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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.

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  • $\begingroup$ 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. $\endgroup$ – Ferra Nov 13 '19 at 22:43
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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.

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  • $\begingroup$ 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) $\endgroup$ – oshill Nov 13 '19 at 23:12
  • $\begingroup$ @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 $F=\oplus_{x\in X}\langle x\rangle$. $\endgroup$ – Arturo Magidin Nov 13 '19 at 23:18

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