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 ...

answered Nov 29 at 18:03

David E Speyer

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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 ...

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