# Categories and the Isomorphism Theorems

I've been trying to work through Mac Lane's "Categories for the Working Mathematician" on my own, but I seem to be struggling with the concept of universality (arrows and elements). In particular, I seem unable to do one of the exercises in the book, which amounts to proving the familiar last 2 isomorphism theorems for groups:

"Use only universality (of projections) to prove the following isomorphisms of group theory:

(a) For normal subgroups $M$ and $N$ of $G$, with $M\subset N$, $(G/M)/(M/N)\cong G/N.$ (I believe there is a typo in the book, as it says $(G/M)/(M/N)\cong G/M$.)

(b) For subgroups $S$ and $N$ of $G$, $N\lhd G$, $SN/N\cong S/(S\cap N)$."

Any help with these two problems (or any info that would shed some light on the whole concept of universality) would be appreciated.

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The key point is to show that one of the objects has the universal property of the other and conclude that they are isomorphic. For example, in point (a) you should show that $(G/M)/(M/N)$ has the universal property of $G/N$. This is: for each map $G\to H$ vanishing on $N$ there is a (unique)map from $(G/M)/(M/N)$ to $H$ making the appropriate diagram commutative. – Quimey Apr 6 '12 at 19:33
@Quimey hmm, I don't think I understand. The definition talks about a "universal arrow" for a fixed element to a functor. I'm having trouble even seeing what the categories I'm dealing with are, and what the functor between them is. – FPP Apr 6 '12 at 20:30
You are correct about the typo in (a), by the way... – Arturo Magidin Apr 6 '12 at 21:34
@CMath Ok, Mac Lane uses a different notation. The functor to consider is the functor from $F$ Groups to Sets such that $F(K)$ is the set of group morphisms from $G$ to $K$ vanishing on $N$. The pair $\langle G/N,p\rangle$ is an universal element of $F$. Now, you can show that $\langle (G/M)/(M/N),\overline{p}\rangle$ is also an universal element. – Quimey Apr 7 '12 at 4:41
Oops! Re-reading the original question, it should be $(G/M)/(N/M) \cong G/N$. I caught my mistake too late to edit, sorry, folks. – David Wheeler Apr 8 '12 at 12:58

What is meant by: "the universal property of the canonical projection $p:G \to G/N$", is that if we have $f \in \mathrm{Hom}(G,G')$ such that $f(N) = \{e_{G'}\}$, then there exists a unique $f'$ with $f = f'\circ p$. This is just the First Isomorphism Theorem in disguise, the $f'$ in question is: $f'(gN) = f(g)$, which is well-defined since $f(n) = e_{G'}$ for all $n \in N$ (so if $gN = g'N$ then $g' = gn$ for some $n \in N$, thus $f(g') = f(gn) = f(g)f(n) = f(g)e_{G'} = f(g)$). This is often paraphrased as "$f$ factors through $p$".

So the "normal way" of showing $G/N \cong (G/M)/(N/M)$ is to show that $f(gN) = (gM)(N/M)$ is a well-defined isomorphism. But let's look at this another way:

The map $p_N: G \to G/N$ is a group morphism that kills $M$ (since $M \subset N$), so $p_N$ factors through the map $p_M: G \to G/M$ so that $p_N = f \circ p_M$, for some morphism $f$. Note that this morphism $f$ goes from $G/M$ to $G/N$ and kills $N/M$, so it in turn factors through $p_{N/M}:G/M \to (G/M)/(N/M)$, that is $f = f' \circ p_{N/M}$. But we also have the map $p_{N/M} \circ p_M$, which kills $N$, so there is a morphism $k$ such that $p_{N/M} \circ p_M = k \circ p_N$.

So $k \circ f' \circ p_{N/M} \circ p_M = k \circ f \circ p_M = k \circ p_N = p_{N/M} \circ p_M$, that is: $k \circ f' = \mathrm{id}_{(G/M)/(N/M)}$.

Similarly, $f' \circ k \circ p_N = f' \circ p_{N/M} \circ p_M = f \circ p_M = p_N$, so that $f' \circ k = \mathrm{id}_{G/N}$ (using implicitly the fact that the projections are epimorphisms to justify the cancellation).

These two facts together imply that $k$ and $f'$ are inverses, and thus isomorphisms. The important thing about all of this, is that we never mentioned any of the elements of $G$, or even any cosets, just homomorphisms between various groups.

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Great, thanks a lot. Just two questions: Why does $f$ kill $N/M$ and why does $p_{N/M}p_M$ kill $N$? – FPP Apr 8 '12 at 18:06
$f$ maps $gM$ to $gN$. An element of $N/M$ is $nM$, which goes to $nN = N$, see? and $p_{N/M}p_M$ sends $N$ to $N/M$ first (via $p_M$), which $p_{N/M}$ then obviously kills. – David Wheeler Apr 9 '12 at 1:28

For every group $H$, we have

$\hom((G/M)/(N/M),H) \cong \{\phi \in \hom(G/M,H) : \phi |_{N/M} = 1\}$ $\cong \{\phi \in \hom(G,H) : \phi|_M = 1, \phi|_N = 1\}$ $= \{\phi \in \hom(G,H) : \phi|_N = 1 \}$ $\cong \hom(G/N,H)$.

The Yoneda lemma implies $(G/M)/(N/M) \cong G/N$. Actually the proof also shows that this isomorphism makes the obvious commutative diagram over $G$ commute. In other words, it gives you the usual description in terms of elements, if you need them at all. Category theory tells us that elements are not important: Morphisms are.

By the way, if you repeat the proof of the Yoneda lemma im this special case, you get precisely the proof by David Wheeler. But as you can see, this is long. And it is a waste of time to repeat the proof of the Yoneda lemma every time; unfortunately this is done in almost every lecture and textbook on abstract algebra.

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