I'm trying to do this methodically. Let $\displaystyle G$ be a group of order 42. The Sylow Theorem says that $\displaystyle G$ 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 $\displaystyle Z_3\trianglelefteq G$, so $\displaystyle Z_3Z_7\cong Z_3\times Z_7$ and is normal because its index is 2. We now look for homomorphisms $\displaystyle \phi_i:Z_2\to\mbox{Aut}(Z_3\times Z_7)$. We note that $\displaystyle \mbox{Aut}(Z_3\times Z_7)\cong Z_2\times Z_6$ and write $\displaystyle Z_2$ as $\displaystyle \langle~x~\rangle$, and $\displaystyle Z_2\times Z_6$ as $\displaystyle \langle~y~\rangle\times\langle~z~\rangle$. (Both have identity $\displaystyle 1$.)

1. $\displaystyle \phi_0:x\mapsto(1,1)$ This is the trivial mapping and gives the only abelian group $\displaystyle G=Z_{42}$.

2. $\displaystyle \phi_1:x\mapsto(1,z^3)$

3. $\displaystyle \phi_2:x\mapsto(y,1)$

4. $\displaystyle \phi_3:x\mapsto(y,z^3)$

The last three mappings result in three non-abelian groups of order 42, namely $\displaystyle Z_{21}\rtimes_{\phi_i}Z_2$ for $\displaystyle i\in\{1,2,3\}$. My guess is that $\displaystyle \phi_1$ produces $\displaystyle Z_3\times D_{14}$, $\displaystyle \phi_2$ produces $\displaystyle Z_7\times D_6$, and $\displaystyle \phi_3$ produces $\displaystyle D_{42}$, 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 $\displaystyle \varphi:Z_3\to\mbox{Aut}(Z_7)$. We note that $\displaystyle \mbox{Aut}(Z_7)\cong Z_6$ and write $\displaystyle Z_3$ as $\displaystyle \langle~x~\rangle$ and $\displaystyle Z_6$ as $\displaystyle \langle~y~\rangle$. With this presentation, $\displaystyle \varphi:x\mapsto y^2$ is a valid homomorphism, so $\displaystyle Z_7\rtimes_{\varphi}Z_3\cong Z_7Z_3$. I will call this group $\displaystyle F_{21}$ because I think groups like this are called Frobenius groups. So now I want to look for homomorphisms $\displaystyle \psi_i:Z_2\to\mbox{Aut}(F_{21})$. Again writing $\displaystyle Z_2$ as $\displaystyle \langle~x~\rangle$, we have:

1. $\displaystyle \psi_0:x\mapsto1$ This is the trivial mapping and gives $\displaystyle G=F_{21}\times Z_2$.

2. $\displaystyle \psi_1:x\mapsto\,?$

I don't know whether $\displaystyle ?$ can be anything valid because I don't understand the structure of $\displaystyle F_{21}$.How can I determine whether its automorphism group has any elements of order 2?

Thanks in advance!