Question

Let \( G \) be a nonabelian group, \( y \in G \), and let the maps \( f, g, h \) from \( G \) to itself be defined by \[ f(x) = yxy^{-1}, \quad g(x) = x^{-1} \quad \text{and} \quad h = g \circ f \circ g. \] Then

• A. \( g \) and \( h \) are homomorphisms and \( f \) is not a homomorphism
• B. \( h \) is a homomorphism and \( g \) is not a homomorphism
• C. \( f \) is a homomorphism and \( g \) is not a homomorphism
• D. \( f, g \) and \( h \) are homomorphisms

Hint:

Conjugation and inversion are both homomorphisms in group theory. The composition of homomorphisms is also a homomorphism.

Answer 1

The Correct Option is B, C

Step 1: Verify if \( f \) is a homomorphism.
The map \( f(x) = yxy^{-1} \) is a conjugation map, which is a homomorphism. This is because: \[ f(xy) = y(xy)y^{-1} = (yx)(y^{-1}y) = yx y^{-1} = f(x)f(y). \]
Step 2: Verify if \( g \) is a homomorphism.
The map \( g(x) = x^{-1} \) is a homomorphism, since: \[ g(xy) = (xy)^{-1} = y^{-1} x^{-1} = g(y) g(x). \]
Step 3: Verify if \( h \) is a homomorphism.
The map \( h = g \circ f \circ g \) is also a homomorphism because it is the composition of homomorphisms: \[ h(xy) = g(f(g(xy))) = g(f(g(x)g(y))) = g(f(g(x))f(g(y))) = h(x) h(y). \]
Step 4: Conclusion.
Thus, the correct answer is \( \boxed{(D)} \).