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...
 A: 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.
A: Here are two approaches, both through the Sylow $11$-subgroup.

*

*Let $|G|=308$. Since $308=2^2\cdot 7\cdot 11$, we see that the number of Sylow $11$-subgroups is $1$ (it's one of $1$, $2$, $4$, $7$, $14$ and $28$, and only $1$ is $1$ modulo $11$). Let $P$ denote the unique, normal Sylow $11$-subgroup of $G$. The quotient group $G/P$ has order $28$, and has a unique Sylow $7$-subgroup, $R/P$. If $Q$ is any Sylow $7$-subgroup of $G$, then $|QP/P|=7$, so $QP=R$. Thus every Sylow $7$-subgroup of $G$ is contained in $R$. But $|R|=77=7\cdot 11$, so has a unique Sylow $7$-subgroup, as needed.


*This starts as before, to obtain a unique Sylow $11$-subgroup $P$. Then $N_G(P)/C_G(P)$ is (isomorphic to) a subgroup of the automorphism group of $P$, which has order $10$. Since $7\nmid 10$, $7\mid |C_G(P)|$, so there is a Sylow $7$-subgroup $Q$ of $G$ centralizing $P$. Thus $11\mid |C_G(Q)|\mid |N_G(Q)|$, so in particular $|G:N_G(Q)|\neq 22$.
Since there is a unique Sylow $7$- and $11$-subgroup, they centralize one another and $G$ is a semidirect product $C_{77}\rtimes C_4$ or $C_{77}\rtimes (C_2\times C_2)$.
To complete the proof, we need to count maps from the Sylow $2$-subgroup $R$ into $\mathrm{Aut}(C_7\times C_{11})\cong C_6\times C_{10}$. If $R\cong C_4$ then the kernel of the map is at least $C_2$, so there are four possible images, according as the image is $1$, the $C_2$ in the first fact, the second fact, or both diagonally.
For $C_2\times C_2$ there are five maps: the unique faithful map, the unique trivial map, and then the three previous maps with kernel of order $2$.
This yields nine groups in total.
