha$ of composite odd order. If such a group exists, we may set $\beta = \alpha^{-1}$. Then the group $G = X \times X$ has the desired three self-normalizing subgroups $\Delta(X)$, $\Delta^{\alpha}(X)$ and $\Delta^{\beta}(X)$. For notice that if $x \alpha = x \beta$, then $\alpha^{2}$ fixes $x$, so that $\alpha$ fixes $x$ as $\alpha$ has odd order. But then $x$ is the identity by hypothesis. Hence any two of the three subgroups have trivial intersection, and have product $X \times X$ by order considerations.
There does exist a group $X$ of order $7^{6} .2^{4}$ which admits a fixed point free automorphism $\alpha$ of order $9$, and which has trivial center. Hence the construction of
the construction above does work in this case, and $G = X \times X$ has three self-normalizing subgroups of the required form. We find a subgroup $Y$ of order $144$ of ${\rm GL}(6,7)$ which has an elementary Abelian normal subgroup $U$ of order $16$, acted on by an element $a$ of order $9$ whose cube centralizes $U$ but such that $a$ itself acts fixed-point freely on $U$, and such that, furthermore, $a$ does not have the eigenvalue $1$ in the given representation. We then take $X$ to be the semidirect product $VU$, where $V$ is a $6$-dimensional vector space over ${\rm GF}(7)$ and $U$ acts as the given elementary Abelian subgroup of ${\rm GL}(6,7)$. Then allowing $a$ to act on $U$ as it does within ${\rm GL}(6,7)$, and to act on $V$ as theh the given matrix yields an action of $a$ on $VU$ as a fixed-point free automorphism of order $9$. In terms of matrices, $a$ is the matrix
$ ( [0,1 , 0 ,0 , 0 , 0],[0 ,0, 1 , 0 , 0 , 0], [2 , 0 , 0 , 0, 0, 0],
[0,0, 0 , 0 ,1 , 0],[0 , 0 , 0 , 0 , 0 , 1],[ 0 , 0 , 0 , 4, 0, 0])$. The group $U$
is the set of diagonal matrices with diagonal entries $\pm 1$ such that the product of the
first three diagonal entries is $1$ and the product of the last three diagonal entries is $1$.
The subgroup $U$ is $a$-invariant, $a^3$ centralizes $U$, but $C_{U}(a) = I$. (Could not get
latex right for matrix). The group $VU\langle a \rangle$ is the semidirect product $VY$.
A similer similar construction works for other odd primes $p$ by considering ${\rm GL}(2p,q)$, where $q$
is a prime congruent to $1$ (mod p).
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edited Jun 26 2011 at 6:58 |
Here's I revise earlier edits to give a strategy coherent account of the construction which might merit some thought (later edit- and it works)shows that such subgroups can exist. The underlying idea of the strategy is as follows: Let $X$ be a non-trivial finite group with trivial center which admits an automorphism $\alpha$ which fixes fixing only the identity (note (informally, and by a slight abuse, known as fixed-point free automorphism). Note that $\alpha$ can't have prime order, for otherwise $X$ would be nilpotent by Thompson's theorem)theorem, and hence would have non-trivial center. Then the direct product $X \times X$ is a product of two self normalizing subgroups of order $|X|$ which intersect trivially. One is $\Delta(X) = \{ (x,x): x \in X \}$. The other is since $Z(X) = 1$. For notice that if $(x,x)^{(a,b)} \in \Delta(X)$ for each $x \in X$, and the then $ab^{-1} \in Z(X)$. The argument for the other subgroup is similar. I look for Now we seek such a group $X$ with trivial center admitting a fixed point free automorphism $\alpha$ of composite odd order. Set If such a group exists, we may set $\beta = \alpha^{-1}$. Then I the group $G = X \times X$ has the desired three self-normalizing subgroups $\Delta(X)$, $\Delta^{\alpha}(X)$ and $\Delta^{\beta}(X)$. Notice For notice that if $x \alpha = x \beta$, then $\alpha^{2}$ fixes $x$, so that $\alpha$ fixes $x$ as $\alpha$ has odd order. But then $x$ is the identity by hypothesis. Hence any two of the three subgroups have trivial intersection, and have product $X \times X$ by order considerations. (Second edit deleted as now redundant). Third edit: There is does exist a group $X$ of order $7^{6} .2^{4}$ which admits a fixed point free automorphism $\alpha$ of order $9$, and which has trivial center. Hence the construction ofthe first edit construction above does work in this case, and $G = X \times X$ has three self-normalizing subgroups of the required form. We find a subgroup $Y$ of order $144$ of ${\rm GL}(6,7)$ which has an elementary Abelian normal subgroup $U$ of order $16$, acted on by an element $a$ of order $9$ whose cube centralizes $U$ but such that $a$ itself acts fixed-point freely on $U$, and such that, furthermore, $a$ does not have the eigenvalue $1$ in the given representation. We then take $X$ to be the semidirect product $VU$, where $V$ is a $6$-dimensional vector space over ${\rm GF}(7)$ and $U$ acts as the given elementary Abelian subgroup of ${\rm GL}(6,7)$. Then allowing $a$ to act on $U$ as it does within ${\rm GL}(6,7)$, and to act on $V$ as theh given matrix yields an action of $a$ on $VU$ as a fixed-point free automorphism of order $9$. In terms of matrices, $a$ is the matrixlatex right for matrix). The group $VU\langle a \rangle$ is the semidirect product $VY$.A similer construction works for other odd primes $p$ by considering ${\rm GL}(2p,q)$, where $q$is a prime congruent to $1$ (mod p).
