3

Let G be a group. The function $f:G \rightarrow G$ defined by $f(x)=x^2$ is a homomorphism iff G is abelian.

I am having trouble with the 2nd part of the proof.

Proof:

Assume the function $f:G \rightarrow G$ is defined by $f(x)=x^2$ then:

  1. Assume $f$ is a homomorphism then $f$ satisfies the property: $$ f(xy)=f(x)f(y) $$

where $x,y \in G$

Notice how: $$f(xy)=(xy)^2=xyxy$$ $$f(x)f(y)=x^2y^2=xxyy $$

Since $f(xy)=f(x)f(y)$ then by substitution:

$$xyxy=xxyy$$

Multiplying the right by $y^{-1}$ and the left by $x^{-1}$ then:

$$yx=xy$$ which means G is abelian

  1. Assume G is abelian, we want to show $f$ is a homomorphism

For the 2nd part of the proof, would I just work backwards from the first part of the proof? Here is an outline:

then for $x,y \in G$

$$yx=xy$$ by definition of abelian groups.

Multiple the right by y and the left by x, we have:

$$xyxy=xxyy$$

which is equal to: $$(xy)^2=(x)^2(y)^2$$

since we are given $f(x)=x^2$ then translating this to the above statement: $$f(xy)=f(x)f(y)$$

which satisfies the property of homomorphism.

Does that look right?

Urban PENDU
  • 2,409
Mark
  • 427

1 Answers1

4

Your solution is indeed correct and just to put it in a nice way what you can do for the 2nd part is the following

You need to show that $f(xy) = f(x)f(y)$. Proceed as follows \begin{align*} f(xy) &= (xy)^2\\ &=xyxy\\ &=xxyy & & (\text{because $xy=yx$})\\ &=x^2y^2\\ &=f(x)f(y). \end{align*}

Urban PENDU
  • 2,409