# Subgroup of finitely generated abelian group is finitely generated (methods)

Relearning math to prep for applying to grad school (have taken time off), I got to this exercise and am trying to work through following my intuition rather than proofs as much as I can. Can someone confirm or deny ideas, examples would be well appreciated too.

Since we are dealing with an abelian group, say $$G$$, I believe we can consider the generators $$\{g_1, \cdots ,g_k\}$$ and the cyclic subgroups they give should be a direct sum of the group (as each subgroup of an abelian group is normal), so $$G\cong G_1\oplus \cdots \oplus G_k$$ where $$G_i=\langle g_i\rangle$$. We can consider an inclusion map $$H\hookrightarrow G \cong \bigoplus G_i$$ should induce an isomorphism $$H\cong H_1\oplus \cdots \oplus H_k$$ where $$H_i$$ is a subgroup of $$G_i$$. So $$H_i=\langle g_i^{\alpha_i}\rangle$$ for some positive integer $$\alpha_i$$, and $$H$$ is generated by the $$g_i^{\alpha_i}$$.

Please let me know if this is a flawed logic. I'm noting an approach like this (I think) gives the same number of generators and should probably be careful about properly describing the generators of $$H$$, and even $$G$$. (as a generator of $$G$$ described as a direct sum should probably not be identified with a generator $$g_i$$ of $$G$$)

Further, the text says this is not true for nonabelian groups. I am trying to thing of how to construct an example. At the least, I am thinking to construct arbitrarily large #'s of generators, consider the dihedral groups $$D_n$$ as $$n\rightarrow \infty$$. We can consider the subgroup that are just the geometric reflections of the $$n$$-gon. These are all independent requiring $$n$$ generators to construct this subgroup. But unsure of how to get to infinitely many generators for a subgroup. Could one suggest a proper group but not its subgroup as a pointer?

Thanks! Also suggestions of study topics/books are greatly appreciated, I am currently rereading Artin, Munkres (point set and algebraic books), Rudin (complex and real), Royden, and occasional recent arxiv papers.

• for sure you cannot argue that way, since nobody tells you that $\langle g_i\rangle\cap \langle g_j\rangle=\{1\}$ whenever $i\neq j$. For example, $\mathbb Z/4\mathbb Z$ is generated by $\{1,3\}$ but certainly has only 1 cyclic factor. For a classical counterexample in the non-abelian case, try the free group on 2 generators. Nov 13, 2019 at 22:43

There are several problems with what you write, although it can be made to work by taking a few steps "back."

First: it is not true that if you pick an arbitrary generating set for a finitely generated abelian group $$G$$, say $$g_1,\ldots,g_n$$, then you will necessarily have that $$G$$ is the direct sum of the cyclic groups generated by the $$g_i$$; even if you pick your set to be minimal. For example, in $$G=\mathbb{Z}$$, then $$g_1=2$$ and $$g_2=3$$ generate, no proper subset of $$\{g_1,g_2\}$$ generate, but $$G$$ is not isomorphic to $$\langle g_1\rangle \oplus \langle g_2\rangle$$.

It is true that one may select a suitably chosen generating set with that property, but this fact is not immediate or immediately obvious.

Second, even if you know that $$G=\langle g_1\rangle\oplus \cdots \oplus \langle g_n\rangle$$, it does not follow that if $$H$$ is a subgroup of $$G$$ then you can write $$H=H_1\oplus \cdots \oplus H_n$$ with $$H_i$$ a subgroup of $$\langle g_i\rangle$$. For example, the diagonal subgroup $$H=\{(n,n)\in\mathbb{Z}\oplus\mathbb{Z}\mid n\in\mathbb{Z}\}$$ of $$\mathbb{Z}\oplus\mathbb{Z}$$ is not equal to the direct sum of a subgroup of $$\{(n,0)\mid n\in\mathbb{Z}\}$$ and a subgroup of $$\{(0,m)\mid m\in\mathbb{Z}\}$$.

The following is true, however:

Theorem. Let $$F$$ be a finitely generated free abelian group. If $$H$$ is a subgroup of $$F$$, and $$H\neq\{0\}$$, then there exists a basis $$x_1,\ldots,x_n$$ of $$F$$, an integer $$r$$, $$1\leq r\leq n$$, and integers $$d_1,\ldots,d_r$$ such that $$d_i\gt 0$$, $$d_1|\cdots|d_r$$, and $$d_1x_1,\ldots,d_rx_r$$ is a basis for $$H$$.

Taking this for granted, let $$G$$ be a finitely generated abelian group. Let $$X$$ be a generating set. Then $$G$$ is a quotient of a free abelian group $$F$$ of rank $$n=|X|$$, $$G\cong F/N$$.

If $$H$$ is a subgroup of $$G$$, then $$H$$ corresponds to a subgroup $$K$$ of $$F$$ that contains $$N$$, with $$H\cong K/N$$. By the theorem, $$K$$ is generated by $$r\leq n$$ elements, and therefore so is $$K/N$$. So $$H$$ is finitely generated.

As for examples in the nonabelian case, I'm not sure if your idea with $$D_n$$ will work; notice that composing reflections can yield a rotation! For instance, in $$D_4$$, the relection of the square about the $$x$$ axis composed with the reflection about the $$y$$ axis results in a rotation, not a reflection. So you aren't just going to get "the reflections", you are going to get the whole of $$D_{2n}$$.

For an example you can get your hands on, consider the group $$G$$ of the $$2\times 2$$ invertible matrices generated by $$\left(\begin{array}{cc} 1 & 1\\ 0 & 1 \end{array}\right) \qquad \text{and}\qquad \left(\begin{array}{cc}2 & 0\\ 0&1\end{array}\right),$$ and let $$H$$ be the subgroup of elements of $$G$$ whose main diagonal entries are both equal to $$1$$. Verify that $$H$$ is a subgroup of $$G$$ that is \textit{not} finitely generated.

• Perfect examples, thanks a lot! Is the argument for breaking $G$ into direct sums true for a basis? (As $g_1=2,g_2=3$ in $\mathbb{Z}$ isn't a basis as $3g_1=2g_2$.) A basis should have trivial intersections in the generated cyclic subgroups. I'll think about it more. Thanks again! (and oops! I totally overlooked a composition of reflections reorients the vertices) Nov 13, 2019 at 23:12
• @oshill: The only abelian groups that have bases are free abelian groups. Rather, you can find (using the theorem I quote) a generating set $g_1,\ldots,g_k$ for $G$ such that if $a_1g_1+\cdots+a_ng_n=0$, then $a_ig_i=0$ for all $i$ (which is not quite a basis, as that would require $a_1=\cdots=a_n=0$). But guaranteeing that such a set exists is not trivial or obvious without the theorem quoted. You can't just say "take a basis" without proving that there is a basis. It is true that if $X$ is a basis for the abelian group $F$, then $F=\oplus_{x\in X}\langle x\rangle$. Nov 13, 2019 at 23:18