Let $H$ and $K$ be subgroups of finite order of group $G$ such that $|H|> \sqrt{|G|}$ and $|K|> \sqrt{|G|}$. Show that $|H \cap K|>1$.
Since $H$ and $K$ are subgroups of $G$. Hence $e\in H $ as well as $e\in K $(,where $e\in G$ is the identity element of $G$). So clearly $e\in H\cap K$. Thus $|H\cap K |\geq 1$. But how to show, $|H\cap K |> 1$? I am not quite getting it. Also, there might be similar posts concerning the same topic on this site, but I can't seem to find it either . . .
We know that :$$\lvert G\rvert \geq \lvert HK \rvert=\frac{\lvert H \rvert \lvert K \rvert }{\lvert H\cap K\rvert }> \frac{\lvert G\rvert}{\lvert H\cap K\rvert}$$ So, $1>\frac{1}{\lvert H\cap K\rvert} \cdots $
By Lagrange's theorem in a finite group, the order of a subgroup divides the order of the group.
Thus: $|H|, |K|$ divide $|G|$
Now $e\in H\cap K\implies |H\cap K|\geq 1$.
Since $|H||K|>|G|$ by assumption: let us take the right cosets of $S=H\cap K$
Index of a subgroup is defined $p=[G:S]=\frac{n}{m}$ and it is the number of right cosets of $S$.
Thus: we have partition of $G$:
$\{Sg_1, Sg_2,...Sg_p\}, \{g_1,...g_p\}\in G$.
Where $n,m$ are the orders of $G,S$ respectively.
Now each coset of $H\cap K$ is also a coset of $H$ and a coset of $K$.
So the indices of $H,K$ are smaller than $\frac{|G|}{\sqrt{|G|}}$.
This implies at least 2 cosets of $S$. i.e. $Se, Sg$, $e$ is the identity.
this implies $g\in S$.
