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Question:

Show that the kernel of the group action of $G$ acting on set $A$ is equal to the kernel of the corresponding permutation representation of this action.

I'm lost in this definition as I am only familiar with the definition of a kernel of a homomorphism as $\{g \in G~|~\varphi(g) = I_A\}$ (the elements of the domain in which the image is the identity of the target group)

How is this definition applicable to a map $f: G\times A \to A$ in which the target domain is a set?

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  • $\begingroup$ To me this is the definition of the kernel of a group action. Is there some (other) definition in the book you're working from? $\endgroup$ – Inactive - avoiding CoC Jun 13 '16 at 20:11
  • $\begingroup$ @Servaes The definition is $\{ g \in G ~|~ga = a,$ for all $a\in A\}$. The previous definition would not make sense since a set cannot have an identity element. I don't really understand how I was supposed to know this without the definition for a kernel of a group action :( $\endgroup$ – obliv Jun 13 '16 at 20:18
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Let $G$ be a group acting on $A$.

The kernel of the action is the set $K =\{g \in G; g \cdot a = a , \forall a \in A\}$. Now the corresponding permutation representation is a group homomorphism $\psi : G \to S_A$ given by $\psi (g)(a) = g \cdot a $.

  • $K \subseteq \ker \psi$

Let $k \in K$, then for all $a \in A$ we have that $\psi (k) (a) = k \cdot a = a$ thus, $\psi (k) = id_A$ and then $k \in \ker \psi$.

  • $\ker \psi \subseteq K$

Let $k \in \ker \psi$ be given. Then for all $a \in A$ we have $$ k \cdot a = \psi (k) (a) = id_A(a) = a $$

thus $k \in K$.

Edit: The kernel of $\psi$ is given by $\ker \psi = \{g \in G ; \psi (g) =id_A\}$.

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  • $\begingroup$ Don't you mean $K = \{g \in G; ga = a, \forall a \in A\}$? How do you assert that $K \subset \ker \psi$? Can you define the kernel of $\psi$? $\endgroup$ – obliv Jun 13 '16 at 20:38
  • $\begingroup$ Sure, and you're right about K. $\endgroup$ – Aaron Maroja Jun 13 '16 at 20:41
  • $\begingroup$ Oh I see now. Thank you very much! $\endgroup$ – obliv Jun 13 '16 at 20:48
  • $\begingroup$ Glad I could help! $\endgroup$ – Aaron Maroja Jun 13 '16 at 20:53
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The kernel of a group action is defined as the set of all group elements which act as the identity. The problem is asking you to show that this definition is related to the kernel of a homomorphism by showing that the kernel of the group action is isomorphic to the kernel of the homomorphism from $G$ onto the permutation group of the set.

Please correct me on anything that might be wrong or ambiguous, since I haven't done this in a while.

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  • $\begingroup$ If the group action is 'faithful' (injective?) then does this mean the kernel of the group action is simply the identity set of the group? $\endgroup$ – obliv Jun 13 '16 at 20:23
  • $\begingroup$ What do you mean by identity set of the group? $\endgroup$ – thkim1011 Jun 13 '16 at 20:29
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    $\begingroup$ @Obliv The kernel of a faithful group action is simply the trivial subgroup (the subgroup containing the identity only). $\endgroup$ – thkim1011 Jun 13 '16 at 20:33

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