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Questions about the branch of algebra that deals with groups.
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 …
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 …
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 …
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, …
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 …
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 …
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} …
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 …
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 …
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 …
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 …
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 …
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 …
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 …