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Dec
2
comment If $G $ is a group of order $12$ not isomorphic to $A_4$ then does $G$ have an element of order $6$ ?
Which direction do you have problems with?
Dec
2
answered Is there a group which has precisely all finite groups as subgroups?
Dec
2
comment Generating a group by its $q$-elements.
How about $G=S_4$ and $q=3$?
Dec
2
comment If $G $ is a group of order $12$ not isomorphic to $A_4$ then does $G$ have an element of order $6$ ?
@user123733: Please tell me which step you have problems with.
Dec
1
comment If $G $ is a group of order $12$ not isomorphic to $A_4$ then does $G$ have an element of order $6$ ?
Assuming that $G$ has no element of order 6 is equivalent to the centralizer of a 3-Sylow having order 3. The normalizer of a 3-Sylow has order either 3 or 12 (Sylow). As $Z_3$ has only one nontrivial automorphism, we know that $G$ has four 3-Sylows and hence four elements not of order 3. These are the elements of the (therefore unique and hence normal) 2-Sylow subgroup $V$. As any 3-Sylow acts non-trivially on $V$, we can exclude $V=Z_4$ and get $V=Z_2\times Z_2$ with the elements of order 3 permuting the elements of order $2$. That's $A_4$.
Nov
29
comment Permutation group of a set
@DerekHolt: Ups, sorry. You're right, of course.
Nov
28
comment Permutation group of a set
@DerekHolt: Isn't $|G| = \sum_{g\in G^\#} |X^g|$ and every element $g\ne 1$ has at least one fixed point? ($G^\#$ is $G$ without 1)
Nov
27
comment Permutation group of a set
"no cycles of length one" I'd call "no fixed points". Burnside's Lemma tells you that the elements of your group have in average 2 fixed points. Looking at the identity you get exactly one element $x$ with two fixed points (all others have one). How many fixed points do the conjugates of this element $x$ have? Now consider that the centralizer of $x$ in $G$ acts on the set of fixed points of $x$.
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!)