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edited Jun 25 2011 at 22:37 |
Here's a strategy which might merit some thought (later edit- and it works). Let $X$ be a finite group with trivial center which admits an automorphism $\alpha$ which fixes only the identity (note that $\alpha$ can't have prime order, for otherwise $X$ would be nilpotent by Thompson's theorem). Then the direct
product $X \times X$ is a product of two self normalizing subgroups of order $|X|$ which intersect
trivially. One is $\Delta(X) = \{ (x,x): x \in X \}$. The other is
$\Delta^{\alpha}(X) = \{ (x,x\alpha): x \in X\}$. That $\Delta(X)$ is self-normalizing is clear,
since $Z(X) = 1$, and the argument for the other subgroup is similar. I look for such a group $X$ with trivial center admitting a fixed point free automorphism $\alpha$ of composite odd order. Set $\beta = \alpha^{-1}$. Then I the group
$G = X \times X$ has the desired three self-normalizing subgroups $\Delta(X)$, $\Delta^{\alpha}(X)$ and $\Delta^{\beta}(X)$. Notice that if $x \alpha = x \beta$, then $\alpha^{2}$ fixes $x$, so that $\alpha$ fixes $x$ as $\alpha$ has odd order. But then $x$ is the identity by hypothesis. Hence any two of the three subgroups have trivial intersection, and have product $X \times X$ by order considerations.
(Second edit deleted as now redundant).
Third edit: There is a group $X$ of order$7^{6} .2^{4}$ which admits a fixed point free automorphism $\alpha$ of order $9$, and which has trivial center. Hence the construction of
the first edit does work in this case, and $G = X \times X$ has three self-normalizing subgroups of the required form. We find a subgroup $Y$ of order $144$ of ${\rm GL}(3,7)$ GL}(6,7)$ which has an elementary Abelian normal subgroup $U$ of order $16$, acted on by an element $a$ of order $9$ whose cube centralizes $U$ but such that $a$ itself acts fixed-point freely on $U$, and such that, furthermore, $a$ does not have the eigenvalue $1$ in the given representation. We then take $X$ to be the semidirect product $VU$, where $V$ is a $6$-dimensional vector space over ${\rm GF}(7)$ and $U$ acts as the given subgroup of ${\rm GL}(6,7)$. Then allowing $a$ to act on $U$ as it does within ${\rm GL}(3,7)$GL}(6,7)$, and to act on $V$ as theh given matrix yields an action of $a$ on $VU$ as a fixed-point free
automorphism of order $9$. In terms of matrices, $a$ is the matrix
$ \left( \begin{array}{clcrcc} 0 & 1 & 0 & ( [0,1 , 0 & ,0 & 0\\0 & , 0 & , 0],[0 ,0, 1 & , 0 & , 0 & 0\\ , 0], [2 & 0 & 0 & , 0 & , 0 & 0\\
, 0& , 0& , 0],
[0,0, 0 & , 0 & ,1 & 0\\0 & , 0],[0 , 0 & , 0 & , 0 & , 0 & 1\\ , 1],[ 0 & , 0 & , 0 & , 4& , 0& 0\\ \right)$, 0])$. The group $U$
is the set of diagonal matrices with diagonal entries $\pm 1$ such that the product of the
first three diagonal entries is $1$ and the product of the last three diagonal entries is $1$.
The subgroup $U$ is $a$-invariant, $a^3$ centralizes $U$, but $C_{U}(a) = I$. (Could not get
latex right for matrix).
