You learned this a long time ago:
"Positive times positive is positive",
"Positive times negative is negative",
"Negative times positive is negative",
"Negative times negative is positive".
The quotient group is of order 2. $N$ is automatically normal because $G$ is abelian. By definition:
$aN = bN$ if $a^{-1}b \in N$, which happens if and only if either both $a,b \in N$ or $a,b \not\in N$.
The two cosets are, explicitly: $N$ and $-N$, and these can be regarded as $[1]$ and $[-1]$.
This uses properties of the real numbers, which you are expected to know. The relevant property, here, is:
$\forall x \in \Bbb R-\{0\},\ x^2 > 0$ (that is: $x^2 \in N$).
This shows that for ANY coset $xN$, we have $(xN)(xN) = x^2N = N$.
So our cosets can only have order 1 ($N$ itself), or order 2.
Now if $y \not\in N$ (that is $y < 0$, see below) we have:
$(-1)^{-1}y = (\frac{1}{-1})y = (-1)y = -y \in N$, so that $yN = (-1)N$ which shows that the only OTHER coset is: $(-1)N = -N\ $ (this uses the trichotomy rule of the order on the real numbers).
Hint: in particular, $H$ is an element (the zero element) of $G/H$. What does it look like geometrically? Can you graph it? Once you know that, the other cosets are just translates of $H$, so they should look the same, just moved around. For instance, what does the coset $(0,1) + H$ look like?
Best Answer
Hints: (1) For $\lambda \in \def\Cs{\mathbb C^\times}\Cs$, can you compute $\lambda H =\{\lambda h \mid h \in H\}$, the coset of $\lambda$?
(2) For using the isomorphism theorem, you need an epimorphism $p \colon \Cs \to \Cs$ with kernel $H$. Can you think of one? As you write, $H$ consists of the forth roots of unity, so $h^4 = 1$ exactly for $h \in H$.