In studying for an upcoming prelim, I came across this problem:
Classify all groups of order $182 = 2*7*13$.
Now, the standard tricks here are to look at Sylow's theorems or semi-direct products. Let $n_p$ denote the number of Sylow $p$-subgroups. We can conclude that $n_7 =1$ since $n_7 |26$ and $n_7 \equiv 1 (\mod 7)$. Now, the standard element counting tricks seem to fail me here. Frequently, supposing that one of the other Sylow subgroups is not normal will lead to some sort of contradiction, but I think even assuming $n_{13} = 14$ (which is the next smallest thing it can be), there is just barely enough room in the group for $n_2$ to be $7$. Any ideas on where to proceed from here?
In a related question, does anyone have on hand a good resource for qualifying/prelim exam questions of the flavor “ Classify all groups of order…'' where the order given is in some regards manageable? I've been working through the examples on Stack exchange, but this is such a standard type of problem I'm inclined to think there might be a good resource for these types of questions.
Best Answer
Here is my comment with some added detail:
Let $G$ be a group of order $182$. You already concluded that you have a normal $7$-Sylow subgroup, so let us call this subgroup $P$ and look at $G/P$.
$G/P$ has order $26$ so the $13$-Sylow subgroup has index $2$. This gives us a subgroup of $G$ of index $2$, which is normal due to the index. We will call this subgroup $N$. So now we have that $G$ is a semidirect product of $N$ and a $2$-Sylow subgroup which we will call $Q$.
Let us consider $N$ a bit more carefully. It is a group of order $91 = 7\cdot 13$ and we easily see from the Sylow theorems that both its Sylow subgroups are normal, and hence $N$ is cyclic.
So we now know that our group is a semidirect product of a cyclic group of order $91$ with one of order $2$, so we need to look for subgroups of $\rm{Aut}(C_{91})$ of order $1$ or $2$ (the one of order $1$ just gives us a direct product, which will give the cyclic group of order $182$ which is the only abelian possibility).
Now, $\rm{Aut}(C_{91})$ is isomorphic to $C_6\times C_{12}$ so it has $3$ distinct subgroups of order $2$ and we need to determine which of these yield distinct semidirect products.
To do this, it is probably a good idea to see precisely how these automorphisms of order $2$ act on $C_{91} \simeq C_7\times C_{13}$. The three possible automorphisms are then $(g,h)\mapsto (g^{-1},h)$, $(g,h)\mapsto (g,h^{-1})$ and $(g,h)\mapsto (g^{-1},h^{-1})$.
To see that these give three distinct semidirect products, we can count the number of elements of order $2$ in each.
An arbitrary element can be written as $(g,h)\varphi$ where $\varphi$ is either the identity or the given automorphism of order $2$. Clearly if $\varphi$ is the identity, the element cannot have order $2$, so let us now compute the square of such an element in each of the three cases (remember that $\varphi$ is its own inverse and that conjugation by $\varphi$ corresponds to applying $\varphi$).
In the first case, we get $(g,h)\varphi(g,h)\varphi = (g,h)(g^{-1},h) = (1,h^2)$ which is the identity if and only if $h^2 = 1$ which means if and only if $h = 1$, so we get precisely $7$ elements of order $2$ in this group.
Similarly, in the second case, we get that the element has order $2$ if and only if $g = 1$ so we get $13$ element of order $2$ in this case.
In the final case, we see that all elements of the given form have order $2$, so we have $91$ such elements (this case gives us the dihedral group).