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So I'm proving that a group $G$ with order $112=2^4 \cdot 7$ is not simple. And I'm trying to do this in extreme detail :)

So, assume simple and reach contradiction. I've reached the point where I can conclude that $n_7=8$ and $n_2=7$.

I let $P, Q\in \mathrm{Syl}_2(G)$ and now dealing with cases that $|P\cap Q|=1, 2^2, 2^3$ or $2^4$.

I easily find contradiction when $|P\cap Q|=2^4$ and $2$.

Um, got stuck REAL bad on the case $|P\cap Q|=2^3$ and $2^2$.

If $|P \cap Q |=2^3= 8$ and $|P|=|Q|=16$, is there any relationship between $P,Q$ and their intersection that can help me?

TMM
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Akaichan
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3 Answers3

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If $G$ is a simple group, it must have exactly $7$ Sylow $2$-subgroups. Thus $G$ embeds into $S_7$, and in particular into $A_7$ since $G$ does not have a subgroup of index $2$. But the order $A_7$ is not divisible by $112$.

If you want to go along the lines of your original idea, you can rule out the case $|P \cap Q| = 2^3$ by noticing that then $P \cap Q$ is normal in $P$ and $Q$ (as a subgroup of index $2$), so $N_G(P \cap Q)$ contains $P$ and $Q$, which implies that $N_G(P \cap Q) = G$.

ADDED: I'm not sure if there is an easy way to deal with rest of the cases. However, there is a nice argument which also works for proving that every group of order $p^n q$ ($p$, $q$ distinct primes) is nonsimple. I believe the idea of the proof goes back to G. A. Miller (around 1900-1910). Here's an illustration of it in this case.

Suppose that $G$ is a simple group of order $112$. Then $G$ has exactly $7$ Sylow $2$-subgroups. Let $P, Q \in Syl_2(G)$ be such that $P \neq Q$ and that $D = P \cap Q$ has largest possible order. Steps for the proof:

  1. Using the fact that $D < N_P(D)$ and $D < N_Q(D)$ (proper inclusion), prove that $N_G(D)$ cannot be a $2$-group.

  2. Thus $D$ is normalized by an element $g \in G$ of order $7$. Prove that $P, gPg^{-1}, \ldots, g^6Pg^{-6}$ are distinct. Conclude that $D$ is contained in every Sylow $2$-subgroup.

  3. Since the intersection of all Sylow $2$-subgroups is normal, $D$ is trivial.

  4. By counting elements in Sylow $2$-subgroups, prove that $G$ contains exactly one Sylow $7$-subgroup.

This same argument works for proving the statement for groups of order $p^n q$.

Mikko Korhonen
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    Could you expand on your statement "Thus G embeds into S7". Thanks! – Ziggy Apr 24 '13 at 00:50
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    @Ziggy: If $G$ is a simple group and $[G:H] = n$, then the left coset action gives us an injective homomorphism $f: G \rightarrow S_n$ (in this case $H = N_G(P)$ for $P$ some Sylow $2$-subgroup). With Sylow subgroups you could apply the conjugation action as well. – Mikko Korhonen Apr 24 '13 at 08:07
  • @Mikko: In your first arguement, I couldn't see why G embeds into $A_7$. Could you explain a little more?Thanks! – Ergin Süer Sep 02 '14 at 16:33
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    @ErginSuer: Let $H \leq S_n$. If $H$ has order $> 2$, then you can see that $H$ must contain an even permutation $\neq 1$. Thus $H \cap A_n$ is a nontrivial normal subgroup of $H$. In the case where $H$ is simple this implies $H \cap A_n = H$, ie. $H \leq A_n$. – Mikko Korhonen Sep 03 '14 at 06:09
  • @MikkoKorhonen I can't seem to see why $N_G(P\cap Q)$ contains both $P$ and $Q$ implies that $N_G(P\cap Q)=G$. Can you elaborate more on this? Can you also link the paper by G.A. Miller. Can't seem to work out the solution even with the hints. :( – Explorer1234 Jun 21 '21 at 14:42
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    @Explorer1234: You could just use Lagrange's theorem. You know $N_G(P \cap Q)$ has order a multiple of $2^4$ since it contains $P$, so..? Also I don't remember anymore if this is the original reference, but in the old book by Miller/Blichfeldt/Dickson, Miller gives the argument in §73 p. 185 there. – Mikko Korhonen Jun 21 '21 at 15:19
  • @MikkoKorhonen Can you provide more details to points 1 and 2 of the solution by Miller? Cant seem to work it out. – Explorer1234 Jun 23 '21 at 07:09
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    @Explorer1234: For 1., if $N_G(D)$ is a $2$-group, it is contained in some $2$-Sylow $R \neq P$, and then $R \cap P > D$, contradicting maximality of $D$. In 2., you have $D$ contained in $g^i P g^{-i}$ for all $i$. – Mikko Korhonen Jun 23 '21 at 07:14
  • Question could we suppose it’s simple then n7 = 8 and n2=7 thus there’s 48 elements of order 7 and 105 elements of order 2 contradicting order of group being 112? – MyMathYourMath Oct 25 '22 at 15:15
  • @MyMathYourMath: How do you get $105$ elements of order $2$? For $7$-Sylows you can count $48$ since $P \cap Q = 1$ when $P$ and $Q$ are distinct subgroups of order $7$. But for distinct $2$-Sylows of order $2^4$, it is possible that $P \cap Q \neq 1$. Also, it is not always the case that all the elements in a $2$-Sylow have order $2$. – Mikko Korhonen Oct 26 '22 at 02:10
  • I did 7 times 2^4-1 which is 105 since we’re supposing it’s not simple as n2 can’t be 1. Does this not work? – MyMathYourMath Oct 26 '22 at 13:09
  • @MyMathYourMath: No, usually it does not work. You can count $6 \cdot 8 = 48$ elements of order $7$ when $n_7 = 8$. This is because in this case, all $7$-Sylows contain $6$ elements of order $7$, and distinct $7$-Sylows do not have elements of order $7$ in common. – Mikko Korhonen Oct 26 '22 at 13:40
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I wanted to write out Mikko Korhonen's first idea for proof in detail as a separate answer, since it is not trivial at all, and provoked some questions in the comments.

