Proof of transitivity for groups

Let ${\displaystyle H'/H''}$ be subquotient of ${\displaystyle H}$, furthermore ${\displaystyle H:=G'/G''}$ be subquotient of ${\displaystyle G}$ and ${\displaystyle \varphi \colon G'\to H}$ be the canonical homomorphism. Then all vertical (${\displaystyle \downarrow }$) maps ${\displaystyle \varphi \colon X\to Y,\;x\mapsto x\,G''}$

		${\displaystyle G''}$	${\displaystyle \leq }$	${\displaystyle \varphi ^{-1}(H'')}$	${\displaystyle \leq }$	${\displaystyle \varphi ^{-1}(H')}$	${\displaystyle \vartriangleleft }$	${\displaystyle G'}$
	${\displaystyle \varphi \!:}$	${\displaystyle {\Big \downarrow }}$		${\displaystyle {\Big \downarrow }}$		${\displaystyle {\Big \downarrow }}$		${\displaystyle {\Big \downarrow }}$
		${\displaystyle \{1\}}$	${\displaystyle \leq }$	${\displaystyle H''}$	${\displaystyle \vartriangleleft }$	${\displaystyle H'}$	${\displaystyle \vartriangleleft }$	${\displaystyle H}$

are surjective for the respective pairs

${\displaystyle (X,Y)\;\;\;\in }$

${\displaystyle {\Bigl \{}{\bigl (}G'',\{1\}{\bigr )}{\Bigr .}}$

${\displaystyle ,}$

${\displaystyle {\bigl (}\varphi ^{-1}(H''),H''{\bigr )}}$

${\displaystyle ,}$

${\displaystyle {\bigl (}\varphi ^{-1}(H'),H'{\bigr )}}$

${\displaystyle ,}$

${\displaystyle {\Bigl .}{\bigl (}G',H{\bigr )}{\Bigr \}}.}$

The preimages ${\displaystyle \varphi ^{-1}\left(H'\right)}$ and ${\displaystyle \varphi ^{-1}\left(H''\right)}$ are both subgroups of ${\displaystyle G'}$ containing ${\displaystyle G'',}$ and it is ${\displaystyle \varphi \left(\varphi ^{-1}\left(H'\right)\right)=H'}$ and ${\displaystyle \varphi \left(\varphi ^{-1}\left(H''\right)\right)=H'',}$ because every ${\displaystyle h\in H}$ has a preimage ${\displaystyle g\in G'}$ with ${\displaystyle \varphi (g)=h.}$ Moreover, the subgroup ${\displaystyle \varphi ^{-1}\left(H''\right)}$ is normal in ${\displaystyle \varphi ^{-1}\left(H'\right).}$

As a consequence, the subquotient ${\displaystyle H'/H''}$ of ${\displaystyle H}$ is a subquotient of ${\displaystyle G}$ in the form ${\displaystyle H'/H''\cong \varphi ^{-1}\left(H'\right)/\varphi ^{-1}\left(H''\right).}$

Relation to cardinal order

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 ${\displaystyle Y}$ of ${\displaystyle X}$ is either the empty set or there is an onto function ${\displaystyle X\to Y}$. This order relation is traditionally denoted ${\displaystyle \leq ^{\ast }.}$ If additionally the axiom of choice holds, then ${\displaystyle Y}$ has a one-to-one function to ${\displaystyle X}$ and this order relation is the usual ${\displaystyle \leq }$ on corresponding cardinals.