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Let $G = G_1 \ast G_2$. Then $$G/[G,G] \simeq G_1/[G_1, G_1] \oplus G_2/[G_2,G_2].$$

This is Munkres, section 69, exercise 1; with the hint to use the extension condition for free groups and direct sums in each direction, so find 2 homomorphisms, and show that they are mutual inverses.

It is easy to find some homomorphism in either direction:

(i) left-hand-side (LHS) to right-hand-side (RHS): as RHS is the direct sum of two abelian groups, it is abelian. Any homomorphism from $G$ to RHS will then induce a homomorphism from LHS to RHS (eg, from the elementary first isomorphism theorem).

(ii) RHS to LSH: eg, mapping $x[G_1, G_1]$ to $x[G,G]$ is a homomorphism that extends to one on $G$, using the extension condition for direct sums.

However, (i) is just an abstract statement from which I do not see how to make progress in showing that we have mutual inverses (nor does (ii) help me much so far).

Note: at this point in the book, Munkres has only introducted fundamental groups, covering spaces, deformation retracts, etc, plus a few chapters on free groups and direct sums, to the level of showing the fundamental groups of the Torus, some discussion on figure 8 etc. There are probably very abstract ways in which this might follow; but if you could kindly stick to as elementary a proof or hint as possible, I would appreciate! My group theory knowledge is currently at about the level of one class of group theory.

P.S.: This is not homework. I'm reading Munkres for fun, and try to do every exercise. Usually, this goes fine. But I'm really stuck here.

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Hint. $G/[G,G]$ has the 'universal property' of the direct sum of $G_1/[G_1,G_1]$ and $G_2/[G_2,G_2]$, which Munkres paraphrased as 'extension property'. – cjackal Nov 30 '12 at 2:29
Thanks! The way I read your hint though (probably incorrectly), isn't that what I mention in (ii) above? If yes, I get stuck after observing that. – gnometorule Nov 30 '12 at 2:39
the extension property of direct sum, free product, etc in group theory (usually) means not merely the existence of the extended homomorphism, but also the uniqueness of this extension. Hence if two groups satisfy this 'extension property', then this two groups are same up to isomorphism. That's what I mean. – cjackal Nov 30 '12 at 2:53
up vote 2 down vote accepted

I only managed to find a direct proof. Let $$N:= [G, G], \quad N_i = [G_i, G_i].$$ The key fact to use is that $G/N$ and $G_i/N_i$ are abelian. Call this (1).

Note first that in the finite case for abelian groups $G, H$, we have $G \bigoplus H = G \times H$. This is the notation we will use. Let

$$\begin{align} \phi: & G/N \rightarrow G_1/N_1 \times G_2/N_2 \\ &xyN \mapsto (xN_1, yN_2) \\ \end{align}$$ $$(\phi(\prod x_iy_i N) = (\prod x_i N_1, \prod y_i N_2)),$$

which, using (*), is well-defined, and an obvious surjective homomorphism.

To see that $\phi$ is injective, note first that $$e_{G/N} = N, e_{G_i/N_i} = N_i, e_{G_1/N_1 \times G_2/N_2} = N_1 \times N_2.$$

Assume $x y \tilde{x} \tilde{y} N \mapsto N_1 \times N_2$. Then $x\tilde{x} \in N_1 \subset N, y\tilde{y} \in N_2 \subset N$, so, using (1), $$xy\tilde{x}\tilde{y}N = x\tilde{x}y\tilde{y}N= N.$$

Similarly, if $\prod_{i=1}^nx_iy_iN \mapsto N_1 \times N_2$, we have (using (1)) $$\prod_{i=1}^{n}x_i \in N_1 \subset N, \quad \prod_{i=1}^{n}y_i \in N_2 \subset N,$$ so, using (1) again, $$\prod_{i=1}^{n} x_iy_iN = (\prod_{i=1}^{n}x_i)(\prod_{i=1}^{n}y_i)N = N.$$

We conclude that $\phi$ is injective as well, and so defines the desired isomorphism.

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