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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!)

math4
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  • FWIW $$150 = 2 \times 3 \times 5^2$$ – Kenny Lau Aug 13 '18 at 11:12
  • What does "FWIW" mean ? – Peter Aug 13 '18 at 11:15
  • @Peter For What Is Worth. – José Carlos Santos Aug 13 '18 at 11:35
  • For What It's Worth – Nicky Hekster Aug 13 '18 at 11:36
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    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
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    (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
  • @Nicky I have reopened this question as it is not a duplicate of the linked-to question - the OP wants a different method of proof. (On the other hand, I think "unclear what your asking" may be relevant here...) – user1729 Aug 14 '18 at 07:02
  • @user1729 OK, no problem, but for me it is unclear what is being asked "only element counting". An approach through conjugacy classes (every normal subgroup is a disjoint union of a couple of them) ...? – Nicky Hekster Aug 14 '18 at 08:29
  • @NickyHekster I agree that it's unclear. I would have hoped that they would have responded to my comment, above, but they don't seem to have logged on since posting the question. I will vote-to-close as unclear in 3 hrs (so 24 hrs since they asked the question) if they don't respond/make themselves more clear first. – user1729 Aug 14 '18 at 08:47
  • @NickyHekster Did you vote-to-close as unclear, or as a duplicate? I had presumed you had voted to close as a duplicate as that was the close reason, but I can't remember what happens if you give a different close reason. [I would have pinged amWhy to explain myseld as it was their Gold Badge power's which closed the question, but they haven't commented so I wasn't sure if the ping would work.] – user1729 Aug 14 '18 at 08:50
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    I voted-to-close NOT as being a duplicate but "unclear what you are asking". – Nicky Hekster Aug 14 '18 at 09:54
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    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
  • @CMonsour Hmm...I wonder if, on balance, you should post your answer as solution to the other question. We can then close this one. (Don't worry about reputation - the +1 here is mine!) – user1729 Aug 14 '18 at 11:20
  • I can do that also. I think this is a valid answer to a reasonable question, though, so I don't see why people want to get rid of the question. – C Monsour Aug 14 '18 at 11:21
  • I posted a different answer to the other question. Proving the the $5$-Sylow is normal is a different question from just showing that the group is not simple. – C Monsour Aug 14 '18 at 11:44
  • So I still think this question should be kept. – C Monsour Aug 14 '18 at 11:44
  • @user1729 My apologies for not replying to your question. I was cramming for an exam... I have edited my post to clarify what I meant. So sorry for all this! – math4 Aug 14 '18 at 18:45

2 Answers2

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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 $4$.

C Monsour
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  • 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
  • This isn't just true for actions by conjugation. It's a standard result for ANY group action that the set of all group elements that act trivially is a normal subgroup (since it's always the kernel of induced homomorphism to the symmetric group on the set being acted on). – C Monsour Aug 14 '18 at 11:02
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    Ah, right, I see what's going on. It was getting a bit too technical for me for a moment there :-) Basically, $G$ acts transitively by conjugation on the six subgroups so we have a map $G\rightarrow S_6$. As $150$ does not divide $6!$ this map must have a non-trivial kernel, contradicting simplicity. – user1729 Aug 14 '18 at 11:05
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    This is a very useful "counting" technique for showing by elementary means that groups of specific orders are not simple. Apparently it is not as well known as I thought. I hope this answer helps a lot of people with a new idea for solving such problems. – C Monsour Aug 14 '18 at 11:07
  • @CMonsour Thank you for your elegant solution. I am still working on digesting it, but I think it answers the spirit of my original question ("Is there an easy way to solve this problem???"). – math4 Aug 14 '18 at 18:49
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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 $H\le G$.

Ash
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