Editing Group action (section)

=== {{visible anchor|Orbit-stabilizer theorem}} ===
Orbits and stabilizers are closely related. For a fixed {{math|''x''}} in {{math|''X''}}, consider the map {{math|''f'' : ''G'' → ''X''}} given by {{math|''g'' ↦ ''g''&sdot;''x''}}. By definition the image {{math|''f''(''G'')}} of this map is the orbit {{math|''G''&sdot;''x''}}. The condition for two elements to have the same image is
<math display=block>f(g)=f(h) \iff g{\cdot}x = h{\cdot}x \iff g^{-1}h{\cdot}x = x \iff g^{-1}h \in G_x \iff h \in gG_x.</math>
In other words, {{math|1=''f''(''g'') = ''f''(''h'')}} ''if and only if'' {{math|''g''}} and {{math|''h''}} lie in the same [[coset]] for the stabilizer subgroup {{math|''G''<sub>''x''</sub>}}. Thus, the [[Fiber (mathematics)|fiber]] {{math|''f''{{i sup|−1}}({{mset|''y''}})}} of {{math|''f''}} over any {{math|''y''}} in {{math|''G''&sdot;''x''}} is contained in such a coset, and every such coset also occurs as a fiber. Therefore {{math|''f''}} induces a {{em|bijection}} between the set {{math|''G'' / ''G''<sub>''x''</sub>}} of cosets for the stabilizer subgroup and the orbit {{math|''G''&sdot;''x''}}, which sends {{math|''gG''<sub>''x''</sub> ↦ ''g''&sdot;''x''}}.<ref>M. Artin, ''Algebra'', Proposition 6.8.4 on p. 179</ref> This result is known as the ''orbit-stabilizer theorem''.

If {{math|''G''}} is finite then the orbit-stabilizer theorem, together with [[Lagrange's theorem (group theory)|Lagrange's theorem]], gives
<math display=block>|G \cdot x| = [G\,:\,G_x] = |G| / |G_x|,</math>
in other words the length of the orbit of {{math|''x''}} times the order of its stabilizer is the [[Order (group theory)|order of the group]]. In particular that implies that the orbit length is a divisor of the group order.

: '''Example:''' Let {{math|''G''}} be a group of prime order {{math|''p''}} acting on a set {{math|''X''}} with {{math|''k''}} elements. Since each orbit has either {{math|1}} or {{math|''p''}} elements, there are at least {{math|''k'' mod ''p''}} orbits of length {{math|1}} which are {{math|''G''}}-invariant elements. More specifically, {{math|''k''}} and the number of {{math|''G''}}-invariant elements are congruent modulo {{math|''p''}}.<ref>{{Cite book |last=Carter |first=Nathan |title=Visual Group Theory |publisher=The Mathematical Association of America |year=2009 |isbn=978-0883857571 |edition=1st |pages=200}}</ref>

This result is especially useful since it can be employed for counting arguments (typically in situations where {{math|''X''}} is finite as well).

[[File:Labeled cube graph.png|thumb|Cubical graph with vertices labeled]]
: '''Example:'''  We can use the orbit-stabilizer theorem to count the automorphisms of a [[Graph (discrete mathematics)|graph]]. Consider the [[cubical graph]] as pictured, and let {{math|''G''}} denote its [[Graph automorphism|automorphism]] group. Then {{math|''G''}} acts on the set of vertices {{math|{{mset|1, 2, ..., 8}}}}, and this action is transitive as can be seen by composing rotations about the center of the cube. Thus, by the orbit-stabilizer theorem, {{math|1={{abs|''G''}} = {{abs|''G'' &sdot; 1}} {{abs|''G''<sub>1</sub>}} = 8 {{abs|''G''<sub>1</sub>}}}}. Applying the theorem now to the stabilizer {{math|''G''<sub>1</sub>}}, we can obtain {{math|1={{abs|''G''<sub>1</sub>}} = {{abs|(''G''<sub>1</sub>) &sdot; 2}} {{abs|(''G''<sub>1</sub>)<sub>2</sub>}}}}. Any element of {{math|''G''}} that fixes 1 must send 2 to either 2, 4, or 5. As an example of such automorphisms consider the rotation around the diagonal axis through 1 and 7 by {{math|2''&pi;''/3}}, which permutes 2, 4, 5 and 3, 6, 8, and fixes 1 and 7. Thus, {{math|1={{abs|(''G''<sub>1</sub>) &sdot; 2}} = 3}}. Applying the theorem a third time gives {{math|1={{abs|(''G''<sub>1</sub>)<sub>2</sub>}} = {{abs|((''G''<sub>1</sub>)<sub>2</sub>) &sdot; 3}} {{abs|((''G''<sub>1</sub>)<sub>2</sub>)<sub>3</sub>}}}}. Any element of {{math|''G''}} that fixes 1 and 2 must send 3 to either 3 or 6. Reflecting the cube at the plane through 1, 2, 7 and 8 is such an automorphism sending 3 to 6, thus {{math|1={{abs|((''G''<sub>1</sub>)<sub>2</sub>) &sdot; 3}} = 2}}. One also sees that {{math|((''G''<sub>1</sub>)<sub>2</sub>)<sub>3</sub>}} consists only of the identity automorphism, as any element of {{math|''G''}} fixing 1, 2 and 3 must also fix all other vertices, since they are determined by their adjacency to 1, 2 and 3. Combining the preceding calculations, we can now obtain {{math|1={{abs|G}} = 8 &sdot; 3 &sdot; 2 &sdot; 1 = 48}}.