Can one prove that a group of order 150 is not simple using element counting?

Question

I'd like to know whether it's possible to show that a group of order 150 is not simple using only element counting (or mostly element counting). I have seen a solution to this problem here (Every group of order $150$ has a normal subgroup of order $25$), but the method is different.

EDIT: By "element counting," I mean using Sylow's theorems to determine what $n_2$, $n_3$, and $n_5$ can be (other than 1; if any of these were 1, then we'd be done!) and then hopefully use this information to conclude that $G$ is simple. (As user1729 points out below, the only chance seems to be to do something smart with the Sylow 5-subgroups. I guess I am asking if anyone can think of such a smart thing to do!)

Answer 1

There's an easy proof that does involve element-counting, but perhaps not in the way you have in mind. You already know by Sylow that if the group is simple it must have $6$ $5$-Sylow subgroups. If this is the case, then it acts by conjugation on these $6$ subgroups both transitively (by Sylow) and faithfully (since it's simple), and so embeds monomorphically as a subgroup of $S_6$. But, and here comes the "counting" part, since $150$ does not divide $720$ (the order of $S_6$), this is impossible. So no group of order $150$ is simple.

This idea works for a bunch of similar problems as well. In more complex cases, one can also make more sophisticated use of this embedding in a symmetric group to make inferences about the structure of the given group. For example, in some cases the given group could actually have the putative number of Sylow subgroups but would also have an index $2$ subgroup in that event--one can often prove this by showing that some element of the group maps to an odd permutation.

For example, we could also solve this very problem by embedding $G$ in $S_{150}$ using the action of $G$ on itself by right multiplication. Since $G$ has an element of order $2$ and since that element maps to a disjoint product of $75$ transpositions in $S_{150}$, it maps to an odd permutation, i.e., in $S_{150}$ but not $A_{150}$. Thus $G$ has a (necessarily normal) index $2$ subgroup, namely everything that maps to an even permutation (i.e., into $A_{150}$). In fact, one can use this idea to show that any group of order divisible by $2$ but not $4$ cannot be simple. But it feels a lot more tangible to work with $S_6$ than with $S_{150}$, and the action by conjugation on Sylow subgroups is also useful for plenty of group orders divisible by $4$.

Answer 2

It's not by element counting but it might be worth mentioning this answer:

As $G$ has a subgroup of index $6$, which is one of the Sylow $5$-subgroup, and $|G|=150 \nmid 360=6!/2=|A_{6}|/2$. We get that $G$ is non-simple.

The lemma that is used is the following:
Let $G$ be a finite simple group. Then $G$ is isomorphic to a subgroup of $A_{n}$, where $n=|G:H|$, and $H\le G$.