Holy cow, go beyond the first homomorphism theorem! For example, if you have a long exact sequences of vector spaces and linear maps $$ 0 \rightarrow V_1 \rightarrow V_2 \rightarrow \cdots \rightarrow V_n \rightarrow 0 $$ then exactness implies that the alternating sum of the dimensions is 0. This generalizes the "rank-nullity theorem" that $\dim(V/W) = \dim V - \dim W$, which is the special case of $0 \rightarrow W \rightarrow V \rightarrow V/W \rightarrow 0$.
Replace vector spaces and linear maps by finite abelian groups and group homomorphisms and instead you find the alternating product of the sizes of the groups has to be 1.

The purpose of this general machinery is not the small cases like the first homomorphism theorem. Exact sequences and commutative diagrams are about the only way to even think about or formulate large chunks of modern mathematics. For instance, you need commutative diagrams to make sense of universal mapping properties (which is the way many concepts are defined or at least most clearly understood) and to understand the opening scene in the movie "It's My Turn".

Here is a nice exercise. When $a$ and $b$ are relatively prime, $\varphi(ab) = \varphi(a)\varphi(b)$, where $\varphi(n)$ is Euler's $\varphi$-function from number theory. Question: Is there a formula for $\varphi(ab)$ in terms of $\varphi(a)$ and $\varphi(b)$ when $(a,b) > 1$? Yes: $$ \varphi(ab) = \varphi(a)\varphi(b)\frac{(a,b)}{\varphi((a,b))}. $$ You could prove that by the formula for $\varphi(n)$ in terms of prime factorizations, but it wouldn't really explain what is going on because it doesn't provide any meaning to the formula. That's kind of like the proofs by induction which don't really give any insight into what is going on. But it turns out there is a nice 4-term short exact sequence of abelian groups (involving units groups mod $a$, mod $b$, and mod $ab$) such that, when you apply the above "alternating product is 1" result, the general $\varphi$-formula above falls right out. Searching for an explanation of that formula in terms of exact sequences forces you to try to really figure out conceptually what is going on in the formula.

Holy cow, go beyond the first homomorphism theorem! For example, if you have a long exact sequence of vector spaces and linear maps $$ 0 \rightarrow V_1 \rightarrow V_2 \rightarrow \cdots \rightarrow V_n \rightarrow 0 $$ then exactness implies that the alternating sum of the dimensions is 0. This generalizes the "rank-nullity theorem" that $\dim(V/W) = \dim V - \dim W$, which is the special case of $0 \rightarrow W \rightarrow V \rightarrow V/W \rightarrow 0$.
Replace vector spaces and linear maps by finite abelian groups and group homomorphisms and instead you find the alternating product of the sizes of the groups has to be 1.

The purpose of this general machinery is not the small cases like the first homomorphism theorem. Exact sequences and commutative diagrams are the only way to think about or formulate large chunks of modern mathematics. For instance, you need commutative diagrams to make sense of universal mapping properties (which is the way many concepts are defined or at least most clearly understood) and to understand the opening scene in the movie "It's My Turn".

Here is a nice exercise. When $a$ and $b$ are relatively prime, $\varphi(ab) = \varphi(a)\varphi(b)$, where $\varphi(n)$ is Euler's $\varphi$-function from number theory. Question: Is there a formula for $\varphi(ab)$ in terms of $\varphi(a)$ and $\varphi(b)$ when $(a,b) > 1$? Yes: $$ \varphi(ab) = \varphi(a)\varphi(b)\frac{(a,b)}{\varphi((a,b))}. $$ You could prove that by the formula for $\varphi(n)$ in terms of prime factorizations, but it wouldn't really explain what is going on because it doesn't provide any meaning to the formula. That's kind of like the proofs by induction which don't really give any insight into what is going on. But it turns out there is a nice 4-term short exact sequence of abelian groups (involving units groups mod $a$, mod $b$, and mod $ab$) such that, when you apply the above "alternating product is 1" result, the general $\varphi$-formula above falls right out. Searching for an explanation of that formula in terms of exact sequences forces you to try to really figure out conceptually what is going on in the formula.

