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Aug
6 |
awarded | Popular Question |
Jul
8 |
comment |
Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
Here are an estimates (From above and, I think, very imprecise)of number of solutions of $q^{x^{x^x}}=x \mod p$ based on similar constructions. There are some annoying details due to $x+k (\mod p)= x+k (\mod p-1)$ or $x+k(\mod p-1) +1$. So, $2^{x+k}= 2^x2^k$ or $22^x2^k\mod p$. I don't think that there is an easy generalization for points of period 4. I think that the proof goes through due to the group $\langle a,b,w\;|\;b^{-1}ab=b^2,w^{-1}aw=b, w^3=1\rangle$ is finite. (We may define "almost action" of this group, with $xa=x+1$, $xb=2x$, $xw=2^x\mod p$) |
Jul
8 |
comment |
Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$
Here an estimates of |
Dec
14 |
comment |
A special residually finite group
@IanAgol: did you mean some Wilson's construction of just infinte groups of type N(h)? Is it published now? |
Sep
26 |
revised |
The free group $F_2$ has index 12 in SL(2,$\mathbb{Z}$)
edited body |
Sep
24 |
awarded | Necromancer |
Sep
6 |
awarded | Necromancer |
Mar
14 |
awarded | Yearling |
Feb
3 |
answered | Measures idempotent with respect to addition and multiplication. |
Feb
3 |
awarded | Commentator |
Feb
3 |
comment |
The Higman group II
@Ashot. This answers my first question! Thank you. |
Feb
3 |
awarded | Scholar |
Feb
3 |
accepted | The Higman group II |
Feb
2 |
revised |
The Higman group II
added 45 characters in body |
Feb
2 |
comment |
The Higman group II
@Yves and @Derek. Oops, you are right. I will make the corresponding corrections. Thank you. |
Feb
1 |
asked | The Higman group II |
Jan
24 |
comment |
Eigenvalues of the products of a fixed unitari matrix with diagonal unitari matrices
@ Michael I have not notice your comment before. I just consider it as a set... About b). Let $n$ be "very large" Then in example 2) al matrices $DU$ have "very small" spectral gaps. So, the question: for which $U$ all $DU$ have small maximal spectral gap? As $\\{DU\\}$ may be considered as a point in the flag manifold, one could try to relate this gap with a Reimann distance on the flag manifold.... |
Jan
10 |
revised |
Eigenvalues of the products of a fixed unitari matrix with diagonal unitari matrices
added 89 characters in body |
Jan
9 |
asked | Eigenvalues of the products of a fixed unitari matrix with diagonal unitari matrices |
Jan
9 |
comment |
A subgroup intersects every conjugacy class
Yes, a free group seems to have a large subgroup. It may be constructed inductively adding $g$ to $<g_1,g_2,...,g_k>$ for each $g$ with $g^F\cap <g_1,...,g_k>=\emptyset$. (We need to start with a good initial $<g_1,g_2>$ to avoid getting all $F$.) |