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Yes. I'll talk about why elliptic curve discrete log is harder than ordinary discrete log. Suppose we have $g, h$ and want to find $n$ such that $g^n = h$. The usual methods for solving the discrete …

In the penultimate chapter of Sphere Packings, Lattices and Groups, the authors define a $196884$-dimensional real vector space and a faithful representation of the Monster group on that space. Now, …

There's a 12-generator 80-relator presentation for the Monster group. Specifically, we have 78 relators for the Coxeter group Y443: $12$ relators of the form $x^2 = 1$, one for each node in the Coxe …

The Cayley graph is the skeleton of the order-4 octagonal tiling: http://en.wikipedia.org/wiki/Order-4_octagonal_tiling Consequently, we can construct your group $G$ as a (normal) subgroup of the sy …

Let $\mathcal{G}$ be the set of isomorphism classes of finite groups. There is an operation $\mathrm{Aut} : \mathcal{G} \rightarrow \mathcal{G}$ which gives the automorphism group of a given group, u …

Given a finite group $H$, define a norm on $H$ to be a function $f : H \rightarrow \mathbb{R}_{\geq 0}$ satisfying: $f(x) = 0 \iff x = e$ is the identity; $\forall x \in H$, we have $f(x) = f(x^{-1} …

The book Mathematical Evolutions contains the following excerpt: A last, famous, example is the following. It is known that in the space of one hundred and ninety six thousand eight hundred and eight …

The automorphism group of this configuration $C'$ is the Mathieu group $M_{11}$. Firstly, we construct a larger configuration $C$ consisting of a 12-dimensional orthoplex inscribed in a 12-dimensional …

Let $X$ and $Y$ be two copies of $S^2$, and let $A_5$ act on each of them (as a group of rotations). Call these actions $\theta_X$ and $\theta_Y$. Moreover, let $g \in A_5$ be a fixed element of orde …

Take the graph product $G = P \times \mathbb{Z}$ of the Petersen graph with the infinite path graph. This is clearly infinite, finite-degree, and generously vertex-transitive. Then we have two distin …

I have a partial answer, which would be a complete answer (that is to say a complete classification of all such sets $I$) if we assume Dickson's conjecture in number theory. If $G$ is non-Abelian, th …

Although this question already has an accepted answer, which is correct for the question as stated, I posit that the surreal number tree is best viewed as a tree in the order-theoretic rather than the …

The automorphism group is strictly larger than $G$. Note that the automorphism group is a semidirect product of $G$ with the stabiliser of a single vertex, so it suffices to show that the group of aut …

In a similar vein to Geoff Robinson's answer, observe that any proper subgroup of a group of order $n$ can be generated by at most $\Omega(n) - 1$ elements, where $\Omega$ counts the number of prime f …