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Aug
16
comment Abelian subgroup of a group of order $2002$
@Serkan: With full details your approach should be shorter (and easier) than the one given by DonAntonio.
Aug
10
comment Uniqueness of conjugates of a subgroup.
You could also reread Geoff's answer to the question you linked to...
Aug
10
comment Uniqueness of conjugates of a subgroup.
You are trying to prove that $A$ is weakly closed in $B$. Look up this term (or "weak closure") in any group theory book. Often the case $B$ a $p$-Sylow subgroup is of interest.
Jul
30
comment Converting a (signed) permutation to a reduced word
From (9.22) in the book rsp. Claim (a) and (b) in my answer you can easily deduce that $\mathop{des}(\sigma)$ is essentially the same as $\mathop{D}(\sigma)\cap R$ (with $R$ the set of Coxeter generators): just identify $0$ with $(1\; -1)$ and $i>0$ with $(i\; i+1)$.
Jul
29
comment Converting a (signed) permutation to a reduced word
Another good source is chapter 9 of the book The Geometry of the Classical Groups by D.E. Taylor (combined with my answer to math.stackexchange.com/questions/106462/…)
Jun
22
comment Let $f:\mathbb{R}\longrightarrow \mathbb{R}$ a differentiable function such that $f'(x)=0$ for all $x\in\mathbb{Q}$
This question is related: math.stackexchange.com/questions/151931/…
Jun
8
comment A representation is semisimple if its restriction to a subgroup of index prime to Char(F) is semisimple
(12.8) in Aschbacher's book "Finite Group Theory" (directly before Maschke's theorem) proves the statement for $H$ a $p$-Sylow (considering also (12.6)). If someone feels like, please expand it into an answer...
Jun
7
comment On automorphism of some finite 2-group of class nilpotency two
@user1729: I'd call Derek a pro.
Jun
6
comment Set of zeroes of the derivative of a pathological function
@EwanDelanoy: According to mathworld.wolfram.com/MinkowskisQuestionMarkFunction.html the Minkowski question mark function is purely singular, which means its derivative is almost everywhere 0 (according to planetmath.org/encyclopedia/SingularFunction2.html). So it doesn't answer the 1st question.
Jun
2
comment Set of zeroes of the derivative of a pathological function
Take a look at the mean value theorem in J. Dieudonne's "Foundation of Modern Analysis" (8.5.1 in my edition) or at www.ms.unimelb.edu.au/~jjk/doc/MVT.pdf (Theorem 1b). Such a function does not exist...
May
31
comment (Regular) wreath product of nilpotent groups
@SteveD: You should mention your implicit assumption $p\ne q$ ;-). To user31899: To be even more concrete than Steve, take $A$ and $B$ as simple as possible, but violating the condition after "if and only if", e.g., take something like $A=C_3$, $B=C_2$.
May
30
comment If $a_n$ goes to zero, can we find signs $s_n$ such that $\sum s_n a_n$ converges?
@EwanDelanoy: For the greedy construction you get in the proof of your lemma the limit $M\cdot\sqrt{r}$ (if $b_{i+1}$ is perpendicular to the result you got for being greedy on $b_1, \dots, b_i$, and all $b_i$ having the same length).
May
30
comment A representation is semisimple if its restriction to a subgroup of index prime to Char(F) is semisimple
If I remember correctly, you can use the usual averaging argument for proving Maschke's theorem (just take the average over coset representatives of $H$ instead of over all elements of $G$)
May
25
comment Maximal subgroups of a finite p-group
You need the condition $G/U$ not cyclic.
May
24
comment What could the meaning of “invariant of $G$” be?
The lengths of the orbits of a point stabilizer are invariants of the permutation group $G$. So yes, it's an invariant of the group action. Different action have (generally) different subdegrees.
May
21
comment Find the subgroups of index two of this finite semi-direct product
@user1729: No problem, it happened to me before, too (and probably to everybody else).
May
18
comment Find the subgroups of index two of this finite semi-direct product
@user1729: I'm not quite sure what you are wondering about. As you emphasized "no more than three", do you worry about the factor group $G/G'$ being $C_2\times C_2$ or $C_4$? The latter is excluded by the "UPDATE AT 17:30" to the question. If you are worried about your proof $G'$ having index $4$ in $G$ implies "at most three": it's correct.
May
13
comment Nilpotent groups are solvable
@ArturoMagidin: The given definition for nilpotent works only for finite groups (same for solvable).
May
11
comment Find the subgroups of index two of this finite semi-direct product
Hint: $(1 -1)(2 -2)(3 -3)(4 -4)$ is the commutator of $(1 -1)(2 -2)$ and $(1 3)(-1 -3)(2 4)(-2 -4)$. Now generalize.
May
10
comment Find the subgroups of index two of this finite semi-direct product
@EwanDelanoy: The statement about the commutators in my last comment is indeed quite easy to prove. If you'll need details, I'll be back probably tomorrow.