Proving $f(a) = a^{-1}$ is a homomorphism iff $G$ is commutative

Question

Problem.

Let $G$ be a group. Define a function $f : G \longrightarrow G$ by for all $a \in G$, $f(a) = a^{-1}$.

Prove that $f$ is a homomorphism if and only if $G$ is commutative.

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My attempt.

Assume G is commutative;

Then $$f(x * y) = (x * y)^{-1}$$ $$\Rightarrow f(x * y) = y^{-1} * x^{-1}$$ Note: (am I making this up? I remember this from linear algebra and seem to recall it from this course as well.)

$$\Rightarrow f(y) * f(x) = f(x) * f(y)$$ as $G$ is commutative.

Let us now assume that $G$ is homomorphism;

$$f(x * y) = f(x) * f(y)$$ $$\Rightarrow f(x * y) = x^{-1} * y^{-1}$$

At this point I would try to show that $$x^{-1} * y^{-1} \neq y^{-1} * x^{-1}$$

Does not work and therefore it has to be commutative as well, but I don't want to spend more energy on this before knowing that I am doing this right and not just making things up! Thank you.

Answer 1

Hint: if $f$ is a homomorphism, then $$ f((xy)^{-1})=xy $$ by definition, but also \begin{align} f((xy)^{-1})&=f(y^{-1}x^{-1}) &&\text{basic identity}\\ &=f(y^{-1})f(x^{-1}) &&\text{homomorphism}\\ &=\dots \end{align}

For the other direction your reasoning is correct.

Answer 2

• If $a=b$ then $a^{-1}=b^{-1}$ so $f(a)=f(b)$ which means that $f$ is well-defined.

• We have $f(ab)=(ab)^{-1}=b^{-1}a^{-1}$ which equals to $a^{-1}b^{-1}=f(a)f(b)$ if $G$ is abelian and therefor $f$ is a homomorphism.

• If $f(a)=a^{-1}$ is a homomorphism then think about $f(a^{-1}b^{-1})$.

• The same claim is correct if $f\in Aut(G)$.

Answer 3

You can do this in two lines $$y^{-1}*x^{-1}=(x*y)^{-1}=f(x*y)$$

$$x^{-1}*y^{-1}=f(x)*f(y)$$

The terms at the left are equal if and only if $G$ is commutative (since every element can be written as the inverse of another element). The terms at the right are equal if and only if $f$ is a homomorphism.