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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 $P_{D}$

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Doublecheck your first counterexample. Why is it important that $a$ and $b$ are in $H$? And are they? –  Adam Saltz Aug 9 '11 at 2:58
3  
Your specific example of $a=0^\circ,b=90^\circ$ is invalid because $90^\circ\not\in H$. Use instead $45^\circ\in H, 45^\circ+45^\circ\not\in H$ to show $H$ isn't closed under addition. –  anon Aug 9 '11 at 3:05
    
@IAmBrianDawkins: Yes I see my mistake now. My original examples were not in $H$ the first place. –  Student Aug 9 '11 at 3:14
    
What is $P_C$? Is that notation for the power set of $C$? –  Gerry Myerson Aug 9 '11 at 3:18
1  
The second proof looks good to me, with the small omission that you have to note that $P_C$ is non-empty. –  Gerry Myerson Aug 9 '11 at 3:35

1 Answer 1

up vote 3 down vote accepted

For the first question, I am not sure why you computed $\tan a + \tan b$ ? If you want to check that $H$ is closed under addition (which is required to be a subgroup), then you have to check whether the sum two elements still belongs to the set $H$.

If you have two elements $x,y\in H$, then $\tan x \in \mathbb{Q}$ and $\tan y \in \mathbb{Q}$. Now you want to check if $x+y \in H$, which means that you want to check whether $\tan (x+y) \in \mathbb{Q}$. You should be able to use the same identities as you already did to solve this though.

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If I have two elements tan x and tan y in H, don't I want to see if tan x + tan y is in H and not tan(x + y)? –  Student Aug 9 '11 at 4:18
    
You are absolutely right. I originally thought of $x$ representing $\tan x$ instead of viewing it as a real number. –  Student Aug 9 '11 at 4:31

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