Simple example of a short exact sequence of groups

I am new to this and would like to understand $$0 \overset{a0}{\to} B \overset{a1}{\to} A \overset{a2}{\to} A/B \overset{a3}{\to} 0,$$ where $B \subset A$ and they are both Abelian groups. Also maybe it is helpful to know that every subgroup of an Abelian group is normal? This is in the context of a cochain (we were taught starting this direction).

So here is what I understand so far:

If it is a short exact sequence (s.e.s.) (is this given or is it just iff $a_0$ is injective AND $a_3$ is surjective?), then, with the identity element of an additive (for instance) Abelian group being $0$, it is such that,

$$\begin{eqnarray} ker\; a_0 \leq_{sg} 0 \;\Rightarrow ker\; a_0=0 \\ ---------------- {} \\ im\; a_1 = ker\; a_2 \text{ and } ker\; a_2 = B \\ H^2(A) = \frac{ker \; a_2}{im\; a_1} = \frac{B}{B} = 0 \\ ----------------- {} \\ a_2: A \to A/B \hspace{1cm} a_2(a_i) = a_i\,B \\ ker\; a_2 = \{a\in A : a_2 \circ a = 0\} = \{a \in A : a_2\circ a = B\} = B\\ im\; a_2 \cong A/B \\ ---------------- {} \\ a_3\; \text{ zero map } \Rightarrow a_3(a\,B) = B \hspace{1cm}\text{ as the zero of A/B is B ? } \\ H^3(A/B) = \frac{ker\; a_3}{im \; a_2} = B/(A/B) \end{eqnarray}$$

I'm not sure of these results, but I hope it is simple and clear enough to get some feedback. Thanks!

Edit: I forgot to mention that with this construction, $a_1$ induces an isomorphism from $B \to (\underset{s.g.}{{\cdot}}\subset A)$ and $a_2$ induces an isomorphism from $A/im\;B \to A/B$.

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I cannot tell what you are asking. What exactly is it that you want to understand? – Mariano Suárez-Alvarez Sep 14 '12 at 19:52
Well I was wondering if what I wrote down (in the middle) is correct. Any other comments or feedback is appreciated. – nate Sep 14 '12 at 19:56
Saying that a sequence is exact means that all the (co)homology groups vanish, i.e. $\operatorname{ker} a_{i+1}/\operatorname{im}a_i = 0$ for all $i.$ – Andrew Sep 14 '12 at 19:57
All subgroups of abelian groups are normal subgroups. – Thomas Andrews Sep 14 '12 at 20:18

• A priori, this is just a diagram, that is any arrow from $X$ to $Y$ merely denotes a homomorphism.
• It should be noted explicitly, but the linear structure suggests that it is a complex, that is the composition of any two arrows in straight succession is the 0 homomorphism; if there is $X\overset f\longrightarrow Y\overset g\longrightarrow Z$ in the diagram, that this states that $g\circ f=0$ holds.
• If desired, it should also be noted explicitly, that it is an exact sequence (or in general context: which joints are exact and which maybe not), that is $\ker g =\operatorname{im} f$ holds whenever we see $X\overset f\longrightarrow Y\overset g\longrightarrow Z$. So exactness is a claim about that diagram. Note however, that a short exact sequence $$0\overset \alpha \longrightarrow X\overset \beta \longrightarrow Y \overset \gamma \longrightarrow Z\overset \delta \longrightarrow 0$$ is exact iff the following holds

• exactness at $X$: Since $\alpha$ is the only homomorphism $0\to X$ and its image is $0$, we must have that $\ker \beta=0$, aka. $\beta$ is injective. (Formally, the kernel is an arrow; I speak of kernel, image etc. here as subobjects in the object manner throughout)

• exactness at $Z$: Similarly $\delta$ is the only homomorphism $Z\to 0$ and its kernel is all of $Z$; hence we need that $\operatorname{im} \gamma=Z$, aka. $\gamma$ is surjective.
• exactness at $Y$ is maybe the most interesting point: We need $\operatorname{im}\beta=\ker \gamma$.

What (co)homology does is to measure by how much a complex fails to be exact.

Your calculations are in principle correct, however there are a few probable typoes and e.g. $H^2(A)$ would not be used to denote ker/im of this complex that also depends on $B$ etc., but rather what can be computed from e.g. an injective resolution $$0\to A\to I_1\to I_2\to I_3\to\ldots$$ of $A$ by injectives.

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