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seen Apr 9 at 18:27

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.
May
10
comment Find the subgroups of index two of this finite semi-direct product
@EwanDelanoy: I'd guess that there are no other index-2-subgroups of $G$. A possible way to prove this could be to show that all elements of $\{\pm 1\}^n\cap \mathop{Ker}(t')$ are commutators in $G$.
May
10
comment Find the subgroups of index two of this finite semi-direct product
@EwanDelanoy: At least my knowledge about Coxeter groups doesn't help finding all subgroups of index 2. I got this 2nd subgroup by considering $\{\pm 1\}^n$ as $\mathbb{F}_2$ vector space on which $S_n$ acts by permuting a base. As this action is $2$-transitive (for $n\ge 2$), over characteristic $0$ the linear representation would be the sum of two irreducible ones, but not so over characteristic $2$...
May
10
comment Find the subgroups of index two of this finite semi-direct product
@EwanDelanoy: The hint should help you to find a 2nd subgroup of index 2 (which you found).
Mar
16
comment Dimension of cells in the Bruhat decomposition of $GL_3$
The action of $GL_3$ on $V$ induces an action on all subsets of $V$, which maps subspaces to subspaces, preserves dimensions and inclusion.
Mar
16
comment Dimension of cells in the Bruhat decomposition of $GL_3$
Another thing you should understand is the Jordan-Hölder permutation belonging to a pair of maximal flags and - given a maximal flag $f$ and an element $g$ of $GL_3$ - the Jordan-Hölder permutation belonging to the pair $f$, $f^g$.
Mar
16
comment Dimension of cells in the Bruhat decomposition of $GL_3$
I usually prefer to think in terms of maximal flags $(0=V_0<V_1<V_2<V_3=V)$ of subspaces $V_i$ of $V$. $GL_3$ acts transitively on the set $\mathcal{F}$ of maximal flags and $B$ is the stabilizer subgroup of some maximal flag. The double cosets $BwB$ correspond then to orbits on pairs of flags or - if you prefer - to the orbits of $B$ on $\mathcal{F}$. See if you can find some useful meaning of dimension for subsets/subspaces of $\mathcal{F}$.
Feb
27
comment Prove: If $N\lhd G$ and $N$ and $\frac{G}{N'}$ are nilpotent, then $G$ is nilpotent.
Are your groups finite?
Jan
31
comment Are there 16 or 24 automorphisms of $\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}$?
Derek Holt commented in your referred earlier question that automorphism group has order 16. If Derek writes it, I would believe it.
Jan
16
comment Elementary problems with group theoretic solutions
+1: For Fermat's Little Theorem (among others)
Jan
16
comment Elementary problems with group theoretic solutions
Shouldn't a big-list question be community wiki?