Timeline for Groups with triple system of self-normalizing subgroups
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 11, 2014 at 9:40 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
fixed matrix
|
| Jul 1, 2011 at 10:37 | vote | accept | Tom De Medts | ||
| Jun 26, 2011 at 18:52 | comment | added | Geoff Robinson |
Actually, I think an example could be made to work just using ${\rm GL}(3,7)$, with $X$ of order $7^{3}.4$ admitting a fixed-point free automorphism $\alpha$ of order $9$ such that $\alpha^{3}$ centralizes a Sylow $2$-subgroup $U$ (a Klein 4-group) of $X$. The matrix representing $\alpha$ would just be $([0,1,0],[0,0,1],[2,0,0])$, and $U$ would consist of all diagonal unimodular matrices of order $2$. Can't see now why I felt I need to go to $6$-dimensions. Similarly, for other odd primes $p$, ${\rm GL}(p,q)$ should be enough for prims $q \equiv 1$ (mod $p$).
|
|
| Jun 26, 2011 at 9:57 | comment | added | Geoff Robinson |
@jp: Thanks for the smaller example. Second: yes, I should have said in the final write-up that given a fpf automorphism $\alpha$ of $X$ order $p^2$, then we do have $p$ self-normalizing subgroups of $X \times X$, each of order $|X|$, with pairwise trivial intersection. They are the elements of $\{ \Delta_{i}(X): 0 \leq i \leq p-1 \}$, where $\Delta_i(X) = \{(x, x\alpha^{i}): x \in X\}$. The point is that $\alpha^{i-j}$ is fpf for $i \neq j$ in this range. Also, once $\Delta(X)$ is self-normalizing, `$\Delta^{\alpha}(X)$ is, as it is conjugate to $\Delta(X)$ in ${\rm Aut}(X \times X)$.
|
|
| Jun 26, 2011 at 9:08 | comment | added | j.p. | Do I understand it correctly that your construction gives similar factorizations with $p$ different groups given an automorphism of order $p^2$? ($A_i$ self-normalizing, $G = A_i\cdot A_j$ and $A_i\cap A_j = 1$ if $i, j\in \{1,\dots,p\}$ and $i\ne j$) | |
| Jun 26, 2011 at 8:56 | comment | added | j.p. | Another way to construct a group with fixed-point-free automorphism and trivial center is to take $X = Z_{43}\rtimes Z_7 = \langle a, b, c\rangle \rtimes \langle x\rangle$, where $a^x = a^{21}, b^x = b^{11}$ and $c^x = c^{35}$. Then $\alpha \in mathrm{Aut}(G)$ defined by $x^\alpha = x^2, a^\alpha = c^6, b^\alpha = a$ and $c^\alpha = b$ has order $9$ and is fixed-point-free. This construction works for primes $p, q, r$ with $p\mid q-1$ and $p\cdot q\mid r-1$. | |
| Jun 26, 2011 at 8:23 | comment | added | Geoff Robinson |
By the way, another way to see this is that $g \to g^{-1}\alpha(g)$ is an injection from $G$ to $G$, since $\alpha$ is fixed-point free. Since $G$ is finite, it must also be a surjection so every $h \in G$ does have the form $g^{-1}\alpha(g)$ for some $g \in G.$ Note that my edits above have changed $G$ to $X$, because I want to end up with $G$ as the group with the three factorizations.
|
|
| Jun 26, 2011 at 8:18 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
minor typos
|
| Jun 26, 2011 at 6:58 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Rewrote to make coherent account.
|
| Jun 25, 2011 at 22:37 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
typos
|
| Jun 25, 2011 at 22:24 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
corrected matrix (again)
|
| Jun 25, 2011 at 10:21 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Tidying up
|
| Jun 25, 2011 at 9:33 | comment | added | user6976 | Geoff: You are right. | |
| Jun 25, 2011 at 9:13 | comment | added | Geoff Robinson |
These two subgroups each have order $|G|$, so their product is $G \times G$ by order considerations, as I said. If a finite group $X$ has subgroups $A$ and $B$, then the set $AB$ has cardinality $|A| |B|/ |A \cap B|$, but need not be a subgroup. In this case $\Delta(G)$ and $\Delta^{\alpha}(G)$ each have order $|G|$ and have trivial intersection. The set $\Delta(G)\Delta^{\alpha}(G)$ has cardinality $|G|^{2}$, so must be all of $G \times G$. Similarly for other pairs. I know the literature on fpf automorphisms reasonably well. I think these triples should exist.
|
|
| Jun 25, 2011 at 8:44 | comment | added | user6976 | I think the idea is good. But for $G\times G$ to be generated by $\Delta(G)$ and $\Delta^\alpha(G)$ you need that elements of the form $g^{-1}\alpha(g)$ generate $G$. Is it always true? There are many papers (after John Thompson) about groups admitting fixed point free automorphisms. | |
| Jun 24, 2011 at 22:35 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Corrected typo, expanded text.
|
| Jun 24, 2011 at 22:06 | history | edited | Geoff Robinson | CC BY-SA 3.0 |
Simplified example, and expanded explanation.
|
| Jun 24, 2011 at 21:49 | history | answered | Geoff Robinson | CC BY-SA 3.0 |