IIT JAM 2019 Abstract Algebra: Question (2)

Question:

Let $G$ be a non abelian group, $y\in G$. let the maps $f, g, h$ from $G$ to itself is defined by 

$$f(x)=yxy^{-1}, g(x)=x^{-1}\text{ and }h=g•g$$ Then,

(A) $g$ and $h$ are homomorphism and $f$  is not homomorphism.

(B) $h$ is homomorphism, $g$ is not homomorphism.

(C) $f$ is homomorphism, $g$ is not homomorphism.

(D) $f,g,h$ all are homomorphism.

Solution:  Since $y^{-1}y=e\in G$ where $e$ is identity element in $G$. Hence we have,

$f(ab)=yaby^{-1}=yay^{-1}yby^{-1}=(yay^{-1})(yby^{-1})=f(a)f(b)$ for all $a,b\in G$.

Hence by definition of homomorphism $f$ is homomorphism. 

Note that, $g$ is not a homomorphism since  $g(ab)=(ab)^{-1}=b^{-1}a^{-1}$. Whereas,  $g(a)g(b)=a^{-1}b^{-1}$ so that  $g(ab)$ not always equal to $g(a)g(b)$ ( since there are some $a,b\in G$ such that $b^{-1}a^{-1}≠a^{-1}b^{-1}$ as $G$ is non-abelian, otherwise $G$ will be abelian)

Finally as, 

 $h(ab)=g•g(ab)=g(g(ab))=g(b^{-1}a^{-1})=(a^{-1})^{-1}(b^{-1})^{-1}=ab=h(a)h(b)$ for all $a,b\in G$. 

Hence by definition of homomorphism, $h$ is a homomorphism from $G$ to itself. (Note here $h$ is identity map from $G$ to itself)

Hence answer is, option (B), (C).