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I need to show that $A_5$ has no subgroup of order 30. Any ideas?

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3 Answers

up vote 13 down vote accepted

Such a subgroup would have index $2$ and so immediately be normal and non-trivial. It suffices to show that $A_5$ is simple.

Alternatively, show that the class equation is $60=1+15+20+12+12$. One cannot obtain $30$ as a sum of any subcollection of the summands on the right. Since normal subgroups are composed of whole conjugacy classes, there is then no subgroup of $A_5$ of order $30$.

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One can find a proof that $A_5$ is simple as Theorem 3.1 in this handout of Keith Conrad's. –  Dylan Moreland Jun 16 '12 at 14:57
    
H/T for the reference, Dylan. –  Cameron Buie Jun 16 '12 at 15:33
    
Thank you Cameron. I have a little question about the class equation. In my calculations, there are 4! = 24 members in the Conjugacy class of a 5-cycle, can you explain your calaculation please? –  Roy Jun 16 '12 at 16:47
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You are absolutely correct...for the conjugacy class of a $5$-cycle in $S_5$. In $A_5$, that conjugacy class splits into two classes of twelve $5$-cycles, each. If you think about it, it's inevitable that it must split somehow, since $24$ does not divide $60=|A_5|$ (though it does divide $120=|S_5|$). If I recall correctly, the conjugacy classes are those corresponding to $(1\,2\,3\,4\,5)$ and $(2\, 1\, 3\, 4\, 5)$. –  Cameron Buie Jun 16 '12 at 17:13
    
Thank you again! –  Roy Jun 16 '12 at 18:05
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Assume to the contrary it has a subgroup $H$ of order 30. Then, it has an element $g$ of order 2 by Cauchy's theorem. Then considering the cycle type of $g$ and using $g \in A_5$, $g$ must be a product of two transpositions. But since $H$ is normal in $A_5$, $H$ has to contain all products of 2 transpositions: WLOG assume $g = (1 2)(3 4)$. Then setting $x = (1 2 3)$, $xgx^{-1} = (2 3)(1 4)$. You can set $x = (1 3 2)$ to get $xgx^{-1} = (3 1)(2 4)$. So $H$ contains a subgroup ($V_4$) of order 4, contradicting Lagrange's theorem.

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Is it clear that you could still get to any double transposition by conjugating with elements of $A_5$, rather than $S_5$? –  Dylan Moreland Jun 16 '12 at 15:02
    
Thank you. I had a dim recollection of this general result from a representation theory course, and I think I can do it by just playing around with elements in $A_5$. I thought the OP might be confused on this point, though. –  Dylan Moreland Jun 16 '12 at 15:11
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Thanks for the reference. It seems easy enough in this case, and perhaps this idea works in general: you can conjugate your $g$ to $[12][34]$ using an element of $S_n$; if this is not in $A_n$, then note that conjugating further by $[12]$ does nothing. –  Dylan Moreland Jun 16 '12 at 15:17
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@DylanMoreland - and that's why if you have an cycle of even length (like a transposition) as part of the cycle decomposition, you get the whole conjugacy class. You can do this with an element of $S_n$, and if this is not even, you can compose on the correct side with your cycle (which being even in length is an odd permutation and gives you the requisite element of $A_n$). –  Mark Bennet Jun 16 '12 at 15:23
    
Thank you Serkan –  Roy Jun 16 '12 at 18:07
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Here's another approach: A subgroup of order $30$ is, as was noted, normal, and must contain the commutator subgroup (since the quotient would be of order $2$, hence abelian).

First, note that the commutator subgroup of $A_5$ will contain all $3$-cycles; namely, given a $3$-cycle $(ijk)$, let $r$ and $s$ be the two elements of $\{1,2,3,4,5\}\setminus\{i,j,k\}$; then $$[(rs)(ij),(rs)(ik)] = (rs)(ij)(rs)(ik)(rs)(ij)(rs)(ik) = (ij)(ik)(ij)(ik) = (ijk)\in [A_5,A_5].$$

It must also contain all permutations of the form $(ij)(rs)$ with all $i,j,r,s$ distinct. Indeed, given $(ij)(rs)$, we have $$[(irs),(ir)(sj)] = (isr)(ir)(sj)(irs)(ir)(sj) = (ij)(rs).$$ Thus, this subgroup must contain at least all $3$-cycles (there are $20$ of them, $\frac{1}{3}(5\times 4\times 3)$), and all products of two transpositions (there are $3\times\binom{5}{1}=15$ of them). Together with the identity, this gives at least $36$ elements in the commutator subgroup. In particular, there can be no subgroup of order $30$, as it would necessarily have to contain at least 36 elements...

(In fact, every element of $A_5$ is a commutator; the recently proven Ore Conjecture shows that the same is true for any finite nonabelian simple group, though the proof relies on the classification of finite simple groups).

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@user33814: Neat answer ever had. –  B. S. Jun 16 '12 at 20:49
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