If $f:G \rightarrow G$ is a group homomorphism and $f(ab) = f(b)f(a)$ does this show that $f$ is NOT a homomorphism?

Question

Suppose $G$ is a group and $f : G \to G$ is the function $f(x) = x^{-1}$.

Prove that if $f$ is a homomorphism, then $G$ is abelian.

Proceeding by contrapositive suppose $G$ is not abelian then for every $a,b$ in $G$, $$f(ab) = {(ab)^-}^1 = {b^-}^1{a^-}^1 = f(b)f(a);$$

thus $f$ is not a homomorphism since the homomorphism property $f(ab) = f(a)f(b)$ fails.

My question is different from the possible duplicate provided because I want to know if showing that $$f(ab) = {(ab)^-}^1 = {b^-}^1{a^-}^1 = f(b)f(a);$$ is a correct method to show that a function is NOT a homomorphism

Answer 1

No.

It is enough to show that $f(ab)=f(b)f(a)$ for some homomorphism $f$. This holds when the domain group is abelian.

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We have for a homomorphism $f:G\to G, x\mapsto x^{-1}$ that

$$f(ab)=f(a)f(b)\tag{1}.$$

From the LHS of $(1)$, we have $f(ab)=(ab)^{-1}=b^{-1}a^{-1}=f(b)f(a)$.

From the RHS of $(1)$, we have $f(a)f(b)=a^{-1}b^{-1}$.

Hence $b^{-1}a^{-1}=a^{-1}b^{-1}$.

Hence $G$ is abelian.

Answer 2

Answering the title question. If $a,b\in Z(G)$ then clearly $f(b)f(a)=f(a)f(b)$ satisfy the homomorphism property.

But in general it doesn't, for example in $\mathbb{H}$: $f(ik)=(ik)^{-1}=k^{-1}i^{-1}=(-k)(-i)=ki\neq ik=f(i)f(k)$