$H$ acts on $G/H$. Show that $\mathrm{Fix}(H) = G/H$

Question

So $H$ acts on $G/H$. Where $H$ is a subgroup of $G$. We are also given that $|G| = p^2$ and $|H| = p$, where p is a prime.

$$ H \times G/H \rightarrow G/H, \quad (h, gH) \rightarrow h * (gH) = (hg)H $$

The question is to show that $\mathrm{Fix}(H) = G/H$.

My attempts so far keep getting to the point where I need to show that H is normal. However, the next part of the question is to show that H is normal so I'm obviously going wrong somewhere. In the previous parts of the question I have shown that $\mathrm{Stab}(gH) = H \cap gHg^{-1}$ and $H \in \mathrm{Fix}(H)$. Any help would be much appreciated.

Answer 1

My answer builds up from the hint that user297024 gave, which is crucial in my opinion. The "previous part of the question" that you proved $H \in \operatorname{Fix}(x)$ will be needed to finish my proof. I will assume that the Orbit-Stabilizer Theorem can be used.

By the Orbit-Stabilizer Theorem for any element $x$ of a set $X$ acted by $H$, we have:

$$ |H| = |\operatorname{Stab}(x)|. |\operatorname{Orb}(x)| $$

As $|H| = p$ is a prime number, that leaves either $(|Stab(x)| ; |Orb(x)|)$ to be $(1;p)$ or $(p;1)$. In other words, either $\operatorname{Stab}(x) = H$ or $\operatorname{Stab}(x) = {id}$.

We will prove this by contradiction. Suppose $H$ does not fix the whole set $X$: $\exists x_{1} \neq x_{2} \in X$ such that $\operatorname{Stab}(x_{1}) = H$ but $\operatorname{Stab}(x_{2}) = id$.

This implies that $\operatorname{Orb}(x_{2}) = X$ - every element in X can be "matched" from $x_{2}$ by some element of $H$, including $x_{1} \in X$. (Note that this statement is only valid if $|X|=|H|=p$.)

Therefore, $\exists h \in H$ such that: $$ h * (x_{2}) = x_{1} $$

As $h \in H$, then also $h^{-1} \in H.$ Consider the following: $$ (h^{-1}h) * x_{2} = id * x_{2} = x_{2} $$ by the definition of a group action. But $$ h^{-1} * (h * x_{2}) = h^{-1} * x_{1} = x_{1} $$ by the way we construct $x_{1}$.

This leads to a violation of the definition of a group action. Therefore either $H$ fixes the whole set $X$ or $H$ does not fix anything in $X$. Now in the context of the problem, $X = G/H$ with exactly $$ |X|=|G/H|= \frac{|G|}{|H|}=\frac{p^{2}}{p}=p $$ by Lagrange's Theorem, and we know for a fact that $H$ is a coset of $G/H$. So using the result $H \in \operatorname{Fix}(H)$, that means every other cosets must be fixed by $H$. This completes the proof $\operatorname{Fix}(H) = G/H.$

Answer 2

The orbit equation reads: $$|G/H|=\left|\operatorname{Fix}(H)\right|+\sum_i[H:\operatorname{Stab}(g_iH)]$$ where the $g_iH$'s are representatives of the non-sigleton orbits, necessarily each of size $p$ (orbit-stabilizer). But since $|G/H|=p$ and $\left|\operatorname{Fix}(H)\right|\ge1$ (as $H\in \operatorname{Fix}(H)$), the set of such representatives must be empty, and hence $$|G/H|=\left|\operatorname{Fix}(H)\right|$$ namely $$\operatorname{Fix}(H)=G/H$$