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Questions about the branch of algebra that deals with groups.

16 votes
Accepted

Is there a simple description of this group?

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 …
Adam P. Goucher's user avatar
11 votes
0 answers
181 views

Iterated automorphism groups of finite groups

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 …
Adam P. Goucher's user avatar
6 votes
1 answer
116 views

Bijection from $S^2$ to itself interchanging actions of $A_5$

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 …
Adam P. Goucher's user avatar
3 votes

General bound for the number of subgroups of a finite group

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 …
Adam P. Goucher's user avatar
23 votes
Accepted

Presentation of the Monster Group

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 …
Adam P. Goucher's user avatar
5 votes

Iterated Automorphism Groups

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 …
Adam P. Goucher's user avatar
5 votes
Accepted

A generously vertex transitive graph which is not Cayley?

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 …
Adam P. Goucher's user avatar
11 votes
1 answer
402 views

Traveling Salesman Problem on finite group

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} …
Adam P. Goucher's user avatar
4 votes

Automorphisms of Lubotzky–Phillips–Sarnak graphs

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 …
Adam P. Goucher's user avatar
7 votes

What is the symmetry group of this compound of two polytopes?

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 …
Adam P. Goucher's user avatar
10 votes
0 answers
483 views

A lattice with Monster group symmetries

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 …
Adam P. Goucher's user avatar
5 votes

Automorphism of the transfinite rooted binary tree

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 …
Adam P. Goucher's user avatar
23 votes

What is the geometric shape of the Monster sporadic group?

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, …
Adam P. Goucher's user avatar
31 votes

Is there any theory why (for Bitcoin) the discrete logarithm problem is so hard to solve?

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 …
Adam P. Goucher's user avatar