Subquotient

The relation subquotient of is an order relation.

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 ${\displaystyle \varphi \colon X\to Y,\;g\mapsto g\,G''}$

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

with suitable ${\displaystyle g\in X}$ are surjective for the respective pairs

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

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

${\displaystyle ,}$

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

${\displaystyle ,}$

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

${\displaystyle ,}$

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

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 ${\displaystyle H'/H''\cong \varphi ^{-1}\left(H'\right)/\varphi ^{-1}\left(H''\right)}$.

• Homological algebra