Transitive action

Question

A group $G$ acts on a nonempty set $X$. Suppose $H$ is a subgroup of $G$ and $H$ acts transitively on $X$. In a proof, an author claims:

If $H$ acts transitively on $X$, then obviously $G$ acts transitively on $X$.

But why? I don't see this.

Another thing which is said is: Fix $x \in X$. For $g \in G$, $gx=hx$ for some $h \in H$, so $h^{-1}g \in \text{Stab}_x$.

My question concerning this: Why is $gx=hx$ and why can we imply $h^{-1}g \in \text{Stab}_x$? I thought, multiplying with $h^{-1}$ is only allowed in groups and $gx,\ hx$ are no group elements.

Answer 1

For the first question, this comes directly from the definition of a transitive group action.

Second, $gx \in X$ and as $H$ acts transitively on $X$, there is some $h \in H$ such that $gx = hx$. Remember the definition of a group action- we have for every $g_1, g_2 \in G$ and $x \in X$, $g_1 * (g_2 * x) = (g_1g_2) * x $ and $e * x = x$. So if $g*x = h*x$, $$h^{-1}*(g*x) = h^{-1}*(h*x) \Rightarrow (h^{-1}g)*x = (h^{-1}h)*x = x$$ Thus, $h^{-1}g \in \text{Stab}_x$

Edit: Here, $*$ denotes the group action of $G$ on $X$. I've used $*$ so that we can explicitly see what's going on.

Answer 2

Transitively means that for $x,y\in X$ there exists $h\in H$ with $h.x=y$, $h\in H\subset G$, so $h\in G$ and $G$ acts transitively.

$Stab_x=\{g:g.x=x\}$, $h.x=g.x$ implies that $h^{-1}(h.x)=h^{-1}.(g.x)$ this is i.e to $h^{-1}g.x=x$.

Answer 3

I'll answer the last question first. Saying $G$ acts on $X$ means that each $g \in G$ determines a function from $X$ to itself. Instead of giving that function a new name, we call it "$g$" and write its action as $g(x)$, and then perhaps omit the parentheses, to $gx$ does not mean $g \times x$ in the group, or anywhere else.

Then it makes sense to talk about $$ h^{-1}gx = h^{-1}(g(x)) = h^{-1}(h(x)) = h^{-1} h(x)) = x $$ so $h^{-1}g$ fixes $x$.

There are already several answers to the first two questions. Here are mine.

To show $G$ acts transitively, given $x$ and $y$ you have to find a $g$ that sends $x$ to $y$. But you're told that you can find one in $H$, and $H$ is a subset of $G$.

The "another thing": for any $g$ and any $x$. $gx$ is some element of $H$. Since $H$ acts transitively, there is some element $h \in H$ that sends $x$ to $gx$.