What is the connection between $(G/K)(H/K) \cong G/H$ and this commutative diagram? (Lang's Algebra, p. 17)

Question

There are two questions here and here that concern the same diagram and result, but they seem to have to do with verifying the details of the diagram. I think I've handled that on my own.

(I have also checked Dummit and Foote p. 98, Aluffi p. 101, and the Wikipedia page on the isomorphism theorems, but none of them show a comparable diagram.)

My question has to do with the connection between $(G/K)(H/K) \cong G/H$ and the diagram. Specifically, is the diagram supposed to say the same thing as the $\cong$ statement? What is the intuition here? I feel like these two things are supposed to be connected in my mind, but at the moment they exist separately in my mind.

For context, the way I understand the $\cong$ statement is by following Lang's explanation that there is a surjective homomorphism $G/K \to G/H$ that has $H/K$ as its kernel. Applying the fact that $G/\text{ker } \phi \cong \text{im } \phi$ gives the result.

Excerpt:

Edit:

Looking at the diagram again, I see that we have two exact sequences that go $\text{trivial group} \to \text{kernel} \to \text{group} \to \text{image} \to \text{trivial group}$. But I still don't see the connection to the $\cong$ statement.

Answer 1

A sequence of the form, $$ 1 \to G_1 \to G_2 \to G_3 \to 1 $$ being exact (in the category of groups) is exactly the same information as,

(1) an inclusion $G_1 \subset G_2$ making $G_1$ a normal subgroup of $G_2$

(2) an isomorphism $G_2 / G_1 \xrightarrow{\sim} G_3 $

Explicitly, the exactness of the sequence says

(1) $G_1 \to G_2$ is injective

(2) the image of $G_1 \to G_2$ equals the kernel of $G_2 \to G_3$

(3) $G_2 \to G_3$ is surjective

Putting these together with the first isomorphism theorem gives what I claimed.

Lang's diagram shows that $(G / K) / (H / K) \cong (G / H)$ $\textit{via the given map}$ $G/K \to G/H$. So you get a bit more than the isomorphism you also get an explicit description of the isomorphism and the fact that the diagram commutes tells you this isomorphism is compatible with the "tautological" isomorphism $G/H \to G/H$ which is what the top row expresses.

Answer 2

Perhaps it would be more enlightening if you wrote it a bit more abstractly.

You have short exact sequences $H\to G\to Q_1$ and $H/K\to G/K\to Q_2$. This means that $Q_1\cong G/H$ and $Q_2\cong (G/K)/(H/K)$.

Then you have an isomorphism $Q_1\to Q_2$ (denoted by $\operatorname{id}$ on your diagram) and quotient maps $G\to G/K$ and $H\to H/K$ making the diagram commute. This means that $Q_1$ and $Q_2$ are isomorphic and the isomorphism is exactly what you expect it to be, namely, the $H$-coset of $g\in G$ is taken to the $H/K$-coset of $gK$.

Answer 3

So the idea is that this is a commutative diagram and these are something called exact sequences. The gist is that a sequence $H \xrightarrow[]{\phi} G \xrightarrow[]{\psi} K$ is said to be exact if $\ker(\psi) = \text{im}(\phi)$. Here, the maps are group homomorphisms and $G,H,K$ are groups.

Claim: If $0 \xrightarrow[]{\phi} H \xrightarrow[]{\psi} G$ exact, then $\psi$ is injective.

Proof: We have $\ker(\psi) = \text{im}(\phi)$, $\text{im}(\phi) = 0$ (since there's only one way to have a homomorphism from $0$ to $H$), and so $\ker(\psi) = 0$ is trivial.

Claim: If $G \xrightarrow[]{\phi} K \xrightarrow[]{\psi} 0$ exact, then $\phi$ is surjective.

Proof: There's only one homomorphism from $K$ to $0$ (map everything to $0$) so $\ker(\psi) = K$. Since the sequence is exact, $\text{im}(\phi) = \ker(\psi) = K$.

Now we can extend the definition of exact sequences to include multiple terms in the (hopefully) obvious way. Let's look at the exact sequence $0 \rightarrow H \xrightarrow[]{\phi} G \xrightarrow[]{\psi} K \rightarrow 0$, where these are all groups and group homomorphisms. I omit the names for the maps $0 \rightarrow H$ and $K \rightarrow 0$ since, as noted above, there's only one way to do it. Now $\ker(\psi) = \text{im}(\phi) = H$ (here I'm identifying $H$ with its image as a subgroup of $G$) since $\phi$ is injective. We know $\psi$ is surjective by above, so use the first isomorphism theorem to get

$$ G/\ker(\psi) = G/H \cong K.$$

Now you need to verify that your sequences are actually exact sequences (use the commutativity of the diagram) and then you can invoke this result.