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I just turned in an exam today and I wanted to answer this question, but I couldn't so I had to choose another (you could omit one question).

Up to isomorphism, we had to determine all groups of order 308.

I know that I have to use the Sylow theorems. We can have just one Sylow subgroup of order 11, we can have 1 Sylow subgroup of order 7 or 22 Sylow 7-subgroups, and 1,11,7,or 77 sylow 2-subgroups. I know I need to consider the eight cases and break it down from there, but I don't know how...

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    $\begingroup$ Have you seen other examples worked out in detail? $\endgroup$ – Andrea Mori Dec 19 '13 at 13:42
  • $\begingroup$ Hint: If there's just one subgroup of order 11 (call it $H$), that subgroup must be normal. So (writing $G$ for the group of order 308), we have a quotient group $G/H$ of order 28. It would be helpful to know what that group is. So start by classifying groups of order 28. $\endgroup$ – WillO Dec 19 '13 at 14:17
  • $\begingroup$ No, I've never seen examples like this worked out in detail. I have the Gallian textbook, and it has examples, but I am having trouble following similar steps for 308. $\endgroup$ – Laura Dec 19 '13 at 19:53
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Hints for you to understand and/or prove:

There exists only one single Sylow subgroup of order $\;11\;$ and one unique one of order $\;7\;$. Denote them by $\;P_{11}\,,\,P_7\;$ , resp. Then, the group $\;Q:=P_{11}P_7\cong P_{11}\times P_7\;$ is cyclic and normal, from where we get that any group $\;G\;$ of order $\;308\;$ is a semidirect produc $\;T\rtimes Q\;$ , with $\;|T|=2^2=4\;$ .

Since $\;|\text{Aut}\,( Q)|=10\cdot 6=60\;$, there exist non-trivial homomorphisms $\;T\to\text{Aut} (Q)\;$ : at least two (or even three) if $\;T\cong V\cong C_2\times C_2\;$ , and at least two if $\;T\cong C_4\;$, and together with the two non-isomorphic abelian groups of order $\;308\;$ , we already have at least $\;6/7\;$ non-isomorphic groups of order $\;308\;$ . Continue from here.

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    $\begingroup$ Why is the number of $7$-Sylow subgroups 1 and not 22? $\endgroup$ – ronno Dec 20 '13 at 7:19
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    $\begingroup$ I thought that you could have either 1 Sylow 7-subgroup or 22 Sylow 7-subgroups. $\endgroup$ – Laura Dec 26 '13 at 17:39
  • $\begingroup$ Also, DonAntonio, thank you so much! I understand the beginning about semidirect products, but is there another way of determining without using non-trivial homomorphisms? We didn't learn about them in class, and I'm happy to learn about them now to solve the problem, but I'm really curious about how to solve it with the knowledge I had at the time of the take-home exam. $\endgroup$ – Laura Dec 26 '13 at 17:42

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