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Suppose you have disjoint sets $A, B, C, D $, and $F$ with sizes 1, 12, 12, 15, and 20, respectively. Suppose N is a set formed by taking the union of A with one or more of the other sets. List all the possible sizes of N. Which of these answers divides 60?

I know that $A_5$ is a simple group and that the disjoint sets $$|A|=1,|B|=12,|C|=12,|D|=15,|F|=20$$ are its conjugacy classes of a certain size. And normal subgroup must contain the conjugacy class of size 1, and one or more other conjugacy classes. Thus, the order of any normal subgroup must be a sum of some of these numbers, including the 1. So by Lagrange's theorem, the order must also divide the order of the group. But no such sum among these numbers divides 60, other than 1 and 60 themselves. So its only divisible by $A$ and $A\cup B \cup C \cup D \cup F $.

I'm not sure if this correct, also I have some confusion about all the possible sizes of N. Is it

$|N|$= {13,16,21,25,33,60 }? Is that all or did I miss some?

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  • $\begingroup$ Note that the sizes of $A,B,C,D,E$ are divide 60 and the order of $A_5$ is 60. So do you mean $A,B,C,D,E$ are subgroups of $A_5$ ? In the OP, it is confusing why groups are mentioned in second paragraph while the first paragraph only mention sets.. $\endgroup$ – zongxiang yi May 2 at 5:37
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Nothing in the question talks about groups. There may be some more of the question that does, but it doesn't apply here, so your second paragraph is not on point. You missed several sizes for $N: 28, 36, 40, 45, 48$

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A:=1;
Bs:=[*12,12,15,20*];
SS:=Exclude(Subsets({i : i in [1..#Bs]}), {});
N:={<[Bs[i]: i in s], &+[Bs[i]: i in s] + A> : s in SS};
print {s[2] : s in N};
print {s : s in N | s[2] mod 60 eq 0};

Run the above magma program (http://magma.maths.usyd.edu.au/calc/ ) and you will get

{ 13, 16, 21, 25, 28, 33, 36, 40, 45, 48, 60 } { <[ 12, 12, 15, 20 ], 60> }

So your answer is 60, which the size of $A\cup B \cup C \cup D \cup F$.

PS: it is nothing about groups.

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