$H$ is a subgroup of $G$ with $(G:H) = n$. $X$ is the set of all left cosets of $H$ in $G$. $f:X \to X$ and $f(g_i H) = gg_i H$, for all $g_iH \in X, g \in G$.
Show that $f$ is a permutation: For this I just assumed that I need to show that $f$ is a bijection. So for every $g_i H \in X$, I need to show there exists a $g_k H \in X$ such that $f(g_k H) = g_iH$ and then I need to show that $f(g_k H) = f(g_i H)$ implies that $g_k H = g_i H$. I'm stuck here and don't know how to show these to be true.
$|f|$ as an element of $Sym(x)$ is a divisor of the order of $g$ in G: For this I think I have to use Orbit stabilizer theorem but I'm confused on how to do this. I think that $|f|$ is just the orbits of an element $g_iH \in X$ and I am confused as to what the order of $g$ in $G$ is.
Show that $f(H) = H$ if and only if $g \in H$: So for this I thought that if $f(H) = H$ where $H \in X$ then $f(H) = gH = H$ so then $g \in H$. Conversely if $g\in H$ then $gH = H$ so then $f(H) = gH =H$. I don't think this is right and need a bit of help moving forward.
Any help with these would be much appreciated.
First part: Showing that $f$ is a permutation is the same as showing that it is a bijection, which is the same as showing that it has a left inverse and a right inverse. Now, $f$ is the function which left-multiplies cosets by $g$, so it has an inverse, namely the function which left-multiplies cosets by $g^{-1}$.
Middle part: $|f|$ is the order of $f$, meaning the smallest positive integer $n$ for which $f^n$ (the composition of $f$ with itself $n$ times) is the identity map. Also, if $k$ is any integer for which $f^k$ is the identity map, then $k$ is an integer multiple of $n$.
Now suppose that $k$ is the order of $g$ in $G$. Then $g^k = e$ (the identity). If $aH$ is any element of $X$, then $f^k(aH) = g^k aH = eaH = aH$, so $f^k$ is the identity map. Therefore $k$ must be a multiple of the order of $f$, as claimed.
Last part: $f(H) = H$ if and only if $gH = H$, if and only if $gH$ and $H$ are the same coset, if and only if $g \in H$.