As mentioned in the original question, we assumed $n_2=7$. From Sylow's second theorem, we know that all the 2-Sylow subgroups are conjugate, and we can look at $G$'s action on them by conjugation. This action implies a homomorphism: $$f:G\rightarrow S_7$$ $\ker(f)$ cannot be $G$ since the action is a transitive action. Then, if $\ker(f)$ is non-trivial (meaning $ker(f)\neq\{e\}$) then it is a non-trivial normal subgroup of $G$, and therefore $G$ isn't simple.

Otherwise, we get $ker(f)=\{e\}$ and therefore $f$ is injective, meaning that $G$ is isomorphic to a subgroup of $S_7$. For convenience purposes, we'll write $G\leq S_7$. $G$ cannot be contained in $A_7$ because 112 doesn't divide $|A_7|$. In that case $GA_7=S_7$ and using the second isomorphism theorem we get: $$G/(G\cap A_7)\cong GA_7/A_7\cong S_7/A_7\cong\mathbb{Z}_2$$ and therefore $[G:(G\cap A_7)]=2$ meaning that $G\cap A_7$ is a normal subgroup of $G$.

Dean Gurvitz
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Sylow's theorems require that $n_2=1$ or 7, $n_7=1$ or 8, so $G$ has a chance to be simple only if $n_2=7$ and $n_7=8$. Note that the Sylow 7-subgroups can only intersect at the identity, any two Sylow 2-subgroups can share a subgroup of order at most 8, and a Sylow 7-subgroup and a Sylow 2-subgroup can share only the identity.

Hence the union of Sylow 7-subgroups has $1+8\cdot(7-1)=49$ elements, and the union of the Sylow 2-subgroups has at least $8+7\cdot(16-8)=64$ elements, which happens precisely when they all share a subgroup $H$ of order 8. In this case, the union of all Sylow 7-subgroups and 2-subgroups has $64+49-1=112$ elements, so we learn that no other scenario (one with a greater union of the Sylow 2-subgroups) is allowed.

Now notice that $H$ is a normal subgroup of $G$ since conjugation by $g\in G$ permutes the Sylow 2-subgroups and so preserves their intersection. Hence $G$ is not simple.

Edit: This answer is wrong because the union of the Sylow 2-sbgps can be smaller than 112, see the comments below.

D M
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  • how exactly do you know that the intersection of 2 sylow 2 subgp cannot be 8 though? – Akaichan Apr 05 '13 at 03:14
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    If $n_2=7$, $n_7=8$, the intersection of all Sylow 2-subgroups (and hence any two) MUST have order 8, otherwise the elements wouldn't fit in $G$. The point is that when this happens, the intersection is a normal subgroup of $G$. – D M Apr 05 '13 at 03:19
  • I'm sorry that I'm having a hard time with this but how is the intersection is a normal subgp of $G$? – Akaichan Apr 05 '13 at 03:33
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    Let $H_j$, $j=1,\ldots,7$ be the Sylow 2-sbgps and let $H$ be their intersection. Let $g\in G$, then $gH_jg^{-1}=H_{\sigma(j)}$, where $\sigma\in S_7$. If $h\in H_j$, then $g h g^{-1}\in H_{\sigma(j)}$ and so if $h\in H$, then $ghg^{-1}\in H$ since $\sigma$ is a bijection. – D M Apr 05 '13 at 03:47
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    I'm not sure I follow your argument in "the union of the Sylow $2$-subgroups has at least..". For example, let $G = S_3 \times S_3$. Then $G$ has exactly $9$ Sylow $2$-subgroups. Intersections of distinct Sylow $2$-subgroups have order at most $2$. So if I understand your claim, you're saying that $G$ would have at least $2 + 9 \cdot (4 - 2) = 20$ elements in the union of Sylow $2$-subgroups. But this isn't true, there are only $16$ elements in the union of Sylow $2$-subgroups of $G$. – Mikko Korhonen Apr 05 '13 at 07:28
  • @m.k. You're right, this is a bug in my argument. I upvoted your answer. Is it a common practice to delete wrong answers or keep them since they can teach you what not to do? – D M Apr 05 '13 at 16:42
  • It's up to you. If you keep it maybe you should edit in a note about the mistake so it's not hidden in the comments. I'd say keep it, I think this is a common pitfall when doing these sorts of counting arguments related to Sylow subgroups. – Mikko Korhonen Apr 05 '13 at 17:41