# Inequality for Character Degrees of finite groups

Let $G$ be a finite group, and let $d_1,d_2,\dots,d_n$ be the dimensions of the irreducible representations. It is well-known that $\sum_{i=1}^n d_i^2=|G|$. If I am not mistaken, one has the following inequality $$\sum_{i=1}^n d_i^{3/2}\geq \sqrt{|G|}.$$ Is this known or obvious (or false)? For abelian groups the inequality is far from sharp, because we can replace the right hand side with $|G|$. But for the symmetric group it seems pretty sharp. For example, for $S_6$, the left-hand side is 27.2688 and the right-hand side is $26.8328$.

• Hi Harm, welkom bij mathoverflow :) Did you check small examples, say, with GAP? GAP has lots of character tables stored (or it can compute them, too). – Dima Pasechnik Nov 1 '12 at 17:06
• The character degrees of S_6 are {1, 5, 9, 5, 10, 16, 5, 10, 9, 5, 1}. How do you get 27.2688? I get 227.96. – John Wiltshire-Gordon Nov 1 '12 at 17:33
• it was a typo - see Harm's own answer below... – Dima Pasechnik Nov 1 '12 at 17:37

## 2 Answers

Never mind. The inequality that I wrote down is obvious, because $$(\sum_{i=1}^n d_i^{3/2})^2\geq \sum_{i=1}^n d_i^3\geq \sum_{i=1}^n d_i^2=|G|.$$ I accidentally calculated $\sum_{i=1}^n \sqrt{d_i}$ on the left-hand side. So it might be an interesting question whether $$\sum_{i=1}^n \sqrt{d}_i\geq \sqrt{|G|}.$$ I'm pretty sure this fails for large symmetric groups, because the left-hand side is less or equal than $p(n)|S_n|^{1/4}$ where $p(n)$ is the number of partitions. $p(n)$ grows subexponentially, whereas $|S_n|$ grows superexponentially. So we have $p(n)<|S_n|^{1/4}$ for large $n$. So the left-hand side will eventually be smaller than $\sqrt{|S_n|}$.

• the 2nd displayed inequality already fails for $S_7$. One gets $55.1\geq 70$... – Dima Pasechnik Nov 1 '12 at 17:30

Let $G$ have $k$ conjugacy classes. Then Cauchy Schwarz seems to give $\sum_{i=1}^{k} \sqrt{d_i} \leq \sqrt{k}\sqrt{\sum_{i=1}^{k} d_i }$ and this (using C-S) again is at most $k^{\frac{3}{4}} |G|^{\frac{1}{4}}$. So as long as $k^{3} < |G|,$ the inequality you state is violated. There are many groups $G$ for which $k^{3} < |G|.$ One example is the alternating group $A_{6}$ which has $7$ conjugacy classes and order $360 >7^{3}.$

• I am not sure if the text of the question was changed, but this answer now seems tto give counterexamples to the question : "Is $\sum_{i=1}^{k} \sqrt{d_{i}} \geq \sqrt{|G|}$ " asked in the OPs own answer. – Geoff Robinson Jun 13 '19 at 14:21