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3h

comment 
JacobsonMorozov theorem
clear? every theorem is a clear consequence of any stronger theorem :) Ben's argument shows that it follows just from the classification of $\mathfrak{sl}_2$representations 
1d

revised 
Lower bound for $ \sum_{i=1}^n x_i f(x_i)$ when $\sum_{i=1}^{n}x_i = K$
edited tags 
2d

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JacobsonMorozov theorem
Possibly this criterion is not very practical; at least it shows hat for a unipotent element, if we perform the JacobsonMorozov theorem, the kernel of the resulting homomorphism from $SL_2$ (which is trivial, $\pm 1$, or all of $SL_2$) only depends on the unipotent element and not on the choice of homomorphism. 
2d

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JacobsonMorozov theorem
Ben is right. If you have a rep of $SL_2$ which is a sum of $n_i$dimensional irreducibles $V_{n_i}$, $1\le i\le k$, then the unipotent of $SL_2$ acts with Jordan blocks of size $n_i$, $1\le i\le k$. If all $n_i$ are odd then each $SL_2\to GL(V_{n_i})$ has $1$ in the kernel, so the whole rep factors through $PGL_2$, while otherwise (i.e. if at least one block has even size) one of this maps is faithful so the whole rep is faithful (i.e. doesn't factor). The point is that although $SL_2$ has irreducibles in all dimensions, it does not have faithful irreducibles in all dimensions. 
2d

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JacobsonMorozov theorem
Nice. It's interesting that for any choice of faithful representation of the target Lie group you get one way to check. 
Jul 2 
revised 
Group with finite outer automorphism group and large center
fixed minor typos 
Jul 2 
comment 
Outer automorphisms of finite extensions
followup: mathoverflow.net/questions/206585/… 
Jul 2 
comment 
Determining if a matrix is orthogonal
No, first you mean "modulus 1" instead of "roots of unity", second you also need the matrix to be diagonalizable over the complex numbers. 
Jun 29 
revised 
Asymptotic of min(#generators times diameter), for a Cayley graph of Sn
fixed typo 
Jun 28 
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Decomposition of Braid Groups
What do you mean by "the answer is known"? the question is too vague to have a definite answer (unless you describe all semidirect product decompositions) 
Jun 28 
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Decomposition of Braid Groups
To be precise, a homomorphism to $\mathbf{Z}$ yields not one semidirect product decomposition, but plenty of them. 
Jun 28 
answered  Asymptotic of min(#generators times diameter), for a Cayley graph of Sn 
Jun 28 
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Asymptotic of min(#generators times diameter), for a Cayley graph of Sn
OK, I maybe I need expanders instead of just random pairs, and this gives diameter $\log S_n\simeq n\log n$ for generating subsets of bounded size. It's done in this paper by Kassabov ams.org/journals/era/20051106/S1079676205001460/… 
Jun 28 
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Asymptotic of min(#generators times diameter), for a Cayley graph of Sn
It's related but not really a sequel to the linked post. That post concerns the issue of the max over all generating subset, which is conjecturally polynomial. Here you consider the min, which, to make the question nontrivial, you multiply with the size $d$ of the generating subset. Clearly the diameter should be $\succeq\log_d(G)$, which means here $\succeq n\log_d(n)$. I think it's known that a random pair in $S_n\times (S_nA_n)$ generates $S_n$ with diameter $\preceq\log(G)\simeq n\log n$. This means that $s_n\simeq n\log n$. 
Jun 28 
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Largest Set of Special Unitary Matricies With Invariant Subspace For Adjoint Action
The question is not "what is the largest" but "what are the largest", because there's nonuniqueness. And it's clear that any such subset has to be a closed subgroup. Conversely since we precisely have the adjoint action, any closed subgroup (infinite and distinct from the whole group) has an invariant subspace (its own Lie algebra), so the question boils down to the standard question of classifying (infinite) maximal closed subgroups of $SU(n)$, whose main step is the classification of maximal Lie subalgebras of $\mathfrak{su}(n)$, which is due to Dynkin or before. 
Jun 26 
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Permutation covering of a $G$lattice
Oh, I misread the question, namely that the group acting on the tensor product was the product... 
Jun 26 
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Permutation covering of a $G$lattice
PS I did the computation and it seems that in any characteristic the rank of $J(1,k)\otimes J(1,k)$ is $k^2k$, so the dimension of its kernel is $k$, so its number of Jordan blocks is $k$. So even in characteristic $p$ $J(1,p1)\otimes J(1,p1)$ has $p1$ Jordan blocks, which means that tensoring with characteristic $p$ will not provide any obstruction. 
Jun 26 
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Permutation covering of a $G$lattice
It's clear that the invariant subspaces of all nontrivial $C_p$modules in char. $p$ are nontrivial, because the action is given by a unipotent operator (in $P_i$ the invariants are 1dimensional, generated by the class of the vector $(0,1,\dots,p1)$. So the problem in general is about the decomposition of $K=J(1,p1)\otimes J(1,p1)$ as Jordan blocks ($J(1,k)$ being the unipotent Jordan block of size $k$); it's enough here to know the number of blocks, that is the dimension of the kernel of $K1$. 
Jun 25 
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Avoiding countable subgroups of a group homeomorphic to the Cantor space
Two remarks: 1) for a topological group, "homeomorphic to Cantor space" is equivalent to "metrizable profinite". 2) A "metric group" is a group endowed with a metric, not just a metrizable group. 
Jun 25 
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On the number of ends of a countable simple group
Great! I'd guess that it was known before acylindricity that such $G$ would actually be relatively hyperbolic with respect to vertex groups? (relatively hyperbolic groups were known to be SQuniversal, ArzhantsevaMinasyanOsin 2006) 