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Show that the alternating group $A_4$ of all even permutations of $S_4$ does not contain a subgroup of order $6$.

For me am thinking to write all elements of $A_4$ and trying to find every cyclic subgroup generated by each element of $A_4$, then I have to check whether there exist such a subgroup or not! This is a long procedure for me, I ask if there is a short way to do this.

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If the subgroup has order $6$, then what can you say about the order of the quotient group? What can you conclude after that? –  mathmath8128 Dec 25 '11 at 23:48
    
The quotient group has order 2 –  Junior II Dec 25 '11 at 23:57
    
Here is the ML link. –  Ehsan M. Kermani Dec 26 '11 at 0:36
    
Three proofs of this fact are given in Keith Conrad's notes here.. Highly recommended! –  Prism Aug 24 '13 at 14:09

3 Answers 3

Assume $H \le A_4$ is a subgroup of order 6. Then $H$ contains a unique subgroup $C$ of order 3. So $C$ is characteristic in $H.$ And as $H$ is normal in $A_4,$ we obtain $C$ is normal in $A_4.$ On the other hand, if $(a b c)$ is a generator of $C,$ conjugating $(abc)$ by $(ab)(cd) \in A_4$ we obtain $(bad) \not\in C,$ from which we obtain not only a contradiction, but a potential pun - not too shabby.

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i get u,thanks a lot –  Junior II Dec 26 '11 at 0:13
    
@JuniorII It would be nice of you to upvote and accept jspecter's answer since you found it helpful. –  Alex Becker Dec 26 '11 at 5:39

There is a proof in the QUESTION here.

Note that There are twelve elements in $A_4$ : $(1),\ (12)(34),\ (13)(24),\ (14)(23),\ (123),\ (132),\ (124),\ (142),\ (134),\ (143),\ (234),\ (243)$.

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There are only two groups of order 6: the cyclic group of order 6, and a group isomorphic to $S_3$. But the maximum order of permutations in $S_4$ is 4, which excludes a cyclic subgroup of order 6, and $S_3$ includes simple interchanges, which are not in $A_4$.

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