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Let $G_i$ be abelian Groups. A exact sequence of the form $ 0 \to G_1 \to G_2 \to G_3 \to 0$ is called a short exact sequence. Is the following statement true?

If $G_3$ is finitely generated then there is a short exact sequence with $G_2$ and $G_1$ free groups?

I'm unable in producing a counter example.please help.

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  • $\begingroup$ An exact sequence with only two nonzero terms is just an isomorphism. It should have all three $G_i$ in it. $\endgroup$
    – Arthur
    Aug 19, 2015 at 5:44
  • $\begingroup$ Indeed; your question should probably be, "Is there a (split?) short exact sequence $0 \to G_1 \to G_2 \to G_3 \to 0$?" Perhaps to help us answer, you should say what your definition of "finitely generated" is. (You should also probably include the map $f$ in your initial description, since otherwise it makes no sense.) $\endgroup$ Aug 19, 2015 at 5:53
  • $\begingroup$ @user26857 Sorry,fixed now.Regards, $\endgroup$ Aug 19, 2015 at 20:01

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I assume you mean that a sequence $$0\to G_1 \to G_2\to G_3\to 0\tag{*}$$is short exact. Set $$G_3=\Bbb Z_{a_1}\times \Bbb Z_{a_2}\times \cdots \Bbb Z_{a_n}\times \Bbb Z^m$$ If $G_1=\Bbb Z^n, G_2=\Bbb Z^{n+m}$ and $f:G_1\to G_2$ is given by first multiplying each component with the corresponding $a_i$, then adding $m$ zeroes to the end, the sequence $(*)$ is exact.

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  • $\begingroup$ Sorry..How do you define map from $G_2$ to $G_3$? $\endgroup$ Aug 19, 2015 at 6:14
  • $\begingroup$ @Learner I send $(0,0,\ldots,0,1,0,\ldots,0)$ to $(0,0,\ldots,0,1,0,\ldots,0)$, and let linearity of group homeomorphisms take care of the rest. Alternatively, I send any $(b_1, \ldots, b_{n+m})$ to $(\bar b_1, \bar b_2,\ldots, \bar b_n, b_{n+1}, b_{n+2},\ldots,b_{n+m})$ (that's the same thing, only said a bit differently). $\endgroup$
    – Arthur
    Aug 19, 2015 at 7:17
  • $\begingroup$ @Learner I mean, of course, group homomorphisms, not homeomorphisms. $\endgroup$
    – Arthur
    Aug 19, 2015 at 7:35
  • $\begingroup$ Ofcourse thats true! $\endgroup$ Aug 19, 2015 at 20:01

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