TITLE: Existence of a subgroup $H$ of a finite group $G$ with $|H|>\sqrt{|G|}$
QUESTION [7 upvotes]: 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.

REPLY [9 votes]: 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 $\sqrt{|G|}$.