If $G$ is a group of order $96$ and $H,K$ are subgroups of order $12,16$, can $H∩K$ be trivial?

Question

Question: Given that $G$ is a group of order $96$ and $H$ & $K$ are subgroups of order $12$, $16$ respectively. Then, is $H∩K≠\{e\}$ always?

According to me, if $G$ is ablelian then we must have $H∩K≠\{e\}$ because otherwise order of $HK$ will be greater than order of group $G$ and we get contradiction. Hence if $G$ is abelian group of order $96$ then we must have $H∩K≠\{e\}$

But what when $G$ is non-abelian? Is $H∩K≠\{e\}$ also holds if $G$ is non-abelian? OR there is some non abelian group of order $96$ such that $H∩K=\{e\}$? I have been stuck on this problem for hours, can someone please help me?...

Answer 1

Assume that $H\cap K=\{e\}$ is trivial. Consider function

$$f:H\times K\to G$$ $$f(h,k)=hk$$

Then $f$ is injective. Indeed, if $f(h,k)=f(h',k')$ then $h'^{-1}h=k'k^{-1}$ and since both sides belong to $H$ and $K$ then they are both equal to $e$ meaning $h=h'$ and $k=k'$.

Since $f$ is injective then $|\text{im}(f)|=|H\times K|$ and thus

$$|G|\geq|\text{im}(f)|=|H\times K|=|H|\cdot |K|$$

The first inequality simply follows because $\text{im}(f)$ is a subset of $G$. And so the inequality $|H|\cdot|K|\leq |G|$ and the inequality $12\cdot 16>96$ yield contradiction. Therefore $H\cap K\neq\{e\}$.