Question

let x be a g-set with exactly n orbits gx 1,.., gx n, and assume the stabilizers g{ˣ ᶦ} are all equal. call this subgroup of g, h. explain why as a g-set, x is isomorphic to a disjoint union of n

Answer 1

The G-set X with n orbits, each having equal stabilizers, is isomorphic to a disjoint union of n sets through the left coset action of the stabilizer subgroup H on G.

Let's understand the given scenario:

- (X) is a (G)-set with (n) orbits
`\(Gx_1, \ldots, Gx_n\)`.

- The stabilizers
`\(G_(x_i)\)` (stabilizers of the elements in the orbits) are all equal and form a subgroup (H) of (G).

Now, let's explain why, as a (G)-set, (X) is isomorphic to a disjoint union of (n) sets.

The orbits
`\(Gx_1, \ldots, Gx_n\)` partition (X) into disjoint sets. The stabilizer of each element in an orbit is (H). Consider the set (G/H), the set of left cosets of (H) in (G). For each
`\(x_i\)`, the left coset
`\(Gx_i\)` is one of the elements of (G/H).

Define a function
`\(\phi: X \to G/H\)` as
`\(\phi(gx_i) = gH\)`. This function is well-defined and bijective. It respects the action of (G) on (X) since
`\(\phi(g \cdot x_i) = gH\)` and
`\(\phi(gx_i) = gH\)`. Therefore, (X) is isomorphic to the disjoint union of the orbits
`\(Gx_1, \ldots, Gx_n\)` with respect to the action of (G).

In summary, the isomorphism
`\(\phi: X \to G/H\)` reflects the partition of (X) into orbits and establishes an isomorphism between (X) and a disjoint union of (n) sets.