Classify up to isomorphism the groups of order $203$

Question

Classify up to isomorphism the groups of order $203$. Assume the face that the least $ k \geq 1$ such that

$$2^k \equiv 1 \mod{29}$$

is $k = 28$. [HINT: Look at the Sylow subgroups and represent a group $G$ of order $203$ as a semi-direct product of two cyclic groups.]

I know that $203 = 7 \cdot 29$. To find my Sylow subgroups I want

$$P = n_7: n_7 \equiv 1 \mod 7 \hspace{1.5cm} n_7\mid 29$$ $$Q = n_{29}: n_{29} \equiv 1 \mod{29} \hspace{1.5cm} n_{29} \mid 7$$

From the hint, we see that $2^k > 203$ and so the only possibility for $n_{29} = 1$. Writing out all the elements for $n_7$, we see that there are two possibilities $n_7 = 1$ or $29$.

Because there's two possibilites for $n_7$, I'm now stuck on what to do. I know that $Q$ is a normal subgroup and so would if I let $P_1 = n_7 = 1$ and $P_2 = n_7 = 29$, would I have to workout the semidirect products such that $Q \rightarrow \mathrm{Aut}(P_1)$ and $Q \rightarrow \mathrm{Aut}(P_2)$?

EDIT: The reason I'm doing the SDP from $Q$ is because I thought in order to calculate it, you need to go from the normal subgroup to the other subgroup. If both are normal, then it shouldn't make any difference (I think),

Answer 1

Well, since $\,n_{29}=1\,$ we already know $\,Q\triangleleft G\,$ . As $\,\operatorname{Aut}(Q)\cong C_{28}\,$ , we have two options for a homomorphism $\,P\to\operatorname{Aut}(Q)\,$ : either the trivial one, getting thus the direct product $\,Q\times P\,$ , which is then the cyclic (and abelian, of course) group $\,C_{29}\times C_8\cong C_{203}\,$ , or else an embedding $\,P\to\operatorname{Aut}(Q)\,$ , getting a non-direct SDP, and thus non abelian group, $\,Q\rtimes P\,$ ...