2

The notation is the following: $G$ is a group acting on a set, $G_x = \{g \in G\mid gx = x \}$. What does $G/G_x$ look like? relevant

In the above equation, is it an abusive of notation to write $G/G_x$ because $G_x$ is not neccesarily normal? If so what does it actually mean?

Shaun
  • 44,997
tryst with freedom
  • 11,538
  • 3
    It's not a quotient group, because the stabilizer is almost never normal... it is a collection of left cosets. And for any group and subgroup $H$, the left cosets of $x$ and $y$ are equal if and only if $xH=yH$ as sets, if and only if $y^{-1}xH=H$ as sets, if and only if $y^{-1}x\in H$. – Arturo Magidin Mar 11 '22 at 01:15
  • 1
    Two elements $g,h$ belong to the same coset of $G_x$ if and only if their action on $x$ is the same, i.e if $gx=hx$. – Mark Mar 11 '22 at 01:17
  • OK seems like I understood quotient groups totally wrong – tryst with freedom Mar 11 '22 at 01:20
  • 3
    It's not a quotient group! For the cosets to form a group under the operation $(xH)(yH)=(xy)H$ it is necessary and sufficient that $H$ be a normal subgroup. The stabilizer is not usually a normal subroup, so tnere is no "quotient group". Again, what you need to look at here are left cosets of a subgroup, not quotients. – Arturo Magidin Mar 11 '22 at 01:31
  • You have opened my third eye. Is the question right now? @ArturoMagidin – tryst with freedom Mar 11 '22 at 01:36
  • 1
    What makes you think "the final biconditional [is] false"? No, the question still talks about quotients. For the third time, there is no quotient. – Arturo Magidin Mar 11 '22 at 02:17

1 Answers1

3

If $G$ is a group and $H$ its subgroup then

$$G/H:=\{ gH\ |\ g\in G\}$$

is the set of all (left) cosets of $H$ in $G$. This is a well defined set, regardless of whether $H$ is normal or not. Normality only gives us that this set together with $(gH, g'H)\mapsto gg'H$ operation is a well defined group, a.k.a. the quotient group. Nothing else.

Shaun
  • 44,997
freakish
  • 42,851