Proving independence of discrete variables and the product of them Given that P(A) and P (B) are independent and
\begin{equation}
$P(A=1)=\frac{1}{2}$
$P(A=-1)=\frac{1}{2}$
$P(B=1)=\frac{1}{2}$
$P(B=-1)=\frac{1}{2}$
There is a random variable $C = A \cdot B$,
C is independent with A, with B but not with A and B.
I have a couple of questions about this:
1.Will C be
$P(C=1)=\frac{1}{2}$
$P(C=-1)=\frac{1}{2}$
because $P(A=1)P(B=1)+P(A=-1)P(B=-1)=\frac{1}{2}$ and $P(A=-1)P(B=1)+P(A=-1)P(B=1)=\frac{1}{2}$? Is it allowed just to multiply $P(A)$ and $P(B)$ to get $P(C)$?
2.How can I prove that C and A are independent? Do I have to prove that $P(A, C)=P(A)P(C)$? Can I find $P(A, C)$? Is $P(A,C)$ the same as $A \cdot C$?
How can I also prove that C and A and B are NOT independent? Do I show that $P(C,(A,B)) \neq P(C)P(A,B)$?
Will P(A,B) be
\begin{equation}
P(A, B)=
    \begin{cases} 
          \frac{1}{4}, \text{if A = 1, B = 1},\\
          \frac{1}{4}, \text{if A = 1 ,B = -1},\\
          \frac{1}{4}, \text{if A = -1,B = 1},\\
          \frac{1}{4}, \text{if A = -1,B = -1}\\
    \end{cases}
\end{equation}

When working with binary variables (or other variables with small numbers of possible values), it helps to make a table.
A complete table will display all possible combinations of the values along with the probabilities of those outcomes.  (It is a good idea to enumerate the combinations systematically, because that reduces errors and reveals patterns.  Below, I have listed the outcomes in lexicographic order.)  At the outset, we often don't know all the probabilities.  In this case, I will just call them $p,q,r,$ etc.
Here is the most general table describing your situation.  It was constructed by enumerating all possible combinations of $(\mathscr{A},\mathscr{B})$ and computing their products in the $\mathscr{C}$ column.
$$\begin{array}{l|rrr}
\text{Probability} & \mathscr{A} & \mathscr{B} & \mathscr{A}\mathscr{B}=\mathscr{C}\\
\hline
p & -1 & -1 & 1\\
q & -1 & 1 & -1\\
r & 1 & -1 & -1\\
s & 1 & 1 & 1
\end{array}$$
It will always be the case that the probabilities sum to $1.$  Additionally, your information includes four more equations that can be read directly off the table.  For instance, the event $\mathscr{A}=-1$ comprises the first two rows of the table, whence its chance must be the sum of the probabilities in those rows, $p+q.$  Proceeding in this fashion you may instantly write down all equations by inspecting the table:
$$\begin{aligned}
1 &= p + q + r + s\\
1/2 &= \Pr(\mathscr{A}=-1) = p + q\\
1/2 &= \Pr(\mathscr{A}=1) = r + s\\
1/2 &= \Pr(\mathscr{B}=-1) = p + r\\
1/2 &= \Pr(\mathscr{B}=1) = q + s
\end{aligned}$$
Straightforward linear algebra proves the most general solution to this system of linear equations is determined by one parameter.  Let's use $p.$  Here is that general solution.
$$\begin{array}{l|rrr}
\text{Probability} & \mathscr{A} & \mathscr{B} & \mathscr{C}\\
\hline
p & -1 & -1 & 1\\
1/2-p & -1 & 1 & -1\\
1/2-p & 1 & -1 & -1\\
p & 1 & 1 & 1
\end{array}$$
Finally, because probabilities cannot be negative, you must restrict these solutions to $0 \le p \le 1/2.$  As an extreme example, when $p=1/2$ the distribution is
$$\Pr(-1,-1) = \Pr(1,1) = 1/2.$$
You may check this is consistent with all the information in the question.  Extreme examples (on the boundary of the constraints) often are simple and provide useful examples to help you think through the concepts.  I encourage you to do that with this one.
So: to reason about independence, apply the definition of independence to this general solution.  I leave the easy calculations to you.
