# Determining All Groups of Order 308

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...

• Have you seen other examples worked out in detail? – Andrea Mori Dec 19 '13 at 13:42
• 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. – WillO Dec 19 '13 at 14:17
• 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. – Laura Dec 19 '13 at 19:53

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.
• Why is the number of $7$-Sylow subgroups 1 and not 22? – ronno Dec 20 '13 at 7:19