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Let $G$ be a group of order $240=2^4\cdot 3\cdot 5$.

Assume $G$ is simple, then

  • show that there is no subgroup of index 2, 3, 4 or 5
  • show that there is no subgroup of index 6
  • For the first item, the usual method seems to work:

    Let $H\leq G$ where $[G:H] = 5$ then let $G$ act on the set of right cosets of $H$ by right multiplication.

    Assume $\phi: G\to S_5$ to be its permutation representation. Since $G$ is simple this $\phi$ must be one-to-one which implies $G\cong \mathrm{im}(\phi) \leq G$. But this is impossible since $|S_5| = 5!=120$ while $|G|=240$.

  • For the second item. Let once again $H\leq G$ such that $[G:H] = 6$. Repeating the same argument does not work here, since $240 = |G| \leq |S_6| = 720$.

    Any ideas on how to proceed? I prefer hints to complete solutions.

    Perhaps I could use the theorem $[G:H] = [G:K]\cdot [K:H]$ if I would find a $K$ such that $H\leq K\leq G$. (I doubt it though)

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1 Answer 1

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Let $\varphi: G \to S_6$ be the permutation representation on the six cosets. This is an embedding since $G$ is simple, so we identify $G$ with a subgroup of $S_6$. Simplicity implies all elements of $G$ must be even permutation. Hence $G\leq A_6$, but $240\nmid 360$.


An off-topic comment: There is in fact no simple group of order $240$.

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  • $\begingroup$ I don't see why simplicity implies all elements of $G$ to act on $G/H$ as even permutations... Could you elaborate further? $\endgroup$
    – dietervdf
    Jan 16, 2018 at 7:49
  • $\begingroup$ As example, Let $S = \{1,2,3,4,5,6\}$, why is it impossible for a $g$ to act as a permutation $(1, 2)$? $\endgroup$
    – dietervdf
    Jan 16, 2018 at 7:59
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    $\begingroup$ You treat $G$ as a subgroup of $S_6$. If $G$ contains an odd permutation, then half of $G$'s element is even, half is odd. Those are even form a subgroup of index $2$, which is a proper normal subgroup of $G$. $\endgroup$
    – pisco
    Jan 16, 2018 at 8:02
  • $\begingroup$ I think I get it, thanks! Permutations are so mind-bending! $\endgroup$
    – dietervdf
    Jan 16, 2018 at 8:20

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