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edited Jun 25 2011 at 22:24 |
Here's a strategy which might merit some thought (later edit- and it works). Let $G$ X$ be a finite group with trivial center which admits an automorphism $\alpha$ which fixes only the identity (note that $\alpha$ can't have prime order, for otherwise $G$ X$ would be nilpotent by Thompson's theorem). Then the direct product $G X \times G$ X$ is a product of two self normalizing subgroups of order $|G|$ |X|$ which intersecttrivially. One is $\Delta(G) \Delta(X) = \{ (g,g): g x,x): x \in G X \}$. The other is $\Delta^{\alpha}(G) \Delta^{\alpha}(X) = \{ (g,g\alpha): g x,x\alpha): x \in G\}$X\}$. That $\Delta(G)$ \Delta(X)$ is self-normalizing is clear,since $Z(G) Z(X) = 1$, and the argument for the other subgroup is similar. I would look for such a group $G$ X$ with trivial center admitting a fixed point free automorphism $\alpha$ of composite odd order. Set $\beta = \alpha^{-1}$. Then I think the group$G = X \times G$ would have X$ has the desired three self-normalizing subgroups $\Delta(G)$\Delta(X)$, $\Delta^{\alpha}(G)$\Delta^{\alpha}(X)$ and $\Delta^{\beta}(G)$\Delta^{\beta}(X)$. Notice that if $g x \alpha = g x \beta$, then $\alpha^{2}$ fixes $g$, x$, so that $\alpha$ fixes $g$ x$ as $\alpha$ has odd order. But then $g$ x$ is the identity by hypothesis. Hence any two of the three subgroups have trivial intersection, and have product $G X \times G$ X$ by order considerations. Later (Second edit : I am deleted as now more convinced than ever that such subgroups CAN existredundant).Let $p$ be an odd prime, and let $H$ be Third edit: There is a finite group $X$ of order$7^{6} .2^{4}$ which is not nilpotent, but admits a fixed-point fixed point free automorphism $\alpha$ of order $p^2$. Then $|H|$ is not divisible by $p$. Let 9$, and which has trivial center. Hence the construction ofthe first edit does work in this case, and $G = H/Z_{\infty}(H)$, where $Z_{\infty}(H)$ is X \times X$ has three self-normalizing subgroups of the hypercenter required form. We find a subgroup $Y$ of order $H$, that is the smallest 144$ of ${\rm GL}(3,7)$ which has an elementary Abelian normal subgroup $U$ of order $H$16$, acted on by an element $a$ of order $9$ whose cube centralizes $U$ but such that $a$ itself acts fixed-point freely on $U$, and such that, furthermore, $a$ does not have the factor group has trivial centereigenvalue $1$ in the given representation. Since We then take $H$ is not nilpotent, X$ to be the semidirect product $G$ VU$, where $V$ is non-trivial.By standard properties a $6$-dimensional vector space over ${\rm GF}(7)$ and $U$ acts as the given subgroup of coprime automorphisms${\rm GL}(6,7)$. Then allowing $a$ to act on $U$ as it does within ${\rm GL}(3,7)$, and to act on $\alpha$ still acts V$ as theh given matrix yields an action of $a$ on $VU$ as a fixed point fixed-point freeautomorphism of order $p^2$ on 9$. In terms of matrices, $a$ is the matrix$ \left( \begin{array}{clcrcc} 0 & 1 & 0 & 0 & 0 & 0\\0 & 0 & 1 & 0 & 0 & 0\\ 2 & 0 & 0 & 0& 0& 0\\0 & 0 & 0 & 0 & 1 & 0\\0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 4& 0& 0\\ \right)$. The group $G$ which now has trivial center, and U$is the setup set of diagonal matrices with diagonal entries $\pm 1$ such that the previous paragraph product of thefirst three diagonal entries is in place. So an affirmative answer to $1$ and the question would be provided byexhibiting a non-nilpotent finite group product of the last three diagonal entries is $H$ which admits a fixed-point free automorphism 1$.The subgroup $\alpha$of order U$ is $p^2$ for some odd prime a$-invariant, $p$.a^3$ centralizes $U$, but $C_{U}(a) = I$.
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edited Jun 25 2011 at 10:21 |
Here's a strategy which might merit some thought. Let $G$ be a finite group with trivial center which admits an automorphism $\alpha$ which fixes only the identity (note that $\alpha$ can't have prime order, for otherwise $G$ would be nilpotent by Thompson's theorem). Then the direct
product $G \times G$ is a product of two self normalizing subgroups of order $|G|$ which intersect
trivially. One is $\Delta(G) = \{ (g,g): g \in G \}$. The other is
$\Delta^{\alpha}(G) = \{ (g,g\alpha): g \in G\}$. That $\Delta(G)$ is self-normalizing is clear,
since $Z(G) = 1$, and the argument for the other subgroup is similar. I would look for such a group $G$ with trivial center admitting a fixed point free automorphism $\alpha$ of composite odd order. Set $\beta = \alpha^{-1}$. Then I think the group
$G \times G$ would have the desired three self-normalizing subgroups $\Delta(G)$, $\Delta^{\alpha}(G)$ and $\Delta^{\beta}(G)$. Notice that if $g \alpha = g \beta$, then $\alpha^{2}$ fixes $g$, so that $\alpha$ fixes $g$ as $\alpha$ has odd order. But then $g$ is the identity. Hence any two of the three subgroups have trivial intersection, and have product $G \times G$ by order considerations.
