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

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!)

• FWIW $$150 = 2 \times 3 \times 5^2$$ – Kenny Lau Aug 13 '18 at 11:12
• What do you mean by "element counting"? Do you mean using Sylow's theorems to say that "if there are $6$ Sylow-5 subgroups then there must be $x$ elements of order $5$, a contradiction"? Assuming $G$ is simple, and doing the above (and without doing anything additional/clever) I worked out that there are at least $3$ Sylow $2$-subgroups and $10$ Sylow $3$-subgroups, and there are $6$ Sylow $5$-subgroups. Adding up the corresponding elements does not give you a number bigger than $150$: $3$ elements of order $2$, $27$ of order $3$ and at least $24$ of order $5$. So $54$ elements, plus identity. – user1729 Aug 13 '18 at 11:53
• (Possibly you can do more with the Sylow $5$-subgroups (as they have order $25$ not $5$), but as the answer to the linked post points out, they have pairwise non-trivial intersection so you'd have to be careful.) – user1729 Aug 13 '18 at 11:53
• I voted-to-close NOT as being a duplicate but "unclear what you are asking". – Nicky Hekster Aug 14 '18 at 9:54
• I think I found a good interpretation of the question and posted an answer that both uses counting and is much simpler than the answer for the linked-to question, so please vote not to delete this question. Thanks! – C Monsour Aug 14 '18 at 11:04

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 $4$.
• Why faithful? Simplicity does not rule out the possibility of some element $g\in G$ such that $H^g=H$ for all the Sylow $5$-subgroups $H$. What am I missing? [Nice idea, by the way!] – user1729 Aug 14 '18 at 10:35
• On the contrary, simplicity does rule that out, since the set of all such $g$ is a normal subgroup that is not the whole group (as the action is transitive on more than one element, and thus non-trivial). Since $G$ is simple, that normal subgroup is just the identity. – C Monsour Aug 14 '18 at 10:52
• Let $K$ be the set of all such $g$. Then I understand that $K\lneq G$ (you can use Sylow for this as $K\leq N_G(H)$, and $[G:N_G(H)]=6$). I cannot see why $K$ is normal though. – user1729 Aug 14 '18 at 10:59
• You can verify that normality either directly ($k^{-1}gk$ fixes $k^{-1}Hk$ if $g$ fixes $H$, so if $g$ stabilizes all conjugates of $H$, so does $k^{-1}gk$) or because the set of all such $g$ is the kernel of a homomorphism to $S_6$. – C Monsour Aug 14 '18 at 10:59