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The problem comes from the fat that a Factorization object is not hashable:

sage: f = factor(15)
sage: f
3 * 5
sage: hash(f)
TypeError: <class 'sage.structure.factorization_integer.IntegerFactorization'> is not hashable

It should be (i will open a ticket for that).

You can workaround by replacing the factorization with a hashable equivalent : a tuple:

sage: t = tuple(f)
sage: t
((3, 1), (5, 1))
sage: hash(t)
8332132266671271694

You will be able to recover the factorization as follows:

sage: Factorization(t)
3 * 5
sage: Factorization(t) == f
True

So you can do something like:

sage: S = Set([tuple(factor(10)), tuple(factor(15))])
sage: S
{((2, 1), (5, 1)), ((3, 1), (5, 1))}

sage: len(S.subsets())
4
sage: S.subsets()[3]
{((2, 1), (5, 1)), ((3, 1), (5, 1))}

sage: S.subsets()[3][0]
((2, 1), (5, 1))
sage: Factorization(S.subsets()[3][0])
2 * 5

The problem comes from the fat that a Factorization object is not hashable:

sage: f = factor(15)
sage: f
3 * 5
sage: hash(f)
TypeError: <class 'sage.structure.factorization_integer.IntegerFactorization'> is not hashable

It should be (i will open a ticket for that).

You can workaround by replacing the factorization with a hashable equivalent : a tuple:

sage: t = tuple(f)
sage: t
((3, 1), (5, 1))
sage: hash(t)
8332132266671271694

You will be able to recover the factorization as follows:

sage: Factorization(t)
3 * 5
sage: Factorization(t) == f
True

So you can do something like:

sage: S = Set([tuple(factor(10)), tuple(factor(15))])
sage: S
{((2, 1), (5, 1)), ((3, 1), (5, 1))}

sage: len(S.subsets())
4
sage: S.subsets()[3]
{((2, 1), (5, 1)), ((3, 1), (5, 1))}

sage: S.subsets()[3][0]
((2, 1), (5, 1))
sage: Factorization(S.subsets()[3][0])
2 * 5

sage: a = -5
sage: a.factor()
-1 * 5
sage: tuple(a.factor())
((5, 1),)

The problem comes from the fat that a Factorization object is not hashable:

sage: f = factor(15)
sage: f
3 * 5
sage: hash(f)
TypeError: <class 'sage.structure.factorization_integer.IntegerFactorization'> is not hashable

It should be (i will open a ticket for that).

You can workaround by replacing the factorization with a hashable equivalent : a tuple:

sage: t = tuple(f)
sage: t
((3, 1), (5, 1))
sage: hash(t)
8332132266671271694

You will be able to recover the factorization as follows:

sage: Factorization(t)
3 * 5
sage: Factorization(t) == f
True

So you can do something like:

sage: S = Set([tuple(factor(10)), tuple(factor(15))])
sage: S
{((2, 1), (5, 1)), ((3, 1), (5, 1))}

sage: len(S.subsets())
4
sage: S.subsets()[3]
{((2, 1), (5, 1)), ((3, 1), (5, 1))}

sage: S.subsets()[3][0]
((2, 1), (5, 1))
sage: Factorization(S.subsets()[3][0])
2 * 5

EDIT Be careful that the sign is lost when taking the tuple:

sage: a = -5
sage: a.factor()
-1 * 5
sage: tuple(a.factor())
((5, 1),)