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Let H and K be normal subgroup of a group G. The following square is always a pullback square? $$\begin {matrix} G/H\cap K\rightarrow &G/K\\ \downarrow&\downarrow\\ G/H\rightarrow&G/HK\\ \end {matrix} $$

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What have you tried? –  Erik Vesterlund Jul 30 '13 at 21:58
    
Given a group X and two group homomorphisms $ u:X\to G/H$ and $ v:X\to G/K$ such that $ up=vq $ where $ p:G/H\to G/HK $ and $ q:G/K\to G/HK $ we have to find a homomorphism $ w$ such that $ wa=u$ and $ wb=v$ where a and b are the projection. –  Fabio Lucchini Jul 30 '13 at 22:08
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up vote 2 down vote accepted

Yes, it is a pullback square, and this is a nice generalization of the Chinese Remainder Theorem (the same proof works for ideals of a ring, and they don't have to be coprime!). Unfortunately it is not well-known or at least you cannot find it in many books, I only know that it appears as a little Lemma somewhere in EGA. With this you can also compute Galois groups of composite polynomials.

The proof is very easy, using the usual explicit construction of the pullback (as a subgroup of the direct product). Obviously the square commutes, therefore we get a map $G/(H \cap K) \to G/K \times_{G/(HK)} G/K$, namely $[g] \mapsto ([g],[g])$. One checks immediately that it is injective. Now for surjectivity, let $([g],[g'])$ be in the image, i.e. $g=g'hk$ for some $h \in H, k\in K$. Then $[gk^{-1}]$ is a preimage, since $gk^{-1} \equiv g \bmod K$ and $gk^{-1} =g'h \equiv g' \bmod H$. qed

The same proof works when $H$ and $K$ are just subgroups (not assumed to be normal), but then it is a pullback square in the category of $G$-sets.

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Thank's for confirm my think. I showed that this is also a pulation square, that's both a pullback and pushout square, it's right? –  Fabio Lucchini Aug 5 '13 at 22:19
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Yes that's right. –  Martin Brandenburg Aug 6 '13 at 8:02
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