Here is the 1868 paper of Jordan mentioned in the comments: C. Jordan. Note sur les équations modulaires, C. R. Acad. Sci. Paris 66 (1868), 308-312. Here is Jordan's proof from that paper, as I ...

The following papers are relevant:  J. N. S. Bidwell, M. J. Curran, and D. J. McCaughan, Automorphisms of direct products of finite groups, Arch. Math. 86, 481 – 489 (2006).  J. N. S. ...

The main theorem in a paper of G. A. Miller  is the following: THEOREM. If $l, m, n$ are any three integers greater than unity, of which we call the greatest $k$, it is always possible to find ...

Fix $k > 0$. Suppose that $n > 6$ and $\frac{n(n-3)}{2} > k$. If $[S_n : H] \leq n+k$, then $H$ is one of the following: $S_n$, $A_n$, $S_{n-1}$, or $A_{n-1}$. So in particular if $[S_n : H] =... View answer Accepted answer 9 votes Here is a more general statement, see also Lemma 1.2 in . Lemma: Let$Z$be a self-dual$kG$-module which admits a non-degenerate$G$-invariant symmetric (alternating) bilinear form$b$. Suppose ... View answer Accepted answer 9 votes Let$\mathfrak{g}$be a simple Lie algebra over an algebraically closed field of characteristic$0$and denote the fundamental highest weights by$\varpi_1, \ldots, \varpi_l$, where$l$is the rank of ... View answer Accepted answer 9 votes EDIT: Actually, there is an older reference which will give the result below. See Herzog, Marcel Counting group elements of order$p$modulo$p^2$. Proc. Amer. Math. Soc. 66 (1977), no. 2, 247–... View answer 8 votes Since nobody gave any examples of applications of Frobenius conjecture, here's a small one I read about recently. We will only consider finite groups in what follows. For a group$G$, denote the set ... View answer Accepted answer 7 votes (Turning my comments into an answer). A finite group with all Sylow subgroups cyclic is called a$Z$-group. According to review MR0470050 in MathSciNet, in  it is shown that a$Z$-group is ... View answer 7 votes See Kano, Mikio. On the number of conjugate classes of maximal subgroups in finite groups. Proc. Japan Acad. Ser. A Math. Sci. 55 (1979), no. 7, 264--265. link for proofs of the following ($G$... View answer 5 votes Regarding your second question, for every finite simple group the structure of its outer automorphism group is known. For the sporadic groups, there is the following 2011 preprint on arXiv: ... View answer Accepted answer 5 votes For one answer, here is a theorem due to Thompson and Willems (Bilinear forms in characteristic$p$and the Frobenius-Schur indicator, Lecture Notes in Mathematics 1185, pg. 221-230). For an ... View answer 3 votes It seems that for older issues of these journals, there are no scanned copies available online. But depending on your location, a nearby university library might be helpful for getting scans of ... View answer Accepted answer 3 votes I do not know a reference, but I have thought about the same question. Here is a sketch using arguments that are in the literature. For some basics about regular unipotent elements, see for example ... View answer 3 votes To me the answer seems to be yes. Let$\varpi_i$be the$i$th fundamental dominant weight. Recall first that since$\Phi$is irreducible, for all$i$we have$\varpi_i = \sum_{j = 1}^l q_{ji} \...
I think the answer is yes. We can assume that $G$ is semisimple. As you have seen already, and which is is clear from the commutation relations, we have $[U, U] \leq \prod_{\alpha \in \Phi^+ - \Delta}... View answer 2 votes I think your question has a negative answer. For reductive$G$, there are only finitely many conjugacy classes of unipotent elements in$G$. However, for a parabolic subgroup$P < G$with ... View answer 2 votes This is just a small partial result and some comments. I think the following should settle the case$k = 2$. Suppose that$G$is a group with Sylow$p$-subgroups of order$p^n$and that$n_p(G) = 2p +...