# Tag Info

9

Suppose $G$ is a finite simple group of order $n$ with a nontrivial representation of degree $d$. Then $G$ is isomorphic to a subgroup of $U(d)$. By Collins's sharp version of Jordan's theorem (https://www.degruyter.com/document/doi/10.1515/JGT.2007.032/html), $G$ has an abelian normal subgroup of index at most $(d+1)!$, which must be trivial since $G$ is ...

8

Based on Sean Eberhard's comment, it follows that if $f(n)$ is the number of non-isomorphic abelian subgroups of $S_n$, then $f(n)\geq p_{\mathbb{P}}(n)$, where $p_{\mathbb{P}}(n)$ is the number of partitions of $n$ into prime parts. It was proved by Roth and Szekeres that $$\log p_{\mathbb{P}}(n) = \frac{2 \pi}{\sqrt{3}}\Big(\frac{n}{\log n}\Big)^{1/2}\Big(... 7 Claim 1: (same as my comment) Let q_1, \dots, q_k > 1 be prime powers. Then G = C_{q_1} \times \cdots \times C_{q_k} is isomorphic to a subgroup of S_n if and only if q_1 + \cdots + q_k \leq n. Proof: If q_1 + \cdots + q_k \leq n we can choose disjoint cycles of lengths q_1, \dots, q_k, so the condition is clearly sufficient. Conversely ... 6 It is a standard fact that the automorphism group of G=\mathrm{SL}_2(\mathbf{R}) as topological group equals \mathrm{PGL}_2(\mathbf{R}), which is also the automorphism group of the Lie algebra viewed as \mathbf{R}-algebra. In particular, the outer automorphism group as topological group is cyclic of order 2. I claim that this is the same as abstract ... 6 The answer is yes: \text{Rad}(kG)^s is generated as an ideal by (g-1)^s for G an elementary abelian p-group and s \leq p-1. Lemma: Let V be a k-vector space and let s \leq p-1. Then \text{Sym}^s(V) is spanned by the elements v^s for v \in V. Proof: \text{Sym}^s(V) is clearly spanned by products of the form v_1 v_2 \cdots v_s. We ... 3 Under these assumptions, note that \mathrm{Rad}(kG) coincides with the augmentation ideal \omega(kG) of kG. Now, as remarked by Jeremy Rickard, the question is equivalent to whether the elements (g-1)^s with g\in G generate \mathrm{Rad}^s(kG) as an ideal of kG. Theorem 3.7 in Section 5.3 of the book [D.S. Passman: The algebraic structure of ... 3 Yes, there is a combinatorial way to see this. Firstly, it is much easier to show a baby version of this type of phenomenon. A while ago I read a nice blog post by Qiaochu explaining the identity$$\int_{0}^1 (2\cos \pi x)^n (2\sin^2\pi x)dx=\begin{cases} 0 &\text{ if n is odd} \\ C_{n/2} & \text{ if n is even} \end{cases} by interpreting this ...

3

This was not originally an answer to the question you asked but to the converse question about geodesic paths in $X$ mapping to quasigeodesic paths in $\widehat{X}$. But you asked for a reference for that in a comment, so here it is. But note that this result involves extending the given generating set of $G$ to a larger finite generating set by adjoining ...

2

This is a (cw) post to illustrate Claim 2 from the answer by Sean Eberhard. Below is the plot of the list $(\frac1n,\frac{\log f(n)}{\pi\sqrt{2/3}(n/\log n)^{1/2}})$ for $n$ from $1000$ to $10000$. I used the list of values from OEIS A023893. As you see after $n$ about 3000 the sequence sort of changes behavior; at about $n=5000$ (to be precise, at $5226$) ...

2

We denote by $\mathbb{Z}/m$ the cycle group of order $m$. Some examples: The finite groups $\mathbb{Z}/{p_1^{a_1}}\oplus \cdots \oplus \mathbb{Z}/p_k^{a_k}$, where the $p_i$ are pairwise distinct primes and the $a_i$ are positive integers such that $\sum_i p_i^{a_i}\leq n$, can be clearly embedded into $S_n$. Edit: I also thought to be true that whenever $n=... 2 Theorem (Elek–Szabo). Let$G$be an infinite residually finite hyperbolic group with Property (T). Then$P_G\$ is a finitely generated sofic group that is not residually amenable. That's Theorem 3 of Elek, Gábor; Szabó, Endre, On sofic groups., J. Group Theory 9, No. 2, 161-171 (2006). ZBL1153.20040 arXiv:math/0305352. The construction is also somewhat ...

Only top voted, non community-wiki answers of a minimum length are eligible