Holy cow, go beyond the first homomorphism theorem! For example, if you have a long exact sequences of vector spaces and linear maps $$ 0 \rightarrow V_1 \rightarrow V_2 \rightarrow \cdots \rightarrow V_n \rightarrow 0 $$ then exactness implies that the alternating sum of the dimensions is 0. This generalizes the "rank-nullity theorem" that $\dim(V/W) = \dim V - \dim W$, which is the special case of $0 \rightarrow W \rightarrow V \rightarrow V/W \rightarrow 0$.
Replace vector spaces and linear maps by finite groups and group homomorphisms and instead you find the alternating product of the sizes of the groups has to be 1.

The purpose of this general machinery is not the small cases like the first homomorphism theorem. Exact sequences and commutative diagrams are about the only way to even think about or formulate large chunks of modern mathematics. Just look up any book on homological algebra.

Here is a nice exercise. When $a$ and $b$ are relatively prime, $\varphi(ab) = \varphi(a)\varphi(b)$, where $\varphi(n)$ is Euler's $\varphi$-function from number theory. Question: Is there a formula for $\varphi(ab)$ in terms of $\varphi(a)$ and $\varphi(b)$ when $(a,b) > 1$? Yes: $$ \varphi(ab) = \varphi(a)\varphi(b)\frac{(a,b)}{\varphi((a,b))}. $$ You could prove that by the formula for $\varphi(n)$ in terms of prime factorizations, but it wouldn't really explain what is going on because it doesn't provide any meaning to the formula. That's kind of like the proofs by induction which don't really give any insight into what is going on. But it turns out there is a nice 4-term short exact sequence of abelian groups (involving units groups mod $a$, mod $b$, and mod $ab$) such that, when you apply the above "alternating product is 1" result, the general $\varphi$-formula above falls right out. Searching for an explanation of that formula in terms of exact sequences forces you to try to really figure out conceptually what is going on in the formula.

Holy cow, go beyond the first homomorphism theorem! For example, if you have a long exact sequences of vector spaces and linear maps $$ 0 \rightarrow V_1 \rightarrow V_2 \rightarrow \cdots \rightarrow V_n \rightarrow 0 $$ then exactness implies that the alternating sum of the dimensions is 0. This generalizes the "rank-nullity theorem" that $\dim(V/W) = \dim V - \dim W$, which is the special case of $0 \rightarrow W \rightarrow V \rightarrow V/W \rightarrow 0$.
Replace vector spaces and linear maps by finite abelian groups and group homomorphisms and instead you find the alternating product of the sizes of the groups has to be 1.

The purpose of this general machinery is not the small cases like the first homomorphism theorem. Exact sequences and commutative diagrams are about the only way to even think about or formulate large chunks of modern mathematics. For instance, you need commutative diagrams to make sense of universal mapping properties (which is the way many concepts are defined or at least most clearly understood) and to understand the opening scene in the movie "It's My Turn".

Here is a nice exercise. When $a$ and $b$ are relatively prime, $\varphi(ab) = \varphi(a)\varphi(b)$, where $\varphi(n)$ is Euler's $\varphi$-function from number theory. Question: Is there a formula for $\varphi(ab)$ in terms of $\varphi(a)$ and $\varphi(b)$ when $(a,b) > 1$? Yes: $$ \varphi(ab) = \varphi(a)\varphi(b)\frac{(a,b)}{\varphi((a,b))}. $$ You could prove that by the formula for $\varphi(n)$ in terms of prime factorizations, but it wouldn't really explain what is going on because it doesn't provide any meaning to the formula. That's kind of like the proofs by induction which don't really give any insight into what is going on. But it turns out there is a nice 4-term short exact sequence of abelian groups (involving units groups mod $a$, mod $b$, and mod $ab$) such that, when you apply the above "alternating product is 1" result, the general $\varphi$-formula above falls right out. Searching for an explanation of that formula in terms of exact sequences forces you to try to really figure out conceptually what is going on in the formula.