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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),

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When $n_7=1$, you are in the situation of your last question. Judging from the hint you were given, it is entirely possible for $n_7=28$ as well for some nonabelian group, so you'll need to construct one with the SDP. –  rschwieb Jan 9 '13 at 15:03
    
How can $n_7 = 28$? Don't you mean $29$?. So as I was in the same bit, I can do that formula DonAntonio gave? Is the question asking me to write down all the SDP's again? –  Kaish Jan 9 '13 at 15:15
    
Oh yeah, sorry, that was a typo :) 29 is right! The question is asking you to give all possible different shapes of groups of that order. One of the shapes is going to be the abelian group of order 203 (you should see there is only one!) but there is apparently another nonabelian group of that order which you can construct with a SDP. –  rschwieb Jan 9 '13 at 15:21

1 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, $\,Q\rtimes P\,$ ...

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There are 6 nontrivial homomorphisms $P \to {\rm Aut}(Q)$, but the resulting semidirect products are all isomorphic groups (and Kaish needs to prove that). –  Derek Holt Jan 9 '13 at 17:59

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