Let phi be an automorphism of D4 such that phi H D Find phi

Let phi be an automorphism of D_4 such that phi (H) = D. Find phi (V).

Solution

suppose P2 = D4 = <, >. The order 4 subgroups are H1 = <>, H2 = < >, and H3 = ( ² , >, and the order 2 subgroups are the center Z = h 2 i and H0 = h i. The perspective of generators and relations shows that there is an automorphism of D4 carrying H1 to H2, so when considering possibilities for ker we can ignore H2. But H1 is cyclic while H3 is not, so these two options for ker cannot give isomorphic semidirect products (as the structure of ker is intrinsic to G). Once again using that groups of order 2 have no non-trivial automorphisms, we get two more examples, for P2 = D4 and ker = H1, H3. In the case when ker has order 2, then induces an isomorphism of G/ ker onto the cyclic group (Z/5)×. This rules out ker = Z (since D4/Z \' C2 × C2 is non-cyclic), so leave only the possibility ker = H0 = h i, but this is not a normal subgroup.