Prove that the intersection of two subgroups is a subgroup. In more detail, if  $G$  is a group and $H_1$, $H_2$   are subgroups of G   then $H_1 \cap H_2$   is a subgroup of  G.
Next, give an example of a particular group $G$  (any one you like), and two different subgroups  $H_1$, $H_2$ of $G$  , compute the intersection $H_1 \cap H_2$  , and verify it is indeed a subgroup. 
Finally, give three examples showing that $H_1 \cup H_2$   need not be a subgroup of $G$ .

Theorem Let $G$ be a non-trivial finite group. Then the following are equivalent.
(a) For each pair of subgroups $H_1$ and $H_2$ of $G$, $H_1 \cup H_2$ is a subgroup
(b) $G$ is cyclic of prime-power order.
Proof (b) $\Rightarrow$ (a) follows from the fact that in a cyclic group there is a unique subgroup of order $d$ for each divisor $d$ of $|G|$. (If $G$ has order $p^n$ and $G=\langle g \rangle$, then define $H_i=\langle g^{p^{n-i}} \rangle$. Then $|H_i|=p^i$ and the series $\{1\} =H_0 \subset H_1 \dots \subset H_{n-1} \subset G=H_n$ are all the subgroups of $G$.)
(a) $\Rightarrow$ (b). Assume that $G$ is not cyclic. Then we can find a non-identity $x \in G$, with $\langle x \rangle \subsetneq G$, and take $|\langle x \rangle|$ as large as possible. Also, we can find a $y \in G-\langle x \rangle$. The assumption implies $H=\langle x \rangle \cup \langle y \rangle$ is a subgroup. Clearly $x$, $y$ $\in H$, but $xy \notin H$, since $x$ is not a power of $y$ and vice versa: if $xy=x^i$ for some integer $i$, then $y=x^{i-1} \in \langle x \rangle$, a contradiction. If $xy=y^j$ for some integer $j$, then $x=y^{j-1} \in \langle y \rangle$, whence $\langle x \rangle \subseteq \langle y \rangle$, and hence $\langle x \rangle = \langle y \rangle$, by the maximality of $\langle x \rangle$. Again a contradiction. So $G$ must be cyclic. If $|G|$ has two different prime factors $p$ and $q$, the by Cauchy's Theorem we can find elements $a$ and $b$ of that orders respectively, and $a$ and $b$ are powers of a $g$ with $\langle g \rangle = G$. Again, $\langle a \rangle\cup \langle b \rangle$ is a subgroup of $G$, but it does not contain $ab$, that has order $pq$. A contradiction, the order of $G$ must be divisible by a single prime, whence $G$ is of prime-power order.$\square$
One can also use the well-known fact that a group can never be the union of two proper subgroups. Anyway, the theorem provides you a slew of examples for the second part of your question.
