Proving $xRy\iff xy^{-1}\in\ker(f)$ is equivalent relation, where $f:(G,.) \to (H,.)$ is homomorphism of groups $G$ and $H$.

Question

I know for sure it is but I failed to prove how. I tried to use the homomorphism definition where $f(x*y^{-1}) = f(x)*f(y^{-1})$ but I didn't see a pattern.

Answer 1

Note that $ {\rm Ker}(f) = \{x\in G|\, f(x) = 1_H\}.$

$\bullet$ Reflexive property: Since $f$ is a homomorphism, $f(1_G) = 1_H.$ So, for all $x\in G,$ we have $$ f(xx^{-1}) = f(1_G) = 1_H,$$ so $xx^{-1}\in {\rm Ker}(f).$ This implies $xRx.$

$\bullet$ Symmetric property: For all $x, y\in G,$ if $xRy,$ then $xy^{-1}\in {\rm Ker}(f).$ So $f(x) = f(y).$ Then, we have $$ f(yx^{-1}) = f(y)[f(x)]^{-1} = f(x)[f(x)]^{-1} = 1_H,$$ so, $yx^{-1}\in {\rm Ker}(f).$ This implies $yRx.$

$\bullet$ Transitive property: For all $x, y, z\in G,$ if $xRy$ and $yRz,$ then $xy^{-1}\in {\rm Ker}(f)$ and $yz^{-1}\in {\rm Ker}(f).$ So $f(x) = f(y) = f(z).$ Then, we have $$ f(xz^{-1}) = f(x)[f(z)]^{-1} = f(z)[f(z)]^{-1} = 1_H.$$ so, $xz^{-1}\in {\rm Ker}(f).$ This implies $xRz.$

Answer 2

The relation $R$ is equivalent relation iff

(1) $xRx$

(2) $\forall x, y\in G, xRy \implies yRx$

(3) $\forall x, y, z\in G, xRy$ and $yRz\implies xRz$

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Note that $f(y^{-1})=f(y)^{-1}$ by the definition of the homomorphism. Using these conditions,

(1) somehow trivial: $xx^{-1}=e\in \ker(f)$ It is not that hard to show that $e\in\ker(f)$

(2) $xy^{-1}\in\ker(f)$ implies $f(x)f(y)^{-1}=f(e)$. The only thing we have to show is $f(y)f(x)^{-1}=f(e)$, and it is not difficult.

(3) In the same mood, using $f(x)f(y)^{-1}=f(e)$ and $f(y)f(z)^{-1}=f(e)$, show that $f(x)f(z)^{-1}=f(e)$.

Answer 3

The condition $xy^{-1}\in K$ defines an equivalence relation in $G$ for every subgroup $K\le G$, so it does a fortiori for $K=\operatorname{ker}(f) \le G$.

In fact:

a) $xx^{-1}= e\in K$ (reflective);

b) $xy^{-1}\in K \Rightarrow (xy^{-1})^{-1}=yx^{-1}\in K$ (symmetric);

c) $xy^{-1}\in K\wedge yz^{-1}\in K \Rightarrow (xy^{-1})(yz^{-1})=x(y^{-1}y)z^{-1}=xz^{-1}\in K$ (transitive).