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This question is related to Question on groups of order $pq$, but is different.

It references the same exercise, but an earlier part. The exercise is: A group of order $pq$, $p>q$, contains a subgroup of order $p$ and a subgroup of order $q$.

The part I'm having trouble with is showing that assuming there is no subgroup of order $p$ generates a contradiction. It's easy enough to do with Orbit-Stabilizer Theorem, but that hasn't been introduced in the text yet, and I haven't been able to find a way to do it with a different method.

Obviously Cauchy's Theorem trivializes this, but that hasn't been introduced in the text yet either.

So my question is: How can this problem be solved without Orbit-Stabilizer?

Thanks in advance

Edit:

To clarify, I'm searching for a contradiction (without Orbit-Stabilizer or Cauchy's because those are several sections off in the text) to show the nonexistence of a non-abelian group of order $pq$ having no non-trivial normal subgroups and every element of order $q$ where $q<p$.

Edit:

Mostly for educational purposes, it isn't really high priority.

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jgon
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  • I think the easiest way to prove this result is, well, just to prove Cauchy's theorem! (Wikipedia does it twice...) – user1729 Aug 17 '13 at 16:34
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    @user1729 I mean that would work, but I figure if the exercise is put in this section there's a way to do it without resorting to more powerful tools. And since I'm using the book to self study, a solution using the tools introduced so far would be more educational. (Though perhaps there isn't one) Thanks for commenting though :) – jgon Aug 17 '13 at 16:41
  • I figured that was the case, but, you see, Cauchy's theorem isn't so much a "powerful" tool, but a "fundamental" tool. It is such an important result in group theory, and is so deeply ingrained in the theory, and so those who are used to groups, that I wonder if you -as a newcomer to the theory- are the best placed to answer your question! Anyway, I will stop commenting now as someone might have something useful to say, and you should listen to them instead of me...! – user1729 Aug 17 '13 at 16:47
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    @user1729 Alright thanks, yeah I figured that might be an issue :) – jgon Aug 17 '13 at 16:54
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    @Jim You should see this answer, which looks at groups of order $85=17\times 5$. The author does not use Cauchy's theorem, although it may be used secretly somewhere, for example in one of the other results used. My point is, your answer might be okay... – user1729 Aug 17 '13 at 16:59
  • Hmm interesting, I was basically using the Frobenius Class Equation, though I didn't know that was what it was called :P I might just decide my answer is probably as good as it's going to get. – jgon Aug 17 '13 at 17:06

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We can try some pretty elementary ways, yet some basic stuff in group theory must be assumed.

We take a group $\,G\;,\;\;|G|=pq\;,\;\;p>q$ primes, and assume $\,G\;$ has no subgroup (or has no element) of order $\,p\,$ .

Since all the non-unit elements of $\,G\,$ must have order $\,q\,$ by Lagrange's Theorem (observe that the only naturals that divide $\,pq\,$ are $\,1\,,\,q\,,\,p\,,\,pq\,$) , and since $\,G\,$ cannot be cyclic (otherwise it has one (unique) subgroup of order $\,d\,$, for any divisor $\;d\;$ of $\,pq\,$), $\,G\,$ is the union of all its proper non-trivial subgroups, all of which have order $\,q\,$. Say there are $\,k\,$ such subgroups , so

$$pq=k(q-1)+1\implies (p-k)q=1-k$$

But this last equation is impossible in the natural numbers...(why?) , and we're done.

DonAntonio
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