Let's take a short exact sequence of groups $$1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1$$ I understand what it says: the image of each homomorphism is the kernel of the next one, so the one between $A$ and $B$ is injective and the one between $B$ and $C$ is surjective. I get it. But other than being a sort of curiosity, what is it really telling me?

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    $\begingroup$ It tells you that $B$ is an extension of $C$ by $A$ $\endgroup$ Commented Jul 11, 2014 at 14:22
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    $\begingroup$ oh, I like this $\endgroup$
    – Ziofil
    Commented Jul 11, 2014 at 14:25
  • $\begingroup$ As a big picture thing, a short exact sequences are essentially arithmetic calculations where you're doing arithmetic with groups (and homomorphisms) rather than with group elements. To really understand the utility of an arithmetic calculation requires using them to do arithmetic calculations or even algebra. $\endgroup$
    – user14972
    Commented Jul 11, 2014 at 14:25

4 Answers 4


As Jessica said in a previous answer, the set up initially tells us that in some sense $C = B/A$. To see this, since $A \to B$ is injective, and has image precisely the kernel of some map, we may identify $A \unlhd B$ as a normal subgroup. And this is precisely the kernel of a surjective map $B \to C$ so by the first isomorphism theorem, $C = B/A$.

However, the situation quickly becomes more interesting than this. For example, this situation clearly arises when we have a direct product of groups $B = A \times C$. Motivated by the simplicity of this case, we say that the sequence splits if $B \cong A \times C$. For a non-abelian group, this happens if and only if we can find a map $B \to A$ (think projection) such that the composition $A \to B \to A$ is the identity automomorphism on $A$. For an abelian group, we can say even more that the sequence also splits if we can find a map $C \to B$ such that $C \to B \to C$ is the identity. This happens more generally in abelian categories like the modules over a given ring. This Wikipedia page has more info...

In general we also care about such questions because in algebra and related topics we often find ourselves dealing with so-called long exact sequences of forms

$$... \to A_{-1} \to A_0 \to A_1 \to A_2 \to... $$

Where we know some of the elements of the sequence and some of the maps and would like to identify what the others are. In general any such sequence may be broken down into a number of short exact sequences, and in the best case we may split these sequences to nicely describe an unknown term in terms of the kernel and image of the surrounding maps. The first example that most see in my experience is the Mayer Vietoris sequence which arises in algebraic topology as a way to compute algebraic invariants of a space in terms of those of simpler subspaces.


This set up will tell you that, essentially, $C=B/A$. Think of the first isomorphism theorem.

In some cases, you get that $B=A\oplus C$ (see Wikipedia: Splitting lemma).

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    $\begingroup$ Actually, what you've stated is always true. $\endgroup$
    – jxnh
    Commented Jul 11, 2014 at 14:01
  • $\begingroup$ Yes, you're right. I was muddling two ideas. I'll edit my answer. $\endgroup$
    – Jessica B
    Commented Jul 11, 2014 at 14:06

It's getting more interesting as soon the diagrams get more involved. For example: If the following diagram with exact rows commutes, and the outer columns are both mono/epi/iso-morphism, then the middle one is also a mono/epi/iso-morphism: $$ \begin{matrix} 0\to &A&\to &B&\to &C&\to 0\\ &\downarrow&&\downarrow&&\downarrow&\\ 0\to &A'&\to& B'&\to &C'&\to 0\\ \end{matrix}$$ Translating this statement fully into kernel/image lingo, would make it much less graspable.


"The kernel of $B \to C$" is often not a satisfactory description of a group. Having a group $A$ that it is isomorphic to (along with a given isomorphism) is very useful.

"The quotient $B/A$" is often not a satisfactory description of a gruop. Having a group $C$ that it is isomorphic to (along with a given isomorphism) is very useful.


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