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Pyber showed that the number of groups of order $n$ is $\leq n^{\frac{2}{27}\nu(n)^3+C\nu(n)^{3/2}}$, where $\nu$ is the highest power of a prime dividing $n$ and $C$ is an absolute constant. On the o …

You can avoid checking all $\binom{n}{k}$ subsets of the rows by taking $k$ random combinations of the rows, that is, compute the determinant of $$ \left(\sum\xi_{1i}a_i, \sum\xi_{2i} a_i, \ldots, \su …

It is not true anymore that a proof of this conjecture would lead to significant simplifications. Peterfalvi proved in 1984 a weaker version of this conjecture, which suffices to get rid of the chapte …

Let $\Gamma$ be a free group over the alphabet $X$, $|X|\geq 2$, and put $S=X\cup X^{-1}$. Pick an increasing sequence of integers $n_i$, and put $\Lambda=\{g\in\Gamma|\exists i:\ell(g)=n_i\}$. Under …

P. Hall gave a formula for the number of generators of $G^n$ for any finite simple group $G$. One famous example is the fact that $A_5^{19}$ is 2-generated, but $A_5^{20}$ is not. The question of comp …

One area where logic really helped group theory is the theory of zetafunctions of torsionfree nilpotent groups. Define $\zeta_G(s)=\sum_{U\leq G} (G:U)^{-s}$, where summation runs over all finite inde …

Fuchsian groups are important for the theory of dessin d'enfants in arithmetic geometry. Wolfart wrote a survey here: http://www.math.uni-frankfurt.de/~wolfart/Artikel/abc.pdf . One typical applicati …

Looking at the exponential series, we have $\frac{r^k}{k!}\leq e^r$. Hence we have $$ \frac{r^{m-k-r}}{(m-2k-r)!r!k!} = \frac{1}{r!}\frac{r^k}{k!}\frac{r^{m-2k-r}}{(m-2k-r)!} \leq\frac{e^{2r}}{r!}=\ma …

In general not. Consider the homomorphism $\mathrm{Aff}(p)\rightarrow\mathbb{F}_p^*$ mapping $ax+b$ to $a$. The probability that two random elements generate a subgroup for which the induced map is st …

In general the problem is very difficult. There has been quite some work on the diameter of the Cayleygraph of $S_n$, the best results being due to Helfgott-Seress for the general case, and Helfgott-S …

Pick a prime $p$ in the range $\frac{n}{2}<p\leq n-k$. Then $S_{n-k}$ contains $(p-1)!\binom{n-k}{p}$ elements of order $p$, and elements of order $p$ in $S_n$ are $p$-cycles. There are at most $(p-1) …

One method that often works is the following: A primitive group, which contains a $p$-cycle for a prime $p<n-2$ is $A_n$ or $S_n$. If a permutation $\pi$ contains a cycle of length $p$, and no other c …

A chain in $[H,G]$ has length $\leq\Omega(|G:H|)$, where $\Omega$ denotes the number of prime factors counted with multiplicity. If $H=H_0<H_1<\dots<H_k=G$ is a maximal chain, then there are elements …

Suppose that $(S-n)\cap S$ has upper density 0 for all $n\leq N$. Then the density of the set of integers $x$, such that $|[xN, (x+1)N]\cap S|\geq 2$ is also 0, hence the upper density of $S$ is $\leq …

I prefer the proof going via 3-cycles. It is probably the least elegant proof, but it is the one which you probably would have found when considering the problem without any prior knowledge. Also the …