I'm trying to do this methodically. Let be a group of order 42. The Sylow Theorem says that has a normal subgroup of order 7, and the number of Sylow-3 subgroups is either 1 or 7.
If there is only one Sylow-3 subgroup, then , so and is normal because its index is 2. We now look for homomorphisms . We note that and write as , and as . (Both have identity .)
1. This is the trivial mapping and gives the only abelian group .
The last three mappings result in three non-abelian groups of order 42, namely for . My guess is that produces , produces , and produces , but I don't know how to prove it. So my question is: What is the best way to determine whether these (#2, #3, and #4) are all unique up to isomorphism and, furthermore, whether they are isomorphic to more familiar groups?
Next, if there are three Sylow-3 subgroups, then we look for homomorphisms . We note that and write as and as . With this presentation, is a valid homomorphism, so . I will call this group because I think groups like this are called Frobenius groups. So now I want to look for homomorphisms . Again writing as , we have:
1. This is the trivial mapping and gives .
I don't know whether can be anything valid because I don't understand the structure of . How can I determine whether its automorphism group has any elements of order 2?
Thanks in advance!