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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.
Apr
15
comment Covering a group with the conjugates of two subgroups related by an automorphism
You're right, here's a reference: en.wikipedia.org/w/…
Apr
15
comment Covering a group with the conjugates of two subgroups related by an automorphism
Did you check the symmetric group $S_6$ with an outer automorphism? [$H$ being an $S_5$; I'm not sure if the product of a $2$- and a $4$-cycle is contained in the union]
Apr
15
comment How to find all subgroups of a direct product?
For just the normal subgroups you can take a look at mathoverflow.net/questions/23692/… (all your examples are abelian, so all subgroups are normal then).
Apr
13
comment $p$-Group as union of subgroups
Do you happen to know the Frattini subgroup (the intersection of all maximal subgroups)? If you look at the group modulo its Frattini subgroup everything becomes more easy.
Apr
13
comment $p$-Group as union of subgroups
What did you intend to say with your first sentence "It is well known that a group can not be union of proper subgroups."? As every group is the union of its cyclic subgroups your statement holds only for cyclic groups. Did you want to write "of two proper subgroups"?
Mar
21
comment Are there/Why aren't there any simple groups with orders like this?
@caveman: I think it would be nice if you could add your formulas to the question.
Mar
21
comment Are there/Why aren't there any simple groups with orders like this?
@caveman: What exactly did you mean with your remark "ignoring the matrix groups for which the problem is solved"?
Mar
21
comment Are there/Why aren't there any simple groups with orders like this?
@SteveD: The Lie-type groups are probably excluded by the remark "ignoring the matrix groups for which the problem is solved". (I think, the exponent of $p$ in $|PSL_n(p)|$ grows quadratically for $n \to \infty$, but only linear for all other primes.)
Feb
27
comment Normal subgroups of finite index in infinite direct sum
Calling $\mathop{Supp} x = \{i\in I: x_i \ne 1\}$ the support of $x \in$ your infinite product/sum, take a look at the set $N_U$of all $x$ whose support is not an element of some fixed non-principal ultrafilter U on I. It's not too hard to show that $N_U$ is a normal subgroup. If all $S_i$ are isomorphic to each other, I think its index should be finite.
Feb
18
comment How do I understand $e^i$ which is so common?
Take also a look at this answer to a related question at mathoverflow.
Feb
6
comment What is a minimal polynomial of a group element, and why would we care if it was quadratic?
If you can get the book The Theory of Finite Groups by Kurzweil and Stellmacher, take a look at chapter 9 "Quadratic Action".