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Nov
20
comment How to partition a finite vector space into affine subspaces all of the same dimension
Do you happen to know something about the case $q>2$?
Nov
20
accepted How to partition a finite vector space into affine subspaces all of the same dimension
Nov
20
comment How to partition a finite vector space into affine subspaces all of the same dimension
Thanks, this helps even for the case $d>1$ I'm normally considering: One just has to factor out a $(d-1)$-dimensional (common) subspace, and take the preimages of your solution.
Nov
20
comment How to partition a finite vector space into affine subspaces all of the same dimension
@Timbuc: The order of the field.
Nov
20
asked How to partition a finite vector space into affine subspaces all of the same dimension
Nov
13
comment all Sylow subgroups of $GL_n(\mathbb{F}_q)$
If you understand what an irreducible representation is, and how the Sylow subgroups of the symmetric group look like, you can work it out yourself quite easily.
Nov
13
comment About motivation of Lang's Proof $S_n$ is not solvable for $n\geq 5$.
@LeeMosher: What do refer to with "proved by this argument"? Do you speak about the proof given by Lang or about the observation mentioned in the question?
Oct
29
comment SO(3,q) to PGL(2,q)
I don't have the book "The geometry of the classical groups" by Donald E. Taylor at hand now, but according to google books Theorem 11.6 claims this isomorphism (but I cannot see the proof).
Sep
23
comment last part of proof of schur zassenhaus theorem.
The horrible transveral-based proof can be rewritten in a quite elegant and understandable way, see for example section 3.3 in the book "The Theory of Finite Groups" by Kurzweil and Stellmacher.
Sep
5
comment Conjugation (Group vs. Algebraic)
@David: If an element is not contained in any proper normal subgroup (like if the group is simple), what does the last point tell us? Isn't "conjugates" some more general concept than "normal subgroup"? You can define the latter as a subgroup who happens to be the union of conjugacy classes, but how do you define "conjugate" using the concept "normal subgroup"?
Sep
5
comment Higher residuosity problem, but with a known factorization
Most cases you can solve by setting $d=1$. Do you have any further conditions on $x, a, b$ and $n$?
Sep
4
comment Conjugation (Group vs. Algebraic)
@David: Even loosely, I cannot guess either what you intend to express with your last bullet point. Would you be so nice to expand this thought a bit? (Thanks!)
Sep
3
comment Conjugation (Group vs. Algebraic)
@David: I'd be surprised if Geoff doesn't know the definition of a normal subgroup. OK, his user profile doesn't give much hints, but you could take a look at the answers he gave here and at MO.
Aug
25
comment Does every group have a 'cyclization'?
IMHO: Who doesn't think of $Z_p\times Z_p$ immediately should look at some examples of small finite groups first before studying more theory.
Aug
19
comment How do we know the classification of finite simple groups is finished?
Maybe this article is of interest for you.
Aug
18
comment Maximal height of subgroups in $S_n$?
@Nishant: Your last sentence is the wrong way round: "Thus, this subgroup contains $A_6$, which contradicts that it has a smaller order than $A_6$."
Aug
11
comment Maximal height of subgroups in $S_n$?
@Nishant: Let $S_6$ act on the cosets of that subgroup of index 3 and restrict the action to $A_6$. How does this action of $A_6$ look like?
Aug
10
comment Proving that the intersection of a Sylow p-group with a normal subgroup is also a Sylow p-group
Do you know the highest power of $p$ dividing $|PN|$?
Aug
8
comment Maximal height of subgroups in $S_n$?
Adding to @Nishant's comment, one can say that the series stops at $S_6$: The only subgroup of $S_6$ with index smaller than $6$ is $A_6$, and $A_6$ does not have any subgroup of index smaller than $6$ (both follows from $A_6$ being simple). On the other hand the biggest prime dividing the order of $S_6$ is 5. This can be generalized to all non-prime $n>5$. (In particular the chain hoped for in the last sentence of Nishant's comment does not exist for $n>5$.)
Aug
6
comment how is the jordan-holder theorem used in conjunction with short exact sequences to construct groups of certain order?
You could take a look in chapter 11 "The Theory of Group Extensions" of Derek Robinson's "A course in the theory of groups".