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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.

Srivatsan
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Manos
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1 Answers1

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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.

Arturo Magidin
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