If $P \leq G$, $Q\leq G$, are $P\cap Q$ and $P\cup Q$ subgroups of $G$? 
$P$ and $Q$ are subgroups of a group $G$. How can we prove that $P\cap Q$ is a subgroup of $G$? Is $P \cup Q$ a subgroup of $G$?

Reference: Fraleigh p. 59 Question 5.54 in A First Course in Abstract Algebra.


$P$ and $Q$ are subgroups of a group $G$. Prove that $P \cap Q$ is a subgroup. 

Hint 1: You know that $P$ and $Q$ are subgroups of $G$. That means they each contain the identity element, say $e$ of $G$. So what can you conclude about $P\cap Q$? If $e \in P$ and $e \in Q$? (Just unpack that means for their intersection.)
Hint 2: You know that $P, Q$ are subgroups of $G$. So they are both closed under the group operation of $G$. If $a, b \in P\cap Q$, then $a, b \in P$ and $a, b \in Q$. So what can you conclude about $ab$ with respect to $P\cap Q$? This is about proving closure under the group operation of $G$.
Hint 3: You can use similar arguments to show that for any element $c \in P\cap Q$, $c^{-1} \in P\cap Q$. That will establish that $P\cap Q$ is closed under inverses.
Once you've completed each step above, what can you conclude about $P\cap Q$ in $G$?

$P$ and $Q$ are subgroups of a group $G$. Is $P\cup Q $ a subgroup of $G\;$?

Here, you need to provide only one counterexample to show that it is not necessarily the case that $P\cup Q$ is a subgroup of $G$. 


*

*Suppose, for example, that your group $G = \mathbb{Z}$, under addition. Then we know that $P = 2\mathbb{Z} \le \mathbb{Z}$ under addition (all even integers), and $Q = 5\mathbb{Z} \le \mathbb{Z}$ under addition (all integer multiples of $5$). So $P \cup Q$ contains $2\in P,$ and $5 \in Q.\;\;$ But:$\;$ is $\;2 + 5 = 7 \in P\cup Q\;$? 
So what does this tell regarding whether or not $P \cup Q$ is a subgroup of $\mathbb{Z}\;$?

