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Referring to p. 75 in Lang's Algebra, the statement of Goursat's lemma seems unclear to me. What exactly are the projections $p_1$ and $p_2$? Are they the standard projections $(x_1,x_2) \stackrel{p_i}{\mapsto} x_i$, for $i=1, 2$? If yes, then $p_i$ being surjective does it not imply that $H$ is exactly $G \times G^{'}$? Or is it meant that e.g. $p_1 : H \longrightarrow G$ is just a surjective homomorphism without any other assumption about the mapping $p_1$?

Here is a link to the statement of the lemma, exactly as it appears in Lang: Goursat's lemma on Wikipedia.

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Can you please include the full statement or a related link for the people who doesn't have access to the book? – user13838 Sep 10 '11 at 23:21
Thanks for the hint; done :-) – Manos Sep 11 '11 at 12:27
up vote 9 down vote accepted

Yes, the projections are the projections onto the first and second coordinate.

No, the fact that the projections are onto does not imply that $H\leq G\times G$ is $G\times G$. For example, take $G=\mathbb{Z}_2$, $G\times G = \mathbb{Z}_2\times \mathbb{Z}_2$, and $H=\{(0,0), (1,1)\}$. Then $H\neq G\times G$, but $\pi_1(H)=G$ and $\pi_2(H) = G'$.

The fact that the canonical projections from $H\leq G\times K$ to $G$ and $K$ are onto means that for every $g\in G$ there is (at least one) $k\in K$ such that $(g,k)\in H$, and that for every $x\in K$ there is (at least one) $a\in G$ such that $(a,x)\in H$. Of course, the natural projections are group homomorphisms.

One natural way to construct such subgroups is to consider an onto group homomorphism $f\colon G\to K$, and let $H=\{ (g,k)\in G\times K \mid f(g)=k\}$. This is a subgroup, since $(e,e)\in H$; and if $(g,k),(x,y)\in H$ then $f(g)=k$, $f(x)=y$, so $f(gx^{-1}) = ky^{-1}$, hence $(g,k)(x,y)^{-1} = (gx,ky^{-1})\in H$. Moreover, since we are assuming that $f$ is onto, for every $k\in K$ there exists $g\in G$ such that $f(g)=k$, so $(g,k)\in H$, hence $k\in p_2(H)$; and of course, for every $g\in G$, $(g,f(g))\in H$, so $g\in \pi_1(H)$. Thus, $H$, $G$, and $K$ satisfy the hypothesis of Goursat's Lemma.

A slightly more general example is obtained if you have a surjection $f$ from $G$ onto a quotient $K/N$ of $K$, and let $H=\{(g,k)\in G\times K \mid f(g) = kN\}$. I'll leave it to you to convince yourself that such an $H$ also satisfies the hypothesis of Goursat's Lemma.

What Goursat's Lemma essentially says is that every subgroup that satisfies the hypothesis is actually obtained from an example like the latter.

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