Proof of a Lemma for the Sylow theorems

Question

Lemma: If $H$ is a p-subgroup of a finite group $G$, then $[N_G(H):H]\equiv[G:H]\ (\text{mod}\ p)$.

Let $S$ be the set of left cosets of $H$ in $G$ and let $H$ act on $S$ by left translation. Then,

$hxH=xH,\ \forall h\in H\ \Leftrightarrow x^{-1}hxH=H, \forall h\in H \Leftrightarrow x^{-1}hx \in H, \forall h\in H \\ \Leftrightarrow x^{-1}Hx=H ...$

The rest of the proof is not of importance. I don't understand the last two equivalences.

Could someone elaborate these?

Answer 1

You want to understand $$ x^{-1}hxH=H, \forall h\in H \Leftrightarrow x^{-1}hx \in H, \forall h\in H \Leftrightarrow x^{-1}Hx=H $$ Let's break that down into two: $$ x^{-1}hxH=H, \forall h\in H \Leftrightarrow x^{-1}hx \in H, \forall h\in H \tag{1} $$ $$ x^{-1}hx \in H, \forall h\in H \Leftrightarrow x^{-1}Hx=H \tag{2} $$

Let's take a look at the second one. To say that $x^{1} H x = H$ is to say that the set $U = \{ x^{1} h x \mid h \in H\}$ is the same as the set $H$. Now the left side of the double arrow tells you that $U \subset H$, because each $x^{-1}hx$ is in $H$. So the only question is why $U$ contains every element of $H$. I'm going to prove it by showing that the map $H \to H: h \mapsto x^{-1}hx$ is injective. Because an injective map from a finite set to itself is a bijection.

So: is suppose that $$ x^{-1} h x = x^{-1} h' x $$ multiplying on the left by $x$ and the right by $x^{-1}$, we get $h = h'$. Hence the map's injective.

The other parts are similar -- it's all just counting and conjugation maps.

Answer 2

The first equivalence follows from the fact that $aH = H $ if and only if $a \in H$.

For the second one we have that $x^{-1}Hx = \{x^{-1}hx \mid h \in H\}\subseteq H$. But as $H$ is finite and $|x^{-1}Hx| = |H|$ we must have $x^{-1}Hx = H$. The reverse direction is trivially true.

Answer 3

(i) For any subgroup $H$, and $g\in G$, $$gH=H\iff g\in H.$$ Indeed, if $gH=\{gh\mid h\in H\}=H$, then $g=ge\in gH=H$. Conversely, if $g\in H$, then $gh\in H$ for all $h\in H$ because $H$ is closed under multiplication. Hence, $gH=H$.

(ii) $x^{-1}Hx=\{x^{-1}hx\mid h\in H\}$. Therefore, $$x^{-1}Hx=H\iff x^{-1}hx\in H\mbox{ for all }h\in H.$$