The question is:
If $G$ is the group of all nonzero real numbers under multiplication and $N$
is the subgroup of all positive real numbers, write out $G/N$ by exhibiting
the cosets of $N$ in $G$, and construct the multiplication in $G/N$.
Okay I am confused about what is sufficient answer in writing out the quotient group by exhibiting the cosets of $N$ in $G$: for example I wrote $G/N = \{[a] | a \in G\} = \{Na | a \in G\}$. Is that enough?
And for the construction of multiplication do I use $ [a] = [a']$ and $[b] = [b'] $to end up with $NaNb = Nab$?
Final question: we learned that you need a normal subgroup to prove this for for the given problem should I strt out by proving $N$ is a normal subgroup of $G$.
Best Answer
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).