Prove $f(x) = x * a$ is bijective (preferably using inverse)

Question

I have this question that I am stuck at.

Let $(G; * ; I)$ be a group and let 'a' in $G$. Let $f : G \rightarrow G$ be the function defined by $f(x) = x * a$ for all $x$ in $G$. Prove that $f$ is bijective.

This is what I got from my instructor, but I forgot how he came up with this. I mean I cannot trace back which def. and lemma he used, aside from associativity.

$g(f(x)) = f(x) * a^-1 = x * a * a^-1 = x * I = x$

Also if it is not too much trouble... Would this work if the function is slightly different, $f(x) = a * x$

$f(g(x)) = a * g(x) = a * a^-1 * x = I * x = x$

If all of the above is completely wrong, then what is the correct version using inverse? and maybe also how to prove it using injectivity and surjectivity?

Thanks

Answer 1

You never defined what $g$ is, although it is fairly obvious what it is. If $g(x)=xa^{-1}$ then $f(g(x))=g(f(x))=x$ for all $x\in G$. If a function has both a left and right inverse (or just an inverse) then it is bijective.

For a direct proof without defining an inverse,

Injectivity

If $f(x)=f(y)$ then $xa=ya$. Multiplying both sides on the right by $a^{-1}$ will give you $x=y$.

Surjectivity

Let $g\in G$ and let $x=ga^{-1}$. Then verify $f(x)=g$

For the similar function $h(x)=ax$ you can easily modify the above.