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Given $|G|=21$, determine the possible structures without pq theorems or semi products.

I can show there is one Abelian case. I want to show there is only one non-Abelian case with the following method:

Let $H_7$ be the normal Sylow subgroup of order 7, and $H_3$ one of the 7 Sylow subgroups of order 3.

So, $\ H_7=\langle y\rangle$ and $H_3=\langle x \rangle$.

Because $H_7$ is normal, $\ xyx^{-1}=y^{k}$, $\ $ so $\ \ y^{k^{3}-1}=1.$

Solving: $\ k^{3}-1\equiv 0 \ (mod \ 7) $, $\ k=1,2,4$.

I see that $k=1$ represents the case when G is abelian, but then I have two additional cases.

sps
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    You reversed your $x$ and $y$ at some point, but anyway, assuming $x$ has order $3$, note that $H_3$ is also $\langle x^2 \rangle$ and study what the effect of conjugating by $x^2$ is in the two cases. In other words, whether you get $k=2$ or $4$ is just a question of which generator you chose. – John Brevik Dec 11 '17 at 19:17
  • Yeah, the x and y were reversed in the beginning. Also, we have not learned about semidirect products yet. – sps Dec 11 '17 at 19:24
  • "without pq theorems or semi products". I think, this does not make much sense, because the arguments in the end are very similar to pq theorems, and the language of semidirect products just makes things clearer and not more difficult. There are several questions about it with many good answers, so people have given help a lot. So let me mention some links: here, or here. – Dietrich Burde Dec 11 '17 at 19:27

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