Search Results

It is a guess. Possibly what is meant is that the polynomial is palindromic: algebraically this means whenever $\alpha$ is a root $\alpha^{-1}$ is also a root, which translates to $f(x) = x^m f(\frac …

For a prime number $p$ the symmetric group $S_p$ is generated by a $p$-cycle and a transposition. These two elements generate individually minimal cyclic subgroups. I think this is case d) of Verre …

Look at Cayley's embedding Theorem which realises every group as a group of permutations. Permutations are functions (from a set to itself . . .). So an abelian group is a set of commuting permutati …

Early in a course in Algebra the result that every group can be embedded as a subgroup of a symmetric group is introduced. One can further work on it to embed it as a subgroup of a suitable (higher …

Slightly off-topic, a weird subgroup in two dimensions is constructed in this paper: Ryuji Maehara. On a connected dense proper subgroup of ${\bf R}^2$ whose complement is connected . Proc. Amer. Ma …

Can refine Stefan Kohl's suggestion by taking subsets containing the identity element of $G$ and of cardinality dividing $n$. So the upper bound is $\sum_{d>1, d| n}^n {n-1\choose d-1}$

In Galois theory of algebraic number fields while discussing a prime lying above a prime of the base filed the inertial group is defined as one inducing identity at the residue field level. Your d …

There is a paper in arxiv by Robert Guralnick and Gunter Malle that answers your question in a stronger way. Their aim is to prove existence of algebraic surfaces obtained in a specific way as a q …