For the following question related to lagrange's theorem: suppose that $H$ and $K$ are unequal subgroups of a group $G$, each of order $16$. Prove that $24 \leq |H \cup K| \leq 31$.  

There are some confusions I am hoping someone can clarify for me please.

(1) When it says "unequal subgroups", does it mean $|H\cap K|=\{1\}$?  

(2) Am I correct to assume that $H\cup K$ is a subgroup and hence either $K\subset H$ or $H \subset K$. Then either $|H|||K|$ or $|K|||H|$   

(3) There is another theorem that states: if $H$ and $K$ are subgroups of $G$, $<H\cup K>=HK$. I don't think this fact is relevant.   

(4) to get $31$ on the right hand side. We have $|H \cup K|=|H|+|K|-|H\cap K|=16+16-1=31$  

(5) for the left lower bound $24$, I am not sure how to get it. Also could it be even lower like to $3$ for $|H|=|K|=2$ and $|H\cap K|=1$  

Thank you in advance.