Could I get some feedback on my work below? Thanks in advance.
- $G = \langle \mathbb{R}, + \rangle , H = \{ x \in \mathbb{R}: \tan x \in \mathbb{Q} \}$
(i). If $\tan a$ and $\tan b$ $\in H$, then from trigonometry \begin{align*} \tan a + \tan b = \tan(a + b) (1 - \tan a \tan b) \notin H \end{align*} For example, if we let a = 0$^{\circ}$ and b = 90$^{\circ}$, then $\tan a + \tan b = \infty \notin \mathbb{Q}$.
Thus $H$ must NOT be a subgroup of $G$.
- Let $C$ and $D$ be sets, with $C \subseteq D$. Prove that $P_{C}$ is a subgroup of $P_{D}$ where $P_{C}$ and $P_{D}$ are the power sets of C and D respectively
$C \subseteq D$ means that every element in $C$ is an element in $D$. And the operation in this case is the symmetric difference $(\Delta)$ where the nullset is the identity and each element is its own inverse.
In order to show that $P_{C}$ is a subgroup of $P_{D}$:
(i). Let $A$ and $B$ be any two sets in $P_{C}$. The symmetric difference $A \Delta B$ must also be a set in $P_{C}$. (ii). Let $A$ be a set in $P_{C}$. Then the inverse of $A$ must also be in $P_{C}$.
(i). The symmetric difference of two sets $A, B \subseteq P_{C}$ is defined as $A \Delta B = (A \setminus B) \cup (B \setminus A)$. Since $(A \setminus B)$ and $(B \setminus A)$ will always yield sets that are contained in $P_{C}$, we can conclude that $P_{C}$ is closed under the operation of symmetric difference.
(ii). Since the inverse of any set $A$ is itself, then the inverse of every element must also be contained in $P_{C}$. So $P_{C}$ is closed under inverses.
Thus $P_{C}$ is a subgroup of $P_{D}$