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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
awarded  Citizen Patrol
Jan
16
comment Elementary problems with group theoretic solutions
Shouldn't a big-list question be community wiki?
Nov
22
comment Why do elements of coprime orders commute in nilpotent groups?
I'd be interested in how you would prove this. Do you have a concrete idea?
Nov
22
comment Efficiently find the generators of a cyclic group
You write that finding a primitive root takes polynomial time. Do you mean relative to the size of the given prime? (Finding the order of an element of $\mathbb{F}_p^\times$ takes - to my knowledge - as long as factoring $p-1$.)
Nov
22
comment Efficiently find the generators of a cyclic group
If you don't care about finding the smallest primitive root, you can combine several elements to one that generates the subgroup generated by the given elements. You can then speed up the algorithm by iterating only over the small primes $2, 3, 5, 7, 11, \dots$ (instead of all numbers).
Nov
21
comment Efficiently find the generators of a cyclic group
@joriki: No need to delete your comment just because you missed something. Now my comment is dangling in the air without reference. Shall I remove it, too? I think the upvoting of comments serves well for compacting a too long list of comments.
Nov
21
comment Efficiently find the generators of a cyclic group
@joriki: You can do slightly more than just caching. If you found elements of order $m$ and $n$, you can construct an element of order $\mathrm{lcm}(m, n)$.
Oct
31
comment How the center of a non-cyclic free Group is trivial?
@user1729: OK. Have fun!
Oct
31
comment How the center of a non-cyclic free Group is trivial?
@user1729: I still don't understand what part of lhf's proof you are intending to reuse except "Assume the center is non-trivial".
Oct
29
comment How the center of a non-cyclic free Group is trivial?
@lhf: Thanks for leaving your answer and adding the warning!
Oct
29
comment How the center of a non-cyclic free Group is trivial?
@user1729: I did't doubt your proof using one-relator groups with torsions (even if it seems likely that you need the fact that free groups with more than one generator have a trivial center to get the result you cited). But your proof doesn't just fill in details. It only takes over the assumption needed for a proof by contradiction. Then lhf's and your proofs diverge completely ("then every group with the same number..." vs. a concrete quotient of $F_n$). By the way, how difficult is the proof about one-relator groups with torsion?
Oct
28
comment A condition of abelian groups related to an automorphism
Sylow's theorem is a slight overkill. Cauchy's is enough.
Oct
28
comment How the center of a non-cyclic free Group is trivial?
@ineff: No, I mean what I wrote.
Oct
28
comment How the center of a non-cyclic free Group is trivial?
@user1729: The image of a group with non-trivial center can very well have a trivial center.
Oct
28
comment How the center of a non-cyclic free Group is trivial?
@user1729: We both criticize the sentence "If there was a free group with non-trivial center, then every group with the same number of generators would have a non-trivial center.", but IMO ineff's argument is not correct rsp. not correctly phrased.
Oct
28
comment How the center of a non-cyclic free Group is trivial?
@Basil: Why do you accept the wrong solution of lhf instead of the partially correct solution by ZulfiqarIII?