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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?2011-12-25
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    The quotient group has order 22011-12-25
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    Three proofs of this fact are given in Keith Conrad's notes [here.](http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/A4noindex2.pdf). Highly recommended!2013-08-24
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    [Here](http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=447674) is the ML link.2011-12-26

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