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.
 A: $P\cup Q$ need not be a subgroup. For example, $2\mathbb Z$ and $3\mathbb Z$ are subgroups of $\mathbb Z$, the group of integers under ordinary addition, but their union is not a subgroup because $2\in 2\mathbb Z$, $3\in 3\mathbb Z$ but $2+3=5\not\in 2\mathbb Z\cup 3\mathbb Z$.
To show that $P\cap Q$ is a subgroup, note that $e\in P$ and $e\in Q$ so that $e\in P\cap Q$ and $P\cap Q$ is nonempty. If $a\in P\cap Q$ and $b\in P\cap Q$, then $a\in P$, $a\in Q$, $b\in P$ and $b\in Q$, so $ab^{-1}\in P$ and $ab^{-1}\in Q$, so $ab^{-1}\in P\cap Q$.
A: For the last part: $P \cup Q$ is a subgroup if and only if $P \subset Q$ or $Q \subset P$.
$\Leftarrow$ is obvious
$\Rightarrow$ Assume by contradiction that this is not true. Pick $x \in P \backslash Q$ and $y \in Q \backslash P$.
Then $x,y \in P \cup Q$ implies $x+y \in P \cup Q$, hence $x+y$ in either $P$ or $Q$. But then, either $x=(x+y)-y \in Q$ or $y=(a+y)-x \in P$ contradiction.
A: Since $1 \in H_{1} \cap H_{2}$, the intersection is not empty. Now let $x, y \in H_{1} \cap H_{2}$. Then $xy^{-1} \in H_{1}$ and $H_{2}$ since $x, y \in H_{1}$ and $H_{2}$. Thus $H_{1} \cap H_{2} \leqslant G$.
Consider $H_{1} = 2\mathbb{Z}$ and $H_{2} = 3\mathbb{Z}$. Then we have $H_{1} \cup H_{2} = 2\mathbb{Z} \cup 3\mathbb{Z}$, which does not have $3 - 2 = 1$.
A: 
$P$ 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 $\mathbb{Z}\;$?

A: Counter: Letting $x = \left( \textrm{1 2 3}  \right), y = \left( \textrm{1 2}  \right)$. Then $\langle x \rangle , \langle y \rangle$ are subgroups of $S_3$, but their union is not (it is not closed).  
Intersection: if $x, y \in H \cap K$, then $x, y \in H, K$. From this, closedness, existence of inverses and identity in $H, K$ follow easily, and so in $H \cap K$.
