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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".
May
20
comment A question from Isaac's Book
@JackSchmidt: Is your last comment a statement or a question? (The exercise is 8.2.11 in Kurzweil/Stellmacher - and they need solvability of either $G$ or $A$.)
May
15
comment Is group theory useful in any way to optimization?
Please be so polite to give at least links to the other place when posting twice the same question.
May
15
comment Proving facts about groups with representation theory.
@TobiasKildetoft: Purely group-theoretic proofs of Burnside's theorem were given by Goldschmidt (1970, odd), Bender (1972, general) and Matsuyama (1973, even). But Bender provided also a purely group-theoretic proof for Frobenius groups with $H$ of even order.
Jan
31
comment Proving facts about groups with representation theory.
H. Bender found a very simple, purely group-theoretic proof for 2. in case that $H$ as even order. A Fourier-analytic proof can be found here.
Nov
13
comment On direct product
Which direction of the proof do you have problems with?
Nov
7
comment Bijection map from a set of subgroup to another set of subgroup under some condition.
@user1729: Ah thanks, on my other computer I see the sign for normality double and thought $N$ is only subnormal...
Nov
6
comment How to show that $S(n, q)$ is a chamber complex?
For $n = 2$ the vertices of $S(2, q)$ are all $1$-dimensional, so none can contain another one. Walls are empty sets. So the answer to the question at the end is "No". (A good source for a proof of your first question is the book The Geometry of the Classical Groups by D.E. Taylor.)
Nov
6
comment Bijection map from a set of subgroup to another set of subgroup under some condition.
In my addition of Kurzweil/Stellmacher this exercise is given at the end of section 1.3 with $N$ normal in $G$. Which section is your exercise from?
Oct
29
comment Why is conjugation by an odd permutation in $S_n$ not an inner automorphism on $A_n$?
You could look at the kernel of the map $S_n \to \mathrm{Aut}(A_n)$ given by conjugation.