This problem originated from trying to find the group isomorphisms of groups of order 21. I already worked out that there's one subgroup of order 7 that's normal, and one subgroup of order 3. Even though 7 ≡ 1(mod 3), by simple counting I worked out that there can't be seven groups of order 3 and one of order 7.
Let the Sylow 7 subgroup be N, and let the Sylow 3 subgroup be H. I then considered the semi-direct product of N and H, and the function f:H→Aut(N). I know I can do this because |H||N|=21, N is normal, and H ∩ N = {e}, since all elements of H have different orders to all elements of N.
Let H= < a >, N= < b >. There's obviously the trivial map when 1,a,and a^2 go to e, and then a mapping where f(a)=b^2 and f(a^2)=b^4. But what about the mapping where f(a)=b^4 and f(a^2)=b^2? Are these two mappings different?
And I'm pretty new here, so could anyone also tell me how to properly type out the powers of numbers instead of just using ^? Thanks a lot.