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Let $H,K$ be two subgroups of a group $G$ and let $A=H\cap K$ and $B$ be the subgroup of $G$ generated by $H$ and $K$. We know that \begin{array}{rcl} A & \rightarrow & H \\ \downarrow & & \downarrow \\ K & \rightarrow & G \end{array} is a pull-back square (morphisms are inclusions).

Does there exists a similar interpretation for $B$?

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2 Answers 2

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The most blunt way of describing $\langle H \cup K \rangle \le G$ is as the image of the canonical homomorphism $H * K \to G$ (determined by the inclusions $H \hookrightarrow G, K \hookrightarrow G$), where $H * K$ denotes the coproduct (free product) of $H$ and $K$ as abstract groups.

This generalises readily to any cocomplete regular category and any small family of subobjects. (Exercise.)

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$H$ and $K$ are both objects in the category of subobjects of $G$. The category of subobjects is a preorder (exercise), so it behaves like a poset except that some objects are isomorphic. The categorical product in this category, which agrees with the pullback, is the intersection $H \cap K$, and the categorical coproduct in this category is the subgroup of $G$ generated by $H$ and $K$.

The nLab asserts that in a topos, the coproduct of subobjects is their pushout along their product, but this is false for groups.

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