Ive gone through the proof of using semi direct products to show that a possible isomorphism class of a group of order 21 is the following:
Where we follow the below procedure to obtain this presentation:
21 = 7*3 which implies for the non abelian case we have 1 normal subgroup of order 7 and 7 subgroups of order 3.
The let H = C$_7$ and K = C$_3$. Then we have a semi direct product of H and K.
Consider our possible homomorphism from K -> Aut(H). Since K has order 3, the image must divide |Aut(H)| = 6 and must divide 3 as it is the index of the kernal of our homomorphism. Then the image can be of order 1 or 3, 1 gives an abelian group classification so we focus on order 3.
Then we aim to find an element of Aut(H) of order 3 in our image. Well the automorphism that sends y -> y$^2$ has order 3 as Aut(H) is isomorphic to Z$_7$$_X$ and 2 has order 3 in Z$_7$$_X$.
Then we have the relation xyx$^{-1}$ = y$^2$. Then the book, and everyone on the internet it seems will end it here and not search for any other possible relations in this group. How can we be sure there are no other relations that need to be defined in this group?
Note:
Similarly we could arrive at another automorphism of order 3 with the defining map xyx$^{-1}$ = y$^4$ as 4 has order 3 also in Z$_7$$_X$. This would define, presumably, a new group with of course a new presentation.
My try at an answer:
When multiplying two elements in H $\rtimes$ K, we have all the pieces we need. Multiplication in the second component of the 2-tuple (h,k) is already defined as K is already a defined group with a known structure and generator, so one does not need anything additional other than the relation y$^3$ = 1. The generator for H also helps, but we need one more piece of information, namely how our homomorphism maps values in K to conjugation actions on H. Once this information is known, one can compute any value in the group. Since every value in the group is now computable, the relations we have listed are sufficient.
Thanks!
