characterizing groups of order 1960

Question

This problem is from an an Introduction to Abstract Algebra by Derek John that I am solving.

I am trying to prove that any group of order 1960 aren't simple, so I am doing it by contradiction, but I got stuck in the middle.

Suppose that $|G| = 1960 = 2^3 * 5 * 7^2$, by Sylow theory we have 2,5,7 subgroup of G. After some computations I got the following least values for $n_p$. The following are the values $n_2 = 5$, $n_7 = 8$, and $n_5 = 56$, but I don't know how to proceed further.

Answer 1

If the group is not simple, then there actually must be $8$ Sylow-$7$ subgroups. It's the only option give the Sylow theorems.

So $G$ permutes these $8$ subgroups through conjugation. That means there is a map from $G$ to $S_8$. The kernel of this map is a normal subgroup, so the kernel is either all of $G$ or is trivial. The map itself is nontrivial since Sylow-7 subgroups can be permuted to one another. This means the kernel cannot be all of $G$. On the other hand, $7^2$ does not divide $8!$, so the kernel cannot be trivial. So there actually cannot be $8$ Sylow-$7$ subgroups. There must only be $1$, precluding $G$ from being simple.

Answer 2

Posting my chatroom solution. It is more complicated than Alex Jordan's argument, but does give a bit more information about the groups of size $1960$. More precisely, it shows that either a Sylow 5-subgroup, denoted $P_5$, or a Sylow 7-subgroup, denoted $P_7$, must be a normal subgroup. Also, we will see that $G$ always has an abelian subgroup of order $245$.

1. Unless $P_7\unlhd G$ there must be exactly $8$ Sylow 7-subgroups. In that case the normalizer $N_G(P_7)$ has cardinality $245$. So we can assume that $P_5\le N_G(P_7)$.
2. Because $|P_7|=7^2$ we know that $P_7$ must be isomorphic to either $C_{49}$ or to the additive group of the vector space $\Bbb{F}_7^2$. In the former case $Aut(P_7)\cong C_{42}$, in the latter case $Aut(P_7)\cong GL_2(\Bbb{F}_7)$. These have cardinalities $42$ and $42\cdot48$ respectively.
3. So in both cases we see that $P_7$ has no automorphisms of order five. Therefore the conjugation action of the copy of $P_5$ on $P_7$ must be trivial, and $P_5$ centralizes $P_7$.
4. Therefore $C_G(P_5)$ has size at least $245$ for this copy of $P_5$ (and hence for all of them). Thus also $245\mid |N_G(P_5)|$.
5. Hence the number of Sylow $5$-subgroups is a factor of eight. But the only factor of $8$ that is $\equiv1\pmod5$ is $1$, so $P_5\unlhd G$.

Of course, if $P_7\unlhd G$, the argument of steps 2 - 5 still works, and shows that $P_5\le C_G(P_7)$, and $P_5\unlhd G$. So we still have an abelian subgroup $P_5P_7$.