# Finite abelian groups in a exact sequence

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?

• can you solve it if $C=0$? Nov 3, 2017 at 20:32
• 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 $A$ Nov 3, 2017 at 20:38

$B/A$ 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.