Show that the intersection of two subgroups is the identity or that they're both equal Let $G$ be a group and suppose $A,B<G$ with order $7$, i.e. $|A|=|B|=7.$ Show that either $A\cap B=\{1\}$ or $A=B.$ 
My attempt:
Observe that $A\cap B$ is a subgroup of $A$ and $B.$ Then, by Lagrange's Theorem, the intersection must divide $7$. So its either $1$ or $7$. Therefore, either $A\cap B=\{1\}$ or $A=B.$ $\square$
For the last part, would I need to give more of an explanation?

The idea is ok, but I think you should justify the steps $|A\cap B|=1\implies A\cap B=\{1\}$ and $|A\cap B|=7\implies A=B$ a bit more. It is easy but still better to make explicit.
