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Jul
27
comment A group of order 2520
$G$ has a normal subgroup $N$ of order 21, which has a complement $H := C_G(K)$ of order 120. $H$ is a perfect central extension of $A_5$, whose center acts trivially on $N$ (by conjugation), so $A_5$ has to act on $N = C_7\rtimes C_3$. It fixes the characteristic subgroup $C_7$ of $N$, so it is not hard to see that $A_5$ acts trivially. In total you get that $G$ is the direct product of $N$ and $H$.
Jul
19
comment Historical Question about Schur-Zassenhaus Theorem
According to the articles about Hans Zassenhaus on wikipedia, his group theory book was based on lectures of his advisor Emil Artin and van der Waerden's algebra book was based on lectures by Artin and Emmy Noether.
Jul
17
comment Historical Question about Schur-Zassenhaus Theorem
Q1: According to the references in "The Theory of Finite Groups" by Kurzweil and Stellmacher Schur's paper should be "Schur, J.: Untersuchungen über die Darstellungen der endlichen Gruppen durch gebrochen lineare Substitutionen, J. reine u. angew. Math. 132 (1907), 85-137" as they usually cite the original papers (I didn't check). For the extended theorem by Zassenhaus they refer to the book.
Jul
17
comment Historical Question about Schur-Zassenhaus Theorem
Q3: The first proof of the conjecture was given by Feit and Thompson in their odd order theorem (no other proof was found yet).
Jul
8
comment Optimality proof for greedy algorithm
I just saw that I phrased my counterexample wrongly [I thought the gains of all steps add up]. If you don't have any other assumptions about the gains [to me it is not clear if in your model the order of actions matter, but this does not change much anyway], you could construct a example where greedy is not optimal, by assigning a "high value" x to some action $a$ and lower values to all other actions in the first step, and then by assigning values greater x to all pairs of actions (u, v) with $u\ne a$ and the value x to pairs of actions starting with a.
Jul
8
comment Optimality proof for greedy algorithm
Why do you expect the greedy algorithm to be optimal? Do you have any other (hidden) assumptions? [Otherwise take $\mathcal A = \{a, b, c\}$, $\mathcal B = 2$ (I assume this to be the number of actions before stopping, is this correct?), $g_a(S) = 3$, $g_b(S) = g_c(S) = 2$, $g_x(S|a) = 0$ for $x = a$ or $b$ and $g_x(S|y) = 2$ otherwise].
Jul
5
comment Definition question of convex orbit of finite group action
That the orbit is a convex subset.
Jun
28
comment Image of a normal Hall Subgroup under an automorphism
One problem in your reasoning is that you simply state "Further, let us define a subgroup H of order d."! How do you know that such a subgroup exists (it does)? You should either prove its existence or refer to some theorem that proves it [Or better: you don't need such a subgroup as Andreas showed].
Jun
26
comment If I know the Conjugacy classes of a group, do I know the group?
@jwodder: Take the product of Zev's example with a non-abelian group, e.g., $C_4\times S_3$ and $C_2^2\times S_3$.
Jun
25
comment How useful are geometric aspects when studying finite groups?
I can recommend reading the article "Subgroup complexes" by Peter Webb, pp. 349-365 in: ed. P. Fong, The Arcata Conference on Representations of Finite Groups, AMS Proceedings of Symposia in Pure Mathematics 47 (1987).
Jun
18
comment For subspaces, if $N\subseteq M_1\cup\cdots\cup M_k$, then $N\subseteq M_i$ for some $i$?
Are you sure about your inductive step? The first statement also holds over finite fields, but how does the induction look like that it works only in characteristic 0?
Jun
14
comment Covering a group with the conjugates of two subgroups related by an automorphism
Hi Peter, welcome to math.stackexchange! Why does a positive answer require the classification? Shouldn't an example be enough? (Unfortunately I can't access the link to verify if you meant "The answer is no".)
May
31
comment G is finite group such that H is normal and P is a Sylow p-subgroup of H
As $H$ is normal in $G$, $G$ acts on the set $\mathrm{Syl}_p(H)$ of Sylow $p$-subgroup of $H$ (by conjugation). Restricting the action to a $p$-Sylow $Q$ of $G$ you get a $p$-group acting on a set of order coprime to $p$. This implies that $Q$ has a fixed point $P_0$, i.e., $Q$ normalizes $P_0$. Now $P_0 = P^h$ for some $h\in H$ gives the $p$-Sylow $Q^{h^{-1}}$ of $G$ normalizing $P$.
May
17
comment Why are the p-adic integers a linearly ordered group?
The order of the $p$-adic integers does not come from the $p$-adic norm.
May
15
comment What can we say about the size of $HK\cap KH$ when $HK\neq KH$?
I'd be surprised if one cannot generalize this construction by fixing additional a common subgroup $U = H\cap K$ of $H$ and $K$.
May
15
comment What can we say about the size of $HK\cap KH$ when $HK\neq KH$?
Given finite groups $H$ and $K$ you can surely embed them into a finite group $G$ such that $H\cap K = 1$ and $HK\cap KH = H\cup K$: Take $G$ to be the symmetric group on $H\times K$, and embed $H$ by the regular action on $H\times 1$ fixing $H\times K^\#$ pointwise (i.e., $(h, 1)\cdot h' = (hh', 1)$ and $(h, k)\cdot h' = (h, k)$ for $k\ne1$) and $K$ acts on $H\times K$ by exchanging the copies of $H$ ($(h, k)\cdot k' = (h, kk')$). $hk = kh$ is equivalent to $h=h^k$, which implies $h=1$ or $k=1$.
May
1
comment Finite locally groups
What you call "finite locally" is usually called "locally finite".
Apr
23
comment What about non finitely generated groups?
Locally finite groups are groups whose finitely generated subgroups are finite. Finite group theory is quite helpful for investigating locally finite groups, but they do need new techniques (see for example these classnotes).
Apr
19
comment on a group with perfect automorphism group
Hint: Do you know what outer automorphisms are? If yes, look for a simple group with trivial outer automorphisms group.
Apr
16
comment $p$-Group as union of subgroups
@MarshalKurosh: You're right that the Frattini subgroup doesn't help directly for the question. One can maybe get a better picture of what's going on, but it wouldn't shorten a proof.