Congruence Modulo Normal Subgroup is Congruence Relation

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Theorem

Let $\struct {G, \circ}$ be a group.

Let $N$ be a normal subgroup of $G$.


Then congruence modulo $N$ is a congruence relation for the group operation $\circ$.


Proof

Let $x \mathrel {\RR_N} y$ denote that $x$ and $y$ are in the same coset, that is:

$x \mathrel {\RR_N} y \iff x \circ N = y \circ N$

as specified in the definition of congruence modulo $N$.


Let $x \mathrel {\RR_N} x'$ and $y \mathrel {\RR_N} y'$.

To demonstrate that $\RR_N$ is a congruence relation for $\circ$, we need to show that:

$\paren {x \circ y} \mathrel {\RR_N} \paren {x' \circ y'}$


So:

\(\ds \paren {x \circ y} \circ \paren {x' \circ y'}^{-1}\) \(=\) \(\ds \paren {x \circ y} \circ \paren {y'^{-1} \circ x'^{-1} }\) Inverse of Group Product
\(\ds \) \(=\) \(\ds \paren {\paren {x \circ y} \circ y'^{-1} } \circ x'^{-1}\) Group Axiom $\text G 1$: Associativity
\(\ds \) \(=\) \(\ds \paren {x \circ \paren {y \circ y'^{-1} } } \circ x'^{-1}\) Group Axiom $\text G 1$: Associativity


By Cosets are Equal iff Product with Inverse in Subgroup:

$x \circ x'^{-1} \in N$ and $y \circ y'^{-1} \in N$

Thus:

$\paren {x \circ y} \circ \paren {x' \circ y'}^{-1} \in x \circ H \circ x'^{-1}$

But we also have that:

\(\ds x \circ H \circ x'^{-1}\) \(=\) \(\ds H \circ x \circ x'^{-1}\) Definition of Normal Subgroup
\(\ds \) \(\subseteq\) \(\ds H \circ H\) Definition of Subset Product: $x \circ x'^{-1} \in H$
\(\ds \) \(=\) \(\ds H\) Product of Subgroup with Itself

That is:

$\paren {x \circ y} \circ \paren {x' \circ y'}^{-1} \in H$

and so:

$\paren {x \circ y} \mathrel {\RR_N} \paren {x' \circ y'}$

Hence the result, by definition of congruence relation.

$\blacksquare$


Also see


Sources

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