# finitely generated abelian group and its quotient

G is an abelian group with the following property:

(*) If H is any subgroup of G then there exists a subgroup F of G such that G/H is isomorphic to F.

Now I want to prove that If G is finitely generated abelian group and G has the above property then G is finite.

this is my solution and I have a sense that it is not correct but I can't figure out my mistake: Assume that G=<$g_{1},g_{2},....,g_{n}$>.My solution is based on proving that order of every $g_{i}$ is finite. consider the group generated by $g_{i}$ namely . So we have $G/<g_{i}>$ is isomorphic to a subgroup of G so $G/<g_{i}>$ is finitely generated. Now from there $\forall g\in G$ we have $g<g_{i}>=\sum_{j=1}^{n} a_{i_{j}}g_{i_{j}}$ so we have that $g_{i}$ and all its powers can be written as a finite combination of $g_{i}$s.Now since G is finitely generated then there are finite number of combinations of $g_{j},j=1,.,n$. In this case there are k and l such that $g_{i}^{k}=g_{i}^{l}$ which means that $g_{i}$ has a finite order.

• I am pretty sure that $G/\langle g_i\rangle \cong \langle g_1, g_2, ..., g_{i-1}, g_{i+1}, ..., g_n \rangle$. I'm confused about "$g_i$ and all its powers can be written as finite combinations of $g_i$s." What does this mean? Obviously, all powers of $g_i$ are finite products of $g_i$. Also, is this a group or a ring? If it's not a ring, then what is the $\sum$ about? – Noble Mushtak Jun 18 '16 at 16:11

Your solution is not correct: you in fact did not use the hypothesis, as the only occurrence of the hypothesis in your argument is in showing that $G/\langle g_i\rangle$ is finitely generated, but, as a matter of fact, this holds for any finitely generated group: if $G$ is generated by $g_1,\cdots,g_n$ then $G/\langle g_i\rangle$ is generated by the images of $g_j,$ for $j=1,\cdots,n.$
First use the fundamental theorem of finitely generated abelian groups to write $G$ as $\oplus_{i=1}^n\Bbb Z_{n_i}\oplus\Bbb Z^{n}$ and $G$ is finite if and only if $n$ is $0.$
Assume that $n\ne0.$ It follows that $G$ contains an isomorphic copy of $\Bbb Z.$ So $\forall k,G$ contains $\oplus_{i=1}^n\Bbb Z_{n_i}\oplus k\Bbb Z$ where $k\Bbb Z$ means the subgroup in $\Bbb Z$ generated by $k.$ The quotient of $G$ with respect to this subgroup is isomorphic to $\Bbb Z/k\Bbb Z.$ Then the hypothesis tells us that $G$ contains a cyclic subgroup of order $k$ for any integer $k.$
But this is clearly impossible, as shown by looking at the decomposition $G\cong\oplus_{i=1}^n\Bbb Z_{n_i}\oplus\Bbb Z^{n}.$ (Hint: let $k$ be a large enough integer.)
Therefore $n=0$ and $G$ is finite.