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There is a question in Lang that I am having trouble with. Problem 1.41 for reference.

Problem: Let $H$ be a simple group of order $60$.

a) Show that the action of $H$ by conjugation on the set of its Sylow subgroups gives an embedding $H\hookrightarrow A_n$

Question: I already figured out that they are talking about the $5$-Sylow subgroup, of which there are $6$. But I do not know how to show the imbedding. I have an idea that if I can show that $H$ is generated by order $3$ elements then I can map those into order $3$ elements of $A_6$, but I do not know how to show that $H$ is generated by order $3$ elements.

A hint would be greatly appreciated.

Quang Hoang
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Enigma
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    The action is transitive, so gives a non-trivial homomorphism from $H$ to $S_6$. What can you say about its kernel? And about the preimage of $A_6$? – Jyrki Lahtonen Sep 19 '14 at 17:03
  • The kernel would have to be the identity since H is simple. The pre-image of $A_6$ would have to be H because $A_6$ is simple? – Enigma Sep 19 '14 at 17:16
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    Correct about kernel. The pre-image would, indeed, have to be all of $H$. But the reason is a bit more subtle. Namely otherwise it would have index two in $H$, and ... – Jyrki Lahtonen Sep 19 '14 at 17:18
  • I know that if it has index two, then the preimage of $A_n$ is a normal subgroup of $A_n$, a contradiction. But what I don't understand is why when the pre-image of $A_n$ isn't all of H then it must be half of H? – Enigma Sep 19 '14 at 17:31
  • Okay nevermind I think I know why it would have index 2. Any subgroup of $S_6$ has the same number of odd and even permutations. – Enigma Sep 19 '14 at 17:39
  • Correct! Well done! – Jyrki Lahtonen Sep 19 '14 at 19:33

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