## An Isomorphism That Does Not Give Rise to a Congruence Relation

Let
$V=\mathbb{R}^3$
and let
$S= \{(a,0,b):a,b \in \mathbb{R} \}$

$S= \{(0,0,b):a,b \in \mathbb{R} \}$

Now
$S, \: \subset V$
and
$S$
and
$T$
are isomorphic.
$\mathbb{R} \cap T = \emptyset$

Hence
$\mathbb{R}^3$
can be decomposed into the direct sum of disjoint subspaces of
$\mathbb{R}^3$
,
$\mathbb{R}^3 = \mathbb{R} \oplus T$
.
We can not write
$S$
in this way.
Hence
$S, \: T$
are isomorphic but do not obey any congruence relation.