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We know that:

Theorem: If a simple group $G$ has a proper subgroup $H$ such that $[G:H]=n$ then $G\hookrightarrow A_n$.

This fact can help us to prove that any group $G$ of order $120$ is not simple. In fact, since $n_5(G)=6$ then $[G:N_G(P)]=6$ where $P\in Syl_5(G)$ and so $A_6$ has a subgroup of order $120$ which is impossible. My question is:

Can we prove that $G$ of order $120$ is not simple without employing the theorem? Thanks.

azimut
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Mikasa
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  • Just curious: Are you refering to some specific approaches when asking the question? Thanks. – awllower Aug 06 '12 at 03:00
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    Well, you could classify all groups of order 120. Why do you want to avoid the theorem? – Kevin Carlson Aug 06 '12 at 04:54
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    It does seem to me that this probably is the easiest way to prove that there is no simple group of order $120$. There are other ways to do it, of course, but the ones I can think of are more complicated than that. – Geoff Robinson Aug 06 '12 at 07:02
  • @GeoffRobinson: Yes, it does, but I am just curious if there is other way helping us for such this kinds of problems. Thanks for the time. Thanks for all comments. – Mikasa Aug 06 '12 at 07:23
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    You may want to clarify in your question why $A_6$ can have no subgroups of index $3$. – Eric Auld Dec 17 '14 at 02:00
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    @EricAuld: this can be done by using the theorem again: $A_6$ is simple, so if it had a subgroup of index 3, then $A_6$ would embed into $A_3$. – Ravi Fernando Oct 23 '16 at 04:47

2 Answers2

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Well, you can obtain a contradiction to the simplicity of a finite group $G$ of order $120$ by showing that a Sylow $2$-subgroup $S$ of $G$ can't be a maximal subgroup of $G,$ for example (I won't give the details, but they require somewhat more background than the theorem you want to avoid). Hence $G$ has a subgroup of index $3$ or $5$, but then you are using the embedding in a symmetric group to obtain a contradiction in any case. Or you can do a complicated fusion and transfer analysis with the prime $2,$ but there is a perfect group of order $120$, so that is not straightforward either (the perfect group of order $120$ has a center of order $2$).

Geoff Robinson
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  • Thanks Geoff for your time. I think, I'd better use the theorem instead than the other way. Thanks for your great answer and hints. :) – Mikasa Aug 06 '12 at 12:31
  • @Geoff May I ask for the reason that a Sylow 2-subgroup $S$ of $G$ can't be a maximal subgroup of $G$? Thanks in advance. – Steve Jacob Mar 18 '19 at 16:41
  • @SteveJacob : Suppose that $G$ is simple of order $120$ and a Sylow $2$-subgroup $S$ of $G$ is maximal.By Frobenus's normal p-complement theorem, $S$ has a subgroup $V \neq 1$ such that $N_{G}(V)/C_{G}(V)$ is not a $2$-group. Then $|V| >2$ and clearly $|V| \neq 8$, for otherwise $V = S \not \lhd G$, so $N_{G}(V) = S,$ a contradictio. Similarly, if $V$ has order $4,$ then $V \lhd S$ since maximal subgroups of $2$-groups are normal. Again we obtain $N_{G}(V) = S,$ a contradiction. – Geoff Robinson Mar 18 '19 at 17:03
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Here is a good one.

Of course $n_5=6, n_3=10 \text{ or } 40.$ Assuming simplicity of G, we can embed G into $S_6$

$n_3$ cannot be 40 because otherwise it contains all the 3-cycles and hence is larger than $A_6$, contradiction.

Hence $n_3 = 10$

Goal: Find an element in G of order 6. Such element must be an odd permutation by studying its cycle structure. Hence we are done.

Observe that $N_G(P_3)$ have order 12, $P_3$ is Sylow-3 group.

Of course $N_G(P_3)/C_G(P_3)$ embed into $Aut(P_3)$.

Hence $C_G(P_3)$ has even order, and hence has an element of order 2. Such element multiplies and element in $P_3$ gives our desired order 6 element. We are done.

LVMIMAOQ
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  • i think $n_3$ can also be $4$ – e.ad Apr 25 '22 at 15:39
  • Once you know that $G$ embeds into $S_6$, it's immediate that $G$ embeds into $A_6$ (does not require simplicity of $A_6$), which the OP didn't want to use. Anyway, not every element of order $3$ in $S_6$ must be a $3$-cycle. – Brauer Suzuki Jul 16 '22 at 04:50