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