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I'm having trouble solving this question.

What i've managed so far is. Let G be a group and $|G|=105=3*5*7$, $n_p=|Syl_p(G)|$ and let $P\in Syl_3(G)$, $Q\in Syl_5(G)$ and $R\in Syl_7(G)$. We have that

$$n_3=1 \text{ (from assumption.)}\\ n_5=1|21,\\ n_7=1|15.$$

Since the 7-Sylow and 5-Sylow groups are generated by a single element if two 5-Sylow or 7-Sylow must have a trivial intersection, otherwise they are the same group. By a counting the elements we find that at least two of the p-Sylow in G must be normal. This reduces the problem into three cases as follows.

Case 1: $n_7= 15$ and $n_5=1$:

Case 2: $n_7= 1$ and $n_5=21$:

Case 3: $n_7= 1$ and $n_5=1$:

Any suggestions on how to proceed from here?

Strange Brew
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  • Please search for your question before posting. Thanks. – rschwieb Oct 27 '16 at 16:40
  • My apologies. But what if I don't grasp the solution in that thread? I'm not familiar with semi-direct product. – Strange Brew Oct 27 '16 at 19:24
  • If you have a specific question about that solution, you can ask a new question requesting an explanation of the step you don't understand, or else ask if there is an alternative way that avoids semidirect products. – rschwieb Oct 27 '16 at 19:29
  • Ask for an alternative proof-method in that question thread or in this one? – Strange Brew Oct 27 '16 at 19:30
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    You might get a quick answer if you asked that responder directly in the comments to his solution. If you get no response, it would be OK to post an entirely new question about the part you don't get. – rschwieb Oct 27 '16 at 19:31
  • Alright then, I'll do that. Thanks. – Strange Brew Oct 27 '16 at 19:34

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