Proving a particular subgroup problem

Question

The following is from Abstract Algebra by Gallian 8th edition, page 98.

Let $G$ be a group and let $f$ be a function from $G$ to some set. Show that $H=\{g \in G | f(xg) = f (x) ,\forall x \in G \}$ is a subgroup of $G$.

Here is what I have so far:

Clearly $e \in H$ since $f(xe) = f(x), \forall x \in G$. Using the 2-step subgroup test, I can show that $ab \in H$ whenever $a,b\in H$. However, I am having difficulty showing $a\in H$ implies $a^{-1} \in H$.

If I assume $a \in H$, then $f(xa) = f(x)$. How do I show that $f(xa^{-1}) = f(x)?$

Answer 1

$a\in H$ implies that $f(ba)=f(b)$ for all $b\in G$. Take $b=xa^{-1}$ and then $f(xa^{-1})=f(xa^{-1}a)=f(x)$.

Answer 2

Hint:

Set $y=xa^{-1}$. Compute $f(ya)$.

Answer 3

$f(xa) = f(a)$ for all $x \in G$.

Let $x$ be any element of $G$. Let $y = xa^{-1}$.

$f(xa^{-1}) = f(y) = f(ya) = f(xa^{-1}a) = f(x)$ for all $x \in G$.