This is a second part of a two-part question.
In the first part I was asked to prove that any group of order $5784 = 2^3 \cdot 3 \cdot 241$ has subgroups of the following indexes: $3, 6, 8, 12,$ and $24$.
I proved that part by showing:
- G has subgroups of orders $2, 2^2, 2^3, 3, 241$ (by Sylow theorem).
- Moreover, the subgroup of $241$, lets call it H, is normal since by Sylow's third theorem its easy to see its the only subgroup of that order.
- Therefore for any other subgroup of $G$, $K$, $HK$ is a subgroup as well (since HK = KH due to H normal)
- But the order of HK is $o(HK) = \frac{o(H) \cdot o(K)}{o(H \cap K)} = o(H) \cdot o(K)$ since necessarily $o(H\cap K) = 1$ due to order of H being prime and any other subgroup will not have order that is a multiple of that prime (241)
- Therefore we have subgroups of orders: $241$, $241 \cdot 2$, $241 \cdot 2^2$, $241 \cdot 2^3$, $241 \cdot 3$, which give us the subgroups of required indexes (24, 12, 6, 3, 8 respectively) as well, QED.
Now to prove that there is a specific group of such order that does not have a normal subgroup of index 12, I suppose I need to show that there is such a group where there is more than one subgroup of index 12 (which is of order 241*2 = 482) and somehow prove that it is not equal to all of its conjugates in G, e.g. $g^{-1}Hg$?
Perhaps via some arithmetical considerations involving Sylow's third theorem I can show that there is theoretically a possibility for such group?
Best Answer
Posting the answer I arrived at with the kind help of @SteveD in the comment thread above:
Consider the group 𝐺 = $S_4 \times C_{241}$.
As the direct product of $S_4$ (the symmetric group on 4 elements, of order 4! = 24), and the finite cyclic group of order 241, this is a group of order 5784, as needed.
Assume it has a normal subgroup N of index 12.
Then N has order 5784/12 = 482. Note that 482 = 241 * 2 which is of the form $p \cdot q$ for two primes. Therefore (by Cauchy theorem) it has a subgroup of order 241. Further, this must be the one and the same group $C_{241}$ since there is only a single unique subgroup of this order in G (by Sylow's third theorem which constraints the number of sylow subgroups of a given order that can exist, in this case the conditions are easily seen to imply a single subgroup).
So we have $N \unlhd G$, $C_{241} \unlhd G$ and $C_{241} \unlhd N$, therefore by the third isomorphism theorem we have $ N / C_{241} \unlhd G / C_{241}$.
But $G / C_{241} \simeq S_4$ (by construction of G) and $o(N/C_{241}) = [N : C_{241}] = 2$, which is a contradiction since the symmetric group $S_4$ does not have any normal subgroups of order 2. QED.