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.

  • $\begingroup$ I don't suppose you have learned about $Z$-groups (ie, groups all of whose Sylow subgroups are cyclic)? Have you learned that if the smallest prime dividing the groups order divides it only once, then the group has a normal subgroup of index equal to that prime? $\endgroup$ Commented May 28, 2013 at 19:33
  • 1
    $\begingroup$ Actually, you can get that subgroup I mentioned here directly using Sylow, since you already have a normal subgroup of order $7$, and in the quotient by that you have a subgroup of order $13$ which then has index $2$ and is therefore normal, and pulls back to a normal subgroup of index $2$ in the original group, which then becomes a semidirect product. $\endgroup$ Commented May 28, 2013 at 19:37
  • $\begingroup$ I'm not quite following. I see how you got the subgroup of order 13 which is normal in the quotient. Are you pulling it back to the original group using the correspondence theorem? Also, what does knowing that the group of order 13 tell us about the subgroup in the pullback? Once we know a group of index 2 exists, we know that it is normal. Is the first bit of argumentation just to show that some such subgroup exists? $\endgroup$
    – Bill Kay
    Commented May 28, 2013 at 20:05
  • $\begingroup$ Yes, I used that purely to get the existence of such a subgroup (see my answer for more details). $\endgroup$ Commented May 28, 2013 at 20:37
  • $\begingroup$ All groups of order $2\times$ (odd) always have a normal subgroup $N$ of index $2$, and splits as a semidirect product $N\rtimes C_2$. So any group of order $182=2\times 91$ has a normal subgroup of order $91$, and is a semidirect product. Since every group of order $91$ is cyclic, there is not much work left. $\endgroup$
    – user641
    Commented May 28, 2013 at 22:43

1 Answer 1


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).

  • $\begingroup$ Thank you for your detailed comment- this clears it up for me. $\endgroup$
    – Bill Kay
    Commented May 28, 2013 at 21:01

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