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This is probably very simple, but I'm stuck trying to understand the following argumentation I read in a book:

(here, $A,B, C$ are abelian groups)

[...] since $A, C$ are finite, the exact sequence $0\to A\xrightarrow{f}B\xrightarrow{g}C$ nests $B$ between two finite groups, therefore $B$ must be finite.

From the exact sequence, I can only see that $B/A\simeq B/\ker(g)\simeq im(g)<C$, so that $B/A$ is finite. But how do I know that $B$ itself is finite?

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    $\begingroup$ can you solve it if $C=0$? $\endgroup$
    – Asinomás
    Nov 3, 2017 at 20:32
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    $\begingroup$ if $B$ is an infinite group and $A$ is finite then $B/A$ is infinite. Because the cardinality of $B$ is the cardinality of $B/A$ multiplied by the cardinality of $A$ $\endgroup$
    – Asinomás
    Nov 3, 2017 at 20:38

1 Answer 1

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$B/A$ is finite because $B/A = B/\mathrm{Im}f = B/\mathrm{Ker}g \simeq \mathrm{Im}g \subset C$, and $C$ is finite.

Let's choose a (finite) set of representatives in $B$ for $B/A$: $B/A = \{x_1 + A ,..., x_n+A\}$ .

Then $B= \displaystyle\bigcup_{i=1}^n (x_i +A)$

$A$ being finite, each $x_i +A$ is finite and so $B$ is finite as a finite union of finite sets.

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