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Let $G$ be a finite group and $N$ be a proper normal subgroup of $G$. If $N$ is not contained in in the center $Z(G)$, then prove that there exists a proper subgroup $H$ of $G$, such that $|H|>\sqrt{|G|}$.

Attempt: Since $N\not\subset Z(G)$, we get $x\in N-Z(G)$. Since $x\notin Z(G)$, we get $C(x)=\{g\in G:gx=xg\}$, the centralizer subgroup of $x$ is proper. Now $|G|/|C(x)|=[G:C(x)]=|[x]|$, where $[x]$ is the conjugacy class of $g$. Now if we can show that $|[x]|<|C(x)|$, then we will be done. Also note that $x\in N$ implies that $[x]=\{gxg^{-1}:g\in G\}\subset N$ thus we also have $|[x]|\leq|N|$. Now if we can prove that $|N|<|C(x)|$ then we'll be done. If possible let $|N|\geq |C(x)|$.

But I'm stuck after that. Please let me know if i can proceed after that, or you may give me a hint for another process.

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    Can you tell what $\sqrt G$ means? – 2017-02-20
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    @VikrantDesai: That's $\sqrt{|G|}$, square root of the order of $G$ – 2017-02-20

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Hint: You would be done if you knew that $N$ was "small" enough. How small do you need to know $N$ is, exactly, in order to finish your proof? If $N$ is large instead of small, then can you see a different way you can prove the existence of such an $H$?

A more detailed hint is hidden below:

If $|N|>\sqrt{|G|}$, then you can just take $H=N$. If $|N|\leq\sqrt{|G|}$, can you use this to prove $|C(x)|<\sqrt{|G|}$?

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    If $|N|\leq\sqrt{|G|}$, then $|C(x)|=|G|/|[x]|>|G|/|N|\geq\sqrt{|G|}$. Note that we have $|G|/|[x]|>|G|/|N|$ because $[x]\subsetneqq N$ as $e_G\notin [x]$ – 2017-02-20
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    Yep, looks good! – 2017-02-20