I don't understand the proof of the second isomorphism theorem.

Question

I don't understand how the author gets the last couple of lines. I understand, by the homomorphism theorem(perhaps sometimes called "The First Ismoprhism Theorem"), $\bar{G} \cong G/N$, but how do they then conclude $(G/N)/(K/N) \cong G/K$?

Is the idea to show $\bar{K} \cong (K/N)$ first!? If so, how? If this is the case, I would greatly the details. Also, if this is indeed the case, wouldn't they also have to show $A \cong B$ and $C \cong D \implies A/C \cong B/D?$

If the idea is not to show $\bar{K} \cong (K/N)$, please let me know.

Answer 1

Let $f \colon \overline{G} \to G/N$ denote the isomorphism guaranteed by the First Isomorphism Theorem. When the author writes "this identification carries $\overline{K}$ to the image of $K$ in $G/N$, namely $K/N$" what they mean is that $$f(\overline{K}) = K/N \subset G/N.$$

You should check that this is true, and that in fact the restriction $f|_{\overline{K}} \colon \overline{K} \to K/N$ is a group isomorphism as well.

Hint: You should write down explicitly what $f$ does, and be careful checking that various things are well-defined.

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In general it is not true that $A \cong B$ and $C \cong D$ imply $A/C \cong B/D.$ However, in this case, since one isomorphism $f \colon \overline{K} \to K/N$ is the restriction of the other $f \colon \overline{G} \to G/N$, this statement is true. Let's prove this:

Lemma If $A, C$ are groups and $B \subseteq A$, $D \subseteq C$ normal subgroups, and if $f \colon A \to C$ is an isomorphism such that $f|_{B} \colon B \to D$ is also an isomorphism, then $$A/B \cong C/D.$$

Proof: The obvious isomorphism $\overline{f}$ is as follows: you take an element $aB$ of $A/B$, and you send it to $f(a)D.$ But we need to check that this is well-defined.

If you give me $a, a' \in A$ such that $aB = a'B$, is it true that $f(a)D = f(a')D?$ Well, if $aB = a'B$, then $a^{-1}a' \in B,$ and thus $f(a^{-1}a') \in f(B) = D$. Therefore, we have $$f(a')D = f(a)f(a)^{-1}f(a')D = f(a)f(a^{-1}a')D = f(a)D,$$ where $f(a^{-1}a')D$ is the identity because $f(a^{-1}a') \in D.$

So, $\overline{f}$ is indeed well-defined. Why is it an isomorphism?

It is a group homomorphism because given any $a, a' \in A$, we have $$\overline{f}(aBa'B) = \overline{f}((aa')B) = f(aa')D = f(a)f(a')D = f(a)Df(a')D = \overline{f}(aB)\overline{f}(a'B).$$

It's injective because if $aB \in A/B$ is sent to the identity $D$ of $C/D$, then we have $$D = \overline{f}(aB) = f(a)D,$$ hence $f(a) \in D$. Since $f$ restricts to an isomorphism from $B$ to $D$, it is a bijection from $B$ to $D$ as well, so it follows that $a$ must be in $B.$ So, we have $aB = B,$ which means that the only element sent to the identity $D$ is the identity $B$ of $A/B$.

It's surjective because for any $cD$ in $C/D$, there exists $a \in A$ such that $f(a) = c$ (because $f$ is an isomorphism), and thus $\overline{f}(aB) = cD.$

So, we've shown that $\overline{f}$ is an isomorphism, and we're done.

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Now, we know that $f \colon \overline{G} \to G/N$ is an isomorphism, and it restricts to an isomorphism $\overline{K} \to K/N$. By the Lemma, we have $$\overline{G}/\overline{K} \cong (G/N)/(K/N),$$ as desired.

Answer 2

First part of this proof is actually what you mentioned that $\overline{K} \cong (K/N)$. from first isomorphism theorem $\overline{\phi} : G/N \rightarrow \overline{G}$ is isomorphism (identification) and $\overline{\phi}^{-1} :\overline{G} \rightarrow G/N$ is the inverse homeomorphism (isomorphism). the proof shows that $\overline{\phi}^{-1}(\overline{K})=K/N.$ (actually shows the inverse but $ \overline{\phi}$ is isomorphism).
last lines of the proof uses the fact that if $h:H\rightarrow H'$ is a isomorphism and $K \triangleleft H$ then $H/K \cong H'/h(K)$. author first show that $G/N \cong \overline{G}$. and then shows that this isomorphism maps normal subgroup $\overline{K}$ of $\overline{G}$ to normal subgroup $K/N$ of $G/N$. so the proof is completed.

but if you're not comfortable with this proof you can do it in the hard way and use first isomorphism theorem directly.

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In order to proof the theorem using first isomorphism theorem we should introduce a homeomorphism $q:\frac{G}{N} \rightarrow \frac{\overline{G}}{\overline{K}}$. we can do this using $\phi$ and $\psi$ as introduced in proof: $$ \frac{G}{N} \xrightarrow{\overline{\phi}} \overline{G} \xrightarrow{\psi} \frac{\overline{G}}{\overline{K}} \\ xN \mapsto \phi(x) \mapsto \phi(x) \overline{K} $$ so $q=\psi \circ \overline{\phi} :\frac{G}{N} \rightarrow \frac{\overline{G}}{\overline{K}}$. from first isomorphism theorem we get : $$ \frac{(G/N)}{ker(q)} \cong \frac{\overline{G}}{\overline{K}} \tag{1} $$ we should determine $ker(q)$. $$ ker(q) = \{xN \in G/N \,|\, q(xN)= \phi(x) \overline{K}=\overline{K}\} = \{xN\in G/N \, | \, \phi(x) \in \overline{K} \} \\ =\{xN \in G/N \, | \, x \in K\} = K/N \tag{2} $$ final thing you need is $\overline{G}/\overline{K} \cong G/K$. for this isomorphism you can use $\overline{\phi}$ again. (this is the part that author use $\psi \circ \phi$) $$ \overline{\phi}': G/K \xrightarrow{\cong} \overline{G}/\overline{K} \\ xK\mapsto \overline{\phi}(x) \overline{K} \tag{3} $$ put $(2)$ and result of $(3)$ in $(1)$ and you get : $$ \frac{(G/N)}{(K/N)} \cong \frac{\overline{G}}{\overline{K}} \cong \frac{G}{K} $$