It’s exactly the same reasoning as in my answer to the earlier question. In fact, I was simply proving the special case of this proposition that’s needed for the proof of the theorem. To make the connection between this proposition and its application to the theorem, note that the $G,H,f$, and $Y$ of the proposition correspond to the $G,G/H,q$, and $q(H)$ of the theorem.
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).
Best Answer
Mac Lane doesn't just claim this, he cites the book Algebra by Garret Birkhoff and Saunders Mac Lane (not to be confused with their earlier book A Survey of Modern Algebra) wherein the proofs are carried out (more precisely, pp. 80,81,410-412). The only aspect of normality used in the proofs (at least upon my skimming it) is that normal subgroups are exactly the kernels. In other words, if you replace "normal subgroup" with "kernel of a group homomorphism" in the various statements, the proofs should go through word for word. From this point of view, the coset construction is only a characterization of which subgroups are kernels.
For example, if $N$ is the kernel of $\phi\colon G\to L$, then the universal property of the quotient $G\to G/N$ implies that $N$ is in the kernel of $G/N$ and that we have a factorization $G\to G/N\to L$ of $G\to L$ for a unique $G/N\to L$. But then the kernel of $G\to L$, which is $N$, is contains the kernel of $G/N$, which contains $N$, so $N$ is the kernel of $G/N$.
A more categorical understanding of normality does exist. Namely, a monomorphism $N\hookrightarrow X$ is normal if it is contained in an equivalence class of an equivalence relation on $X$. For more details (though pitched at a somewhat advanced level), see the book From Groups to Categorical Algebra - Introduction to Protomodular and Mal'tsev Categories, by Dominuqe Bourn, or for an even more advanced text, the book Mal'cev, Protomodular, Homological and Semiabelian Categories by Francis Borceux and Dominique Bourn. However, unless the category is exact, normal subgroup are not necessarily kernel, e.g. for Hausdorff topological group kernels are closed normal subgroups.