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So this is a problem from Herstein. My attempt is as follows:

Assume p and q are prime and p>q. We know that there exists a unique subgroup of order p ,call it H=(a) and a subgroup of order q,call it K=(b). Also it can be shown that HK=G.So define T:G->G' by T((a^i)(b^j))= (a')^i(b')^j where a' and b' are the corresponding generators of H' and K' in G'. Is it correct? I am unable to prove that this is a homomorphism. Also since Sylow theorems and semi-direct products or direct products have not been covered till this point I don't want to use them. Thanks Note: I meant unique subgroup of order p, not of order q.

RagingBull
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  • How do we know there are these unique subgroups? If $p=2$ and $q=3$ and $G=S_3$, there are three subgroups of order $2$. – Hagen von Eitzen Oct 30 '18 at 06:44
  • i meant unique subgroup of order p and not q.This can be proved. – RagingBull Oct 30 '18 at 06:56
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    If you want there to be a unique subgroup of order $p$, you have to assume $p>q$ before you state such a fact. Order matters, both for actual logical rigor, and for readability. The way you wrote it, you're implying there is a unique subgroup of order $p$, regardless of which is larger, and only after that are you specifying that $p$ is the larger prime before you continue your argument. – Arthur Oct 30 '18 at 07:17
  • Noted.And edited. – RagingBull Oct 30 '18 at 07:23
  • What are you allowed to assume here? A unique subgroup of a given order is normal. Once you have chosen generators for the first group, you have to find elements of the second which behave the same way (satisfy the same relations) - these will not necessarily be the first elements you choose. – Mark Bennet Oct 30 '18 at 07:25
  • All that is given is that we have two non abelian groups of order pq with p>q. we need to show that they are isomorphic. I dont want to use sylows theorems or semi direct products since they have not been covered till this point in herstein. What i have shown is my attempt. I am looking for hints or if possible, answers to this question.Is it even possible to prove this result without using the heavy machinery of sylow and semi direct products? – RagingBull Oct 30 '18 at 07:31
  • But you do seem to be allowed to assume that if $p\gt q$ the is a unique subgroup of order $p$ - which is not such a trivial thing. Semi-direct products are not "heavy machinery" and you will implicitly end up working with semi-direct products simply because a non-abelian group of order $pq$ is a semi-direct product and one key way into its structure is the unique normal subgroup of order $p$. – Mark Bennet Oct 30 '18 at 08:45
  • i have already proved that a unique subgroup of order p exists if p>q. so you mean to say i should take a look at semidirect products? – RagingBull Oct 30 '18 at 09:00
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    You don't say it anywhere, Raging, but surely you mean for $p$ and $q$ to be prime. You should edit that information into the body of your question. Anyway, you might want to see https://math.stackexchange.com/questions/2750121/for-p-q-primes-such-that-qp-1-show-there-is-only-one-non-abelian-group-of – Gerry Myerson Oct 30 '18 at 09:03
  • Good point.Done. – RagingBull Oct 30 '18 at 09:06
  • Had a look at that link I gave, Raging? – Gerry Myerson Nov 01 '18 at 10:50
  • yeah but it uses semidirect products and sylow theorems. – RagingBull Nov 01 '18 at 11:58

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