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