# Manipulating quotients and direct sums of abelian groups

I'm studying homology from Hatcher's Algebraic Topology. I feel that there is a gap in my group theory knowledge that is making me struggle with this chapter. In particular, the book (and material online) use the following without proof:

• If $$A/B \cong C$$, where $$A,B$$ are abelian groups and $$C$$ is abelian and free, then $$A \cong B \oplus C$$.
• If $$\mathbb{Z}^n/A \cong \mathbb{Z}$$, then $$A \cong \mathbb{Z}^{n-1}$$.

These two seem very related. I'm looking for a proof that doesn't use exact sequences or Category Theory (not introduced in the book at that point yet).

I know the fundamental theorem of finitely generated abelian groups, Sylow theorems, and the basics of ring theory (ideals, domains, CRM, etc). Basically, I read most of Artin's Algebra.

Thank you.

For this it is helpful to think of abelian groups as modules over $\mathbb{Z}$, as Artin discusses in Ch. 12 (in the first edition anyway). You didn't mention modules so I'll articulate the proof without explicit use of this idea, but if you have gotten to Ch. 12, then I think it will clarify things to think in those terms.

$A/B\cong C$ means we have a surjective homomorphism from $A$ to $C$ with kernel $B$. Call it $f:A\rightarrow C$. Because $C$ is free, it has a set of generators $G$ such that every element of $C$ is represented uniquely in terms of the generators. Each generator lies in $f$'s image, because $f$ is surjective. Find one preimage in $A$ for each generator of $C$, and consider the subgroup $C'$ of $A$ that they generate.

I claim (i) that $C'$ is isomorphic to $C$, in fact $f$ restricted to $C'$ is an isomorphism $C'\rightarrow C$, and (ii) $A$ is the direct sum of $B$ and $C'$.

I'm inclined to give you the satisfaction of proving the claim yourself, but let me know if you'd like me to write out the proof.

Anyway, this handles the first bullet.

You're right that the second bullet is closely related. It is basically the above argument plus one additional feature, which is that any subgroup of $\mathbb{Z}$ is isomorphic to $\mathbb{Z}$ itself, which comes from the division algorithm. (Take the smallest positive element $p$ in any nontrivial subgroup $A$ of $\mathbb{Z}$; it must divide every element of $A$, or else the remainder would be a smaller positive element in $A$. Then $A$ must be the cyclic group generated by $p$.)

The outline of the argument is this: (1) any subgroup of the free finitely generated abelian group $\mathbb{Z}^n$ is free, finitely generated and abelian - therefore $A$ is all these things, i.e. $A\cong \mathbb{Z}^m$ for some $m$; (2) the first bullet tells us that $\mathbb{Z}^n \cong A\oplus \mathbb{Z}$; and (3) therefore $m=n-1$, otherwise the ranks don't match.

The question is why is (1) true, i.e. why is it that subgroups of free, finitely generated abelian groups are free, finitely generated and abelian. Abelianness is trivial because subgroups of abelian groups are always abelian. The interesting thing is why a subgroup has to be free and finitely generated (in fact, I claim, by at most as many generators as the ambient group has). This is a second application of your first bullet, plus the fact that all nontrivial subgroups of $\mathbb{Z}$ are isomorphic to $\mathbb{Z}$. Here is an argument by induction:

Let $A$ be any subgroup of $\mathbb{Z}^n$. If $n=1$, then either $A=0$ or $A$ is isomorphic to $\mathbb{Z}$ as a group, so it is rank $0$ or $1$, and either way it is free. This handles the base case.

So suppose $n>1$, and for all $m<n$, the induction assumption holds: every subgroup of a rank $m$ free abelian group is free abelian of rank $\leq m$. Now $\mathbb{Z}^n$ has a projection $\pi$ to its first coordinate. Restrict this map to $A$; the image is a subgroup of $\mathbb{Z}$. Again, it is either $0$, or else isomorphic to $\mathbb{Z}$ itself.

In the former case, $A$ is completely contained in the kernel of $\pi$, which is the subgroup of $\mathbb{Z}^n$ generated by the last $n-1$ coordinates. This is obviously free abelian and rank $n-1$, so the induction hypothesis kicks in to tell us that $A$ is free abelian and rank $\leq n-1$.

In the latter case, $\pi$ restricted to $A$ is a surjective homomorphism from $A$ to a group $C$ isomorphic to $\mathbb{Z}$. As in the proof of your first bullet, we construct a subgroup $C'$ of $A$ by finding any preimage of the generator for $C$ under $\pi|_A$ and seeing the subgroup of $A$ that it generates. As in the proof of the first bullet, it follows that $A = C'\oplus B$, where $B$ is the kernel of $\pi|_A$. Now $B$ is contained in the kernel of $\pi$, and is therefore (just as in the last paragraph) free and of rank $\leq n-1$ by the induction hypothesis. Meanwhile, $C'$ is free rank $1$ by construction. Thus $A$ is free and rank $\leq (n-1)+1=n$.

This completes the proof of your second bullet.

Some commentary:

In case you know a little more about Artin's Ch. 11 and Ch. 12, here's some general context to put these arguments into:

Your first bullet is true in great generality. If $R$ is any ring, and $A$ is an $R$-module and $B$ a submodule, and $A/B=C$ is a free $R$-module, then $A\cong B\oplus C$ by the same argument I gave above: let $f$ be the canonical homomorphism from $A$ to $C$. Lift any basis of $C$ to $A$ and consider the submodule $C'\subset A$ that it generates. The case at hand is the case $R=\mathbb{Z}$.

The second bullet requires the ring to be a p.i.d. I.e., if $R$ is a p.i.d. and $R^n/A=R$, then $A\cong R^{n-1}$. The proof is exactly the same as the above, with all references to the fact that nontrivial subgroups of $\mathbb{Z}$ are isomorphic to it replaced with the analogous fact about p.i.d.'s: all nonzero ideals of a p.i.d. $R$ are isomorphic to it as $R$-modules. (If an ideal $I$ in $R$ is principal and generated by $a$, and $a$ is not a zero-divisor, then multiplication by $a$ is an $R$-module isomorphism from $R$ to $I$.) Likewise, it follows that all submodules of a free, rank $n$ $R$-module are free and of rank $\leq n$.

• Amazing answer. It's answers like this that make this site awesome. Thank you very much. My problem was looking at group theory chapters in books. I should have looked at modules. I had passing familiarity with them, not enough to see the relation here. Once you think of it in terms of modules (linear algebra), it's fairly easy! Nov 3, 2013 at 10:30
• I think this answer deserves more than one upvote. I'll create a bounty once I can (in a few hours). Nov 3, 2013 at 10:31