Show that if $A\cup B = A\vee B$ for subgroups $A$ and $B$, then $A\subseteq B$ or $B\subseteq A$ I have that 
$$A\cup B = A  \vee  B$$
My book defines $A  \vee  B$ as being:
$$\cap\{T: \text{T is a subgroup of $G$ and $A\cup B \subseteq T$}\}$$
So, if I take the intersection of all subgroups of $G$ that contains $A\cup B$, I'll have a subgroup, since the intersection of all subgroups is a subgroup. So I end up knowing that $A\cup B$ is a subgroup.
But this, somehow, migth implie that $A\subseteq B$ or $B\subseteq A$.
I tried to start supposing that $A\nsubseteq B$, then there is an $a \in A$ such that $a\notin B$. If we take $b \in B$, then the product $ab$ can end in $A$ or $B$. I remember something like this from my teacher, but I can't find a contradiction. Any helps?
 A: If $a\in A$ and $b\in B$ and $ab\in B$, then $ab=b'$ for some $b'\in B$.  Thus $a=b'b^{-1}$ and $a\in B$.  If $ab\in A$, then $b\in A$.  So, $a\in B$ for all $a\in A$ or $b\in A$ for all $b\in B$.
A: Hint: The general fact is that if the union of $2$ subgroups is a subgroup, then one of them is contained in the other.
To see this, suppose none is contained in the other, and take an element $a\in A\smallsetminus B$, an element $b\in B\smallsetminus A$. If $A\cup B$ is a subgroup, ask yourself where  $ab$ is.
Note: there are examples of the union of 3 subgroups being a subgroup.
A: Here's a continuation of your attempt:
Since $A \not\subseteq B$, there exists some  $a \in A$ such that $a \notin B$. 
Let $b \in B$. We want to show that $b \in A$.
The element $ab$ is in $A\cup B$ (because $A\cup B$ is a subgroup containing both $a$ and $b$) and is not in $B$ (because if it were then $a=(ab)b^{-1} \in B$ by closure).
Hence $ab \in A$. But then $b=a^{-1}(ab) \in A$ too (because $A$ has inverses and closure).

Note that you're not looking for a contradiction; this proof is actually a direct proof by cases where case 1 is "$A\subseteq B$" (then done) and the second case is "$A\not\subseteq B$" (then $B\subseteq A$). 
