Bounding the number of character degrees of a finite group in terms of the order of the group - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T10:55:54Z http://mathoverflow.net/feeds/question/58635 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58635/bounding-the-number-of-character-degrees-of-a-finite-group-in-terms-of-the-order Bounding the number of character degrees of a finite group in terms of the order of the group Tobias Kildetoft 2011-03-16T13:34:36Z 2011-03-18T12:49:09Z <p>Let $cd(G)$ be the set of degrees of irreducible complex characters of the finite group $G$ (so $cd(G) = \{\chi(1) | \chi\in Irr(G)\}$).</p> <blockquote> <p>What bounds are known of the form $|cd(G)|\leq f(|G|)$ (ie, what functions $f$ are known which satisfies such an inequality)?</p> </blockquote> <p>I can show that $|cd(G)|\leq \sqrt[3]{3|G|-7}$ and if $|G|$ is odd that $|cd(G)|\leq \sqrt[3]{\frac{12}{15}|G|}$. On the other hand, one clearly has $|cd(G)|\leq d(|G|)-1$ (where $d(n)$ is the number of divisors of $n$), and this is better asymptotically. There are two reasons for this question. One is that I am looking at certain inequalities which guarantee that a group will be solvable, and having a good bound on $|cd(G)|$ in terms of the order of the group would help. Also, the question is interesting when compared to the Taketa-inequality ($dl(G)\leq |cd(G)|$ which is conjectured to hold for all solvable groups), since clearly the derived length of a solvable group only grows logarithmically in the order of the group (being bounded by the sum of the exponents in the prime factorization of the order of the group).</p> http://mathoverflow.net/questions/58635/bounding-the-number-of-character-degrees-of-a-finite-group-in-terms-of-the-order/58771#58771 Answer by Greg Kuperberg for Bounding the number of character degrees of a finite group in terms of the order of the group Greg Kuperberg 2011-03-17T17:39:25Z 2011-03-18T12:49:09Z <p>The number of divisors at least roughly resembles the best achievable lower bound. For each prime $p$, there is a group $G_p$ with $p^3$ elements which has an irreducible representation of dimension 1 (the trivial representation) and an irreducible representation of dimension $p$ (because it's non-abelian). Now let $p_n$ be the $n$th prime. The number $$P_n = p_1p_2\cdots p_n$$ is a type of number with a lot of divisors. If you likewise let $$H_n = G_{p_1} \times G_{p_2} \times \cdots \times G_{p_n},$$ then $H_n$ has an irreducible representation for every $d$ that divides $P_n$, and it has $P^3_n$ elements.</p> <p>Now, it is not quite true that $P_n$ has more divisors than any $N &lt; P_n$. It is a good strategy for making a number with many divisors, but soon enough it is better to add more factors of $p_1$, then eventually more factors of $p_2$, etc., than to keep adding new prime factors. To understand this situation better, we can make many numbers (but not all numbers) that have more divisors than their predecessors with the "threshold method". The idea is to optimize the ratio $\log(d(N))/\log(N)$ globally by optimizing it locally (with respect to prime factorization). Pick a constant $t > 0$, the threshold, and say that $N$ should have at least $k > 0$ factors of a prime $p$ if and only if $$\frac{\log(k+1) - \log(k)}{\log(p)} \ge t.$$ Then I think that $d(M) &lt; d(N)$ when $M &lt; N$.</p> <p>In fact, finding large values of $cd(G)$ (which I will use to mean the cardinality of the character degrees rather than the set) is a very similar problem when $G$ is nilpotent. A finite group is nilpotent if and only if it is the product of its Sylow subgroups. The main idea of the construction above is that in this case $cd(G)$ is multiplicative, i.e., the product of its values for $p$-groups. Following the comment by Frieder Ladisch, $cd(G)$ is maximized for $p$-groups by $C_{p^m} \ltimes C_{p^{m+1}}$. (In the first version of the answer I used other $p$-groups that aren't as good.) I.e., this group has character degrees $1,p,\ldots,p^m$, and no $p$ group with $p^{2m}$ or fewer elements can have an irrep with $p^m$ elements. So you can find many record values of $cd(G)$ for nilpotent groups using instead the threshold formula $$\frac{\log(k+1) - \log(k)}{\min(4-k,2)\log(p)} \ge t.$$</p> <p>Let's incorporate the concept of a "record value" by defining $d'(N)$ to be the maximum of $d(M)$ with $M \le N$. Likewise define $cd'(N)$ to be the maximum of $cd(G)$ with $|G| \le N$. Then I think that the above constructions show that $d'(N)$ and $cd'(N)$ are at least similar functions, and that $$d'(N) > cd'(N) > \sqrt[3]{d'(N)}$$ when $N$ is large enough. In fact I think that the exponent of the second inequality climbs from $1/3$ to some higher value, although for nilpotent groups one also has $$\sqrt{d'(N)} > cd'_{\text{nil}}(N).$$</p> <p>Let me also mention that the bound $O(\sqrt[3]{|G|})$ follows immediately from the fact that $|G|$ is the sum of the squares of the dimensions of the irreducible representations --- maybe that's what you have in mind with your bound.</p>