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I was reading the following proof of why there is no simple group of order $120$:

A group of order $120$ cannot be simple

And I couldn't understand the following: "so $A_6$ has a subgroup of order $120$ which is impossible".

Why is it impossible? Isn't the order of $A_6$ $6! / 2$, which equals $360$?

And since $120$ divides $360$, I can't see what's the problem in it...

verret
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ChikChak
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2 Answers2

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I will recite the theorem given in the linked question:

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

Now, Assume you have a subgroup $H \le A_6$, $|H|=120$. The group $A_6$ is simple and $[G:H] = 3$. Thus, by the theorem above, there is an embedding $A_6\hookrightarrow A_3$, a contradiction.

yakobyd
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  • Ah, I can see it now. Is that the theorem that @the_fox referred as the "$n!$ theorem"? – ChikChak Jun 24 '18 at 20:45
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    The theorem in @the_fox 's post is a well known result that is often referred to as Cayley's theorem. The corollaries he presented and the theorem above are based on the same idea (Cayley's theorem). – yakobyd Jun 24 '18 at 21:38
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    By your name I assume that you read hebrew. This hebrew wikipedia page has some more information about the theorem and they present an important generalization which is roughly translated as "The refinement of Cayley's theorem" (I could not find a source in english, perhaps some one in here may know its name). – yakobyd Jun 24 '18 at 21:40
  • As simple as that. – Blue Tomato May 03 '23 at 21:58
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I am unable to find a proper reference for the theorem I am referring to, so here's the relevant part from page 4 of Isaacs' "Finite Group Theory" book.

enter image description here

the_fox
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    Here let $G=A_6$ and $H$ be the subgroup of order $120$. $H$ has index $3$ in $G$ and therefore has three right cosets, including itself. $G$ acts faithfully on the cosets of $H$ by right multiplication - since there are $3$ cosets, the permutations of the cosets form a subgroup of $S_3$, which has order $6$. The kernel of this action is a non-trivial normal subgroup of $G$ of index $3$ or $6$ (we know the action is transitive on cosets, hence the factor $3$ is known). But $G$ is simple, and this is a contradiction. – Mark Bennet Jun 24 '18 at 21:07