commutative diagram of groups in Lang's algebra

Question

In Lang's Algebra (section I.3), he says that we can describe the third isomorphism theorem $\frac{G}{K}/\frac{H}{K}\cong G/H$ by the following commutative diagram. \begin{array}{c} 0 & \to & H & \to & G & \to & G/H & \to & 0\\ & & \downarrow \text{can} & & \downarrow\text{can} & & \downarrow\text{id} & & \\ 0 & \to & H/K & \to & G/K & \to & G/H & \to & 0 \end{array} It inspired me to have an idea. We know that there are many important theorem which is stated by an isomorphism. For example, $\frac{H}{H\cap K}\cong\frac{HK}{K}$, $G/Z(G)\cong \text{inn } G$, ...

Question 1: Are there commutative diagrams to describe these two example? That is, \begin{array}{c} 0 & \to & ? & \to & ? & \to & ? & \to & 0\\ & & \downarrow \text{?} & & \downarrow\text{?} & & \downarrow\text{?} & & \\ 0 & \to & H\cap K & \to & H & \to & HK/K & \to & 0 \end{array} and \begin{array}{c} 0 & \to & ? & \to & ? & \to & ? & \to & 0\\ & & \downarrow \text{?} & & \downarrow\text{?} & & \downarrow\text{?} & & \\ 0 & \to & Z(G) & \to & G & \to & \text{inn }G & \to & 0 \end{array}

Question 2: Can we find some rules (or an algorithm) to fill the the following commutative diagram. Then to produce an isomorphism? \begin{array}{c} 0 & \to & H & \to & G & \to & G/H & \to & 0\\ & & \downarrow \text{?} & & \downarrow\text{?} & & \downarrow\text{?} & & \\ 0 & \to & ? & \to & ? & \to & ? & \to & 0 \end{array}

Answer 1

$\DeclareMathOperator{\inn}{inn}$ $\DeclareMathOperator{\Aut}{Aut}$ Ad Question 1: You could write $$ \begin{array}{ccccccccc} 0 & \rightarrow & H\cap K & \rightarrow & H & \rightarrow & HK/K & \rightarrow & 0\\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \rightarrow & K & \rightarrow & HK & \rightarrow & HK/K & \rightarrow & 0 \end{array} $$ where all the arrows are evident. The isomorphism $G/Z(G)\cong \inn(G)$ is more tricky. If you really really want to use commutative diagrams to show this, then consider the exact commutative diagram (all the arrows being the evident ones and $G\rightarrow \Aut(G)$ maps an element to conjugation by that element) $$ \begin{array}{ccccccccc} & & 0 & & 0 & & 0 & & \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \rightarrow & Z(G) & \rightarrow & Z(G) & \rightarrow & G/Z(G) & & \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \rightarrow & Z(G) & \rightarrow & G & \rightarrow & G/Z(G) & \rightarrow & 0\\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \rightarrow & \inn(G) & \rightarrow & \Aut(G) & \rightarrow & \Aut(G)/\inn(G) & \rightarrow & 0\\ & & \downarrow & & \downarrow & & \downarrow & & \\ & & \inn(G) & \rightarrow & \Aut(G)/\inn(G) & \rightarrow &\Aut(G)/\inn(G) & \rightarrow & 0\\ & & \downarrow & & \downarrow & & \downarrow & & \\ & & 0 & & 0 & & 0 & & \\ \end{array} $$ Now, by the Snake Lemma, there exists a homomorphism $\delta\colon G/Z(G) \rightarrow \inn(G)$ connecting the end of the 2nd row with the start of the 5th row and such that $$ 0\rightarrow Z(G)\rightarrow Z(G)\xrightarrow{0} G/Z(G) \xrightarrow\delta \inn(G) \xrightarrow{0} \Aut(G)/\inn(G) \rightarrow \Aut(G)/\inn(G) \rightarrow 0 $$ is exact. In particular, $\delta$ is an isomorphism.

To shortly address your second question: It might help to study the short Five Lemma or, more generally the Five Lemma.