# Subquotient

In the mathematical fields of category theory and abstract algebra, a subquotient is a quotient object of a subobject. Subquotients are particularly important in abelian categories, and in group theory, where they are also known as sections, though this conflicts with a different meaning in category theory.

For example, of the 26 sporadic groups, 20 are subquotients of the monster group, and are referred to as the "Happy Family", while the other 6 are pariah groups.

A quotient of a subrepresentation of a representation (of, say, a group) might be called a subquotient representation; e.g., Harish-Chandra's subquotient theorem.

In constructive set theory, where the law of excluded middle does not necessarily hold, one can consider the relation 'subquotient of' as replacing the usual order relation(s) on cardinals. When one has the law of the excluded middle, then a subquotient $X$ of $Y$ is either the empty set or there is an onto function $Y\to X$ . This order relation is traditionally denoted $\leq ^{\ast }$ . If additionally the axiom of choice holds, then $X$ has a one-to-one function to $Y$ and this order relation is the usual $\leq$ on corresponding cardinals.

## Transitive relation

The relation »is subquotient of« is transitive.

Proof

Let $G,H,J$  groups and $\phi \colon G\to H$  and $\psi \colon H\to J$  be group homomorphisms, then also the composition

$\psi \circ \phi \colon G\to J,g\mapsto (\psi \circ \phi )(g):=\psi (\phi (g))$

is a homomorphism.

If $U$  is a subgroup of $G$  and $V$  a subgroup of $\phi (U)$ , then $U':=\phi ^{-1}(V)$  is a subgroup of $U\leq G$ . We have $\phi (U')\subseteq V$ , indeed $\phi (U')=V$ , because every $v\in V\subseteq \phi (U)$  has a preimage in $U$ . Thus $(\psi \circ \phi )(U')=\psi (V)$ . This means that the image, say $\psi (V)$ , of a subgroup, say $V$ , of $H$  is also the image of a subgroup, namely $U'$  under $\psi \circ \phi$ , of $G$ .

In other words: If $\psi (V)$  is a subquotient of $\phi (U)$  and $\phi (U)$  is subquotient of $G$  then $\psi (V)$  is subquotient of $G$ .  ■