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I'm trying to find an example of a group $G$ such that $|G| = 120$, and a non-normal subgroup $H$ within it. Of course, my first instinct is to let $G = S_5$, but this doesn't work because all subgroups of $S_5$ are normal. Help would be greatly appreciated!

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    Oh, dear: not at all! The only non-trivial normal subgroup $\,S_5\,$ has is $\,A_5\,$...**all** its other non-trivial subgroups are non-normal2012-12-16
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    If I take the all the permutations with cycle type 5, say {(12345), (12354),(12435),...}, why is this not normal?2012-12-16
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    It's a cyclic Sylow sub group of A_5 of order 5, and all the above are conjugate so the conjugacy class is not of order 1, so not normal2012-12-16
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    @CardFlower, conjugation classes are not *usually* subgroups!2012-12-16
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    And oups. I meant <(12345)> obviously, and its 6 (i think) conjugacy classes.2012-12-16
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    I think the question, @CardFlower, is *why is this normal?* :-)2012-12-16
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    I thought that conjugacy by any element in $S_5$ preserves cycle type, so that regardless of what you hit $<(12345)>$ with, it will stay within $<(12345)>$.2012-12-16
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    Conjugating rho=(12345) will preserve its cycle type, but the result will not necessarily be a power of (12345). (Do some computations for yourself to check.) As it is a 5-cycle, the subgroup generated by it will have 5 elements, whereas the conjugacy class associated to it has 5!/5=4!=24 members. As Don noted, conjugacy classes are not the same thing as subgroups, though conjugacy classes of the symmetric groups are always those permutations of a given cycle type as you observe.2012-12-16

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