The kernel and stabilizers of a group action are subgroups

Let G be a group acting on a set A. Show that the following sets are subgroups of G.

  1. The set K = \{ g \in G \ |\ g \cdot a = a\ \mathrm{for\ all}\ a \in A \} (Called the kernel of the action
  2. For a fixed a \in A, the set S = \{ g \in G \ |\ g \cdot a = a \} (Called the stabilizer of a)

By a previous exercise, for each of these sets we need to show that the identity belongs to the set and that each is closed under multiplication and inversion.

  1. Note that 1 \in K since 1 \cdot a = a for all a \in A. Now suppose k_1, k_2 \in K, and let a \in A. Then (k_1 k_2) \cdot a = k_1 \cdot (k_2 \cdot a) = k_1 \cdot a = a, so that k_1 k_2 \in K. Now let k \in K and a \in A; then k^{-1} \cdot a = k^{-1} \cdot (k \cdot a) = (k^{-1} k) \cdot a = 1 \cdot a = a, so that k^{-1} \in K. Thus K is a subgroup of G.
  2. We have 1 \in S since 1 \cdot a = a. Now suppose s_1, s_2 \in S; then we have (s_1 s_2) \cdot a = s_1 \cdot (s_2 \cdot a) = s_1 \cdot a = a, so that s_1 s_2 \in S. Now let s \in S; we have s^{-1} \cdot a = s^{-1} \cdot (s \cdot a) = (s^{-1} s) \cdot a = a, so that s^{-1} \in S. Thus S is a subgroup of G.
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