Later edit: I am now more convinced than ever that such subgroups CAN exist. Let $p$ be an odd
prime, and let $H$ be a finite group which is not nilpotent, but admits a fixed-point free automorphism $\alpha$ of order $p^2$. Then $|H|$ is not divisible by $p$. Let $G = H/Z_{\infty}(H)$, where $Z_{\infty}(H)$ is the hypercenter of $H$, that is the smallest normal subgroup of $H$
such that the factor group has trivial center. Since $H$ is not nilpotent, $G$ is non-trivial.
By standard properties of coprime automorphisms, $\alpha$ still acts as a fixed point free automorphism of order $p^2$ on the group $G$ which now has trivial center, and the setup of the
previous paragraph is in place. So an affirmative answer to the question would be provided by
exhibiting a non-nilpotent finite group $H$ which admits a fixed-point free automorphism $\alpha$
of order $p^2$ for some odd prime $p$.
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edited Jun 24 2011 at 22:35 |
Here's a strategy which might merit some thought. Let $G$ be a finite group with trivial center which admits an automorphism $\alpha$ which fixes only the identity (note that $\alpha$ can't have prime order, for otherwise $G$ would be nilpotent by Thompson's theorem). Then the direct
product $G \times G$ is a product of two self normalizing subgroups of order $|G|$ which intersect
trivially. One is $\Delta(G) = \{ (g,g): g \in G \}$. The other is
$\Delta^{\alpha}(G) = \{ (g,g\alpha): g \in G\}$. That $\Delta(G)$ is self-normalizing is clear,
since $Z(G) = 1$, and the argument for the other subgroup is similar. I would look for such a group $G$ with trivial center admitting a fixed point free automorphism $\alpha$ of composite odd order. Set $\beta = \alpha^{-1}$. Then I think the group
$G \times G$ would have the desired three self-normalizing subgroups $\Delta(G)$, $\Delta^{\alpha}(G)$ and $\Delta^{\beta}(G)$. Notice that if $g \alpha = g \beta$, then $\alpha^{2}$ fixes $g$, so that $\alpha$ fixes $g$ as $\alpha$ has odd order. But then $g$ is the identity. Hence any two of the three subgroups have trivial intersection, and have product $G \times G$ by order considerations.
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edited Jun 24 2011 at 22:06 |
Here's a strategy which might merit some thought. Let $G$ be a finite group with trivial center which admits an automorphism $\alpha$ which fixes only the identity (note that $\alpha$ can't have prime order, for otherwise $G$ would be nilpotent by Thompson's theorem). Then the direct
product $G \times G$ is a product of two self normalizing subgroups of order $|G|$ which intersect
trivially. One is $\Delta(G) = \{ (g,g): g \in G \}$. The other is
$\Delta^{\alpha}(G) = \{ (g,g\alpha): g \in G\}$. That $\Delta(G)$ is self-normalizing is clear,
since $Z(G) = 1$, and the argument for the other subgroup is similar. I would look for such a group $G$ with trivial center admitting a fixed point free automorphism $\alpha$ of composite odd order coprime to $6$. Set $\beta = \alpha^{-2}$. alpha^{-1}$. Then I think the group
$G \times G$ would have the desired three self-normalizing subgroups $\Delta(G)$, $\Delta^{\alpha}(G)$ and $\Delta^{\beta}(G)$. Notice that if $g \alpha = g \beta$, then $\alpha^{2}$ fixes $g$, so that $\alpha$ fixes $g$ as $\alpha$ has odd order. But then $g$ is the identity.
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answered Jun 24 2011 at 21:49 |
Here's a strategy which might merit some thought. Let $G$ be a finite group with trivial center which admits an automorphism $\alpha$ which fixes only the identity (note that $\alpha$ can't have prime order, for otherwise $G$ would be nilpotent by Thompson's theorem). Then the direct
product $G \times G$ is a product of two self normalizing subgroups of order $|G|$ which intersect
trivially. One is $\Delta(G) = \{ (g,g): g \in G \}$. The other is
$\Delta^{\alpha}(G) = \{ (g,g\alpha): g \in G\}$. That $\Delta(G)$ is self-normalizing is clear,
since $Z(G) = 1$, and the argument for the other subgroup is similar. I would look for such a group $G$ with trivial center admitting a fixed point free automorphism $\alpha$ of composite order coprime to $6$. Set $\beta = \alpha^{-2}$. Then I think the group
$G \times G$ would have the desired three subgroups $\Delta(G)$, $\Delta^{\alpha}(G)$ and
$\Delta^{\beta}(G)$.
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