# Existence of a subgroup $H$ of a finite group $G$ with $|H|>\sqrt{|G|}$

The question is as follows:

Let $$G$$ be a finite group of order $$n$$ having a proper normal subgroup $$N$$ that is not contained in the center of $$G$$. Prove that $$G$$ has a proper subgroup $$H$$ with $$|H|>\sqrt{n}$$.

I believe I have most (if not all) the pieces of an argument, but I'm struggling to put them together. I know that since there exists $$N\triangleleft G$$ such that $$N\not\subset Z(G)$$, then $$|Z(G)|<\frac{n}{2}$$. Also, by application of the orbit-stabilizer theorem and Lagrange's theorem, the size of the conjugacy class of an element $$g\in G$$ is $$n$$ divided by the centralizer of $$g$$. That is, $$|cl_g|=\frac{n}{|C_G(g)|}.$$ So I think I'm trying to show that the order of the centralizer of some element $$h\in N$$ is at least $$\sqrt{n}$$, but I'm not sure.

• If it helps (I don’t know), you can assume $Z(G)$ to be the trivial subgroup (that case implies the general case). – Mindlack Sep 27 '19 at 0:31

## 2 Answers

Let $$k \in N \setminus Z(G)$$ and let $$K$$ be the conjugacy class of $$k$$ in $$G$$.

Then $$|G|/|C(k)|=|K|<|N|$$. Therefore $$|G|<|N||C(k)|$$ and at least one of $$N$$ and $$C(k)$$ has order greater than $$\sqrt{|G|}$$.

• Yes, that's much more concise than my version! – Derek Holt Sep 27 '19 at 12:38
• Thanks @Derek Holt. This is a surprisingly simple way to obtain 'large' subgroups. – S. Dolan Sep 27 '19 at 13:09

The conjugation action of $$G$$ permutes the conjugacy classes of $$N$$ and since the classes are all fixed by $$N$$, we get an induced action of $$G/N$$ on these classes. Furthermore, for $$g \in N$$, the index $$|C_G(g):C_N(g)|$$ is equal to the stabilizer of the conjugacy class $${\rm Cl}_N(g)$$ of $$g$$ in $$N$$ under this action of $$G/N$$.

Let $$g \in N \setminus Z(G)$$. Then there are at most $$|N|/|{\rm Cl}_N(g)| = |C_N(g)|$$ conjugacy classes in $$N$$ having the same size $$|{\rm Cl}_N(g)|$$ as the class of $$N$$, and so (using the fact that $$g$$ is not conjugate to $$1$$) the number of such classes in the orbit of the conjugation action of $$G/N$$ is strictly less than $$|C_N(g)|$$.

So the stabilizer of the class of $$g$$ under this action of $$G/N$$ is greater than $$|G||/(|N||C_N(g)|)$$.

So $$|C_G(g)|/|C_N(g)| > |G||/(|N||C_N(g)|)$$ and hence $$|N||C_G(n)| > |G|$$. So at least one of $$N$$ and $$C_N(g)$$ must have order greater than $$\sqrt{|G|}$$.