# f.g. normal subgroup of finite index in abelian groups

Let $G$ be an abelian group. Suppose $G$ has a (normal) subgroup $N$ which is f.g. torsion-free abelian such that $G/N$ is finite. In light of the fundamental theorem for f.g. abelian groups I ask if $G\cong N\oplus F$, for a finite subgroup $F$ of $G$. This is certainly the case if $0\to N\to G\to G/N\to 0$ splits.

My impression is that an abelian group which is a finite extension of a f.g. abelian group, has a structure not far from the structure of the normal subgroup - only torsion will change (that is, the torsion-free rank won't change).

Thank you.

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You are correct that if $G$ is an abelian group, $F$ is a finitely generated torsion free subgroup, and $G/F$ is torsion, then $G$ will be isomorphic to $F\oplus T'$ for some finite subgroup $T'$; however, as noted by Alex, $T'$ is not necessarily isomorphic to $G/F$.

You can even say a bit more:

Theorem. Let $G$ be an abelian group, with a finitely generated normal subgroup $N$ and such that $G/N$ is finitely generated. Then:

1. $G$ is finitely generated;
2. $\text{free rank}(G) = \text{free rank}(N) + \text{free rank}(G/N)$;
3. If $N$ is torsion free and $G/N$ is torsion, then $G$ is isomorphic to $N\oplus T$, where $T$ is a isomorphic to a subgroup of $G/N$.

Caveat: In the isomorphism in part (3), the isomorphism $G\cong N\oplus T$ need not identify $N$ isomorphically with itself. Again, see the example Alex B. gives.

Part 1 is in fact true for any group:

Lemma. Let $G$ be a group, and let $N$ be a normal subgroup of $G$. If $X\subseteq N$ generates $N$, and $\overline{Y}\subseteq G/N$ generates $G/N$, then there is a subset $Y$ of $G$ that bijects with $\overline{Y}$ and such that $\langle X,Y\rangle = G$. In particular, if both $N$ and $G/N$ are finitely generated, then $G$ is finitely generated.

Proof. For each $y\in\overline{Y}$, let $g_y\in G$ be an element such that $g_yN = y$; let $Y=\{g_y\mid y\in\overline{Y}\}$. I claim that $X\cup Y$ generates $G$.

Indeed, let $g\in G$. Then $gN$ can be expressed in terms of elements of $\overline{Y}$, $gN = y_1^{a_1}\cdots y_m^{a_m}$. Since $\pi(g_{y_1}^{a_1}\cdots g_{y_m}^{a_m}) = \pi(g)$, then it follows that $g(g_{y_1}^{a_1}\cdots g_{y_m}^{a_m})^{-1}\in N$. Thus, it can be expressed as a product of elements of $X$, $$g(g_{y_1}^{a_1}\cdots g_{y_m}^{a_m})^{-1} = x_1^{b_1}\cdots x_n^{b_n}.$$ Therefore, $$g = x_1^{b^1}\cdots x_n^{b_n}g_{y_1}^{a_1}\cdots g_{y_m}^{a_m}\in \langle X\cup Y\rangle.$$

Thus, $G\subseteq \langle X\cup Y\rangle$, hence equality holds. $\Box$

For part 2, the simplest way is to tensor up to $\mathbb{Q}$, i.e., consider the sequence $$N\otimes\mathbb{Q} \to G\otimes\mathbb{Q}\to (G/N)\otimes\mathbb{Q}.$$ It is not hard to verify that $N\otimes\mathbb{Q}$ is a subgroup of $G\otimes\mathbb{Q}$, and that the quotient is isomorphic to $(G/N)\otimes\mathbb{Q}$. Now you are dealing with vector spaces, and the free rank corresponds to the dimension of the resulting vector spaces. So part 2 of the theorem becomes the assertion that if $\mathbf{W}$ is a subspace of $\mathbf{V}$, then the dimension of $\mathbf{V}$ equals the dimension of $\mathbf{W}$ plus the dimension of $\mathbf{V}/\mathbf{W}$, which is true.

Finally, for part 3, part 2 guarantees that the free rank of $G$ equals the free rank of $N$, and so when you express $G$ as direct sum of a free part and a torsion part, the free part will be isomorphic to $N$. To show that the torsion part is a subgroup of $G/N$, note that if $H$ is the torsion subgroup of $G$, then $N\cap H=\{1\}$. Thus, $H\cong H/(H\cap N) \cong HN/N$, so $H$ is isomorphic to a subgroup of $G/N$, since $HN/N\lt G/N$.

(In fact, the torsion part of $G$ is also isomorphic to a quotient of $G/N$, since every quotient of a finite abelian group is isomorphic to a subgroup of that group.)

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Thank you very much Arturo. Is there a more elementary argument for part 2? –  APena Oct 6 '11 at 16:59
In part 3, the point is that $N$ is free abelian, right? That's why the free part of $G$ is isomorphic to $N$, correct? –  APena Oct 6 '11 at 17:05
@APena: It's easy to check that $\text{free rank}(G)\geq \mathrm{free rank}(N) + \mathrm{free rank}(G/N)$ by very simple elementary methods (see here); the converse inequality takes more work, and I can't think of any simple way of doing it. –  Arturo Magidin Oct 6 '11 at 17:07
@APena: The reason the free part of $G$ is isomorphic to $N$ is that both $N$ is free abelian, and that $G/N$ is torsion. If either of those conditions fails, then you would not be able to reach the conclusion. –  Arturo Magidin Oct 6 '11 at 17:08
@APena: But of course, part (2) already shows that your intuition is right: If $G/N$ is finite, $G$ abelian, and $N$ finitely generated, with $N\cong \mathbb{Z}^r\times T$, $T$ torsion, then $G\cong \mathbb{Z}^r\times T'$ with $T'$ torsion; in fact, $T'$ contains a subgroup isomorphic to $T$, and has $G/N$ as a quotient, but that's about as far as you can push it in this generality, I think. –  Arturo Magidin Oct 6 '11 at 17:12

Since $N$ is finitely generated and of finite index, $G$ itself is finitely generated (by generators of $N$, moved by the finitely many coset representatives), so $G\cong \mathbb{Z}^r\oplus \Delta$, for a finite group $\Delta$. Since $G/N$ is finite, $N$ has maximal rank $r$, which answers your question affirmatively.

But be careful: it is not true that there is necessarily a subgroup $F$ such that $G=N\oplus F$. A simple counterexample is given by $2\mathbb{Z}\subset \mathbb{Z}$. The quotient is cyclic of order $2$ but the subgroup if not complemented, i.e. the short exact sequence does not split. Of course, $\mathbb{Z}$ is still isomorphic to $2\mathbb{Z}\oplus\{1\}$.

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Many thanks Alex. –  APena Oct 6 '11 at 16:43