Thread: Boolean algebra,isomorphism

Let (B₁,+,*,^c) is Boolean algebra where B₁ is set of every positive dividers of 2310, in which are defined:x+y = lcm(x,y), x*y = gcd(x,y) and x^c = 2310/x.
Let (B₂,∩,∪,^c) is Boolean algebra of every subsets of {a,b,c,d,e} with standard operations.

Let f:B₁ → B₂ is isomorphism where f(2)={a} , f(3)={b} , f(5)={c} , f(7)={d} , f(11)={e}.
- How f(35), f(110), f(210), f(330) are defined?
- Determine f((30 + 5*7)^c).
- How many elements B₁ have?
- Write all atoms of B₁.
- How many different isomorphism can be defined between B₁ and B₂?
thanks for your time

Originally Posted by kljoki

How f(35), f(110), f(210), f(330) are defined?

Since * corresponds to ∪, factor each of these numbers and use the fact that f(x * y) = f(x) ∪ f(y).

hint: 2310 = 2*3*5*7*11

make a conjecture about "primes" in B₁, and "singleton sets" in B₂.

suppose σ is a bijection on (the underlying set of) B₂.

show that σof is also an isomorphism from B₁ to B₂. how many such permutations (bijections) exist?

prove that every bijection from {2,3,5,7,11} to {a,b,c,d,e} can be written as σof, for some permutation σ of {a,b,c,d,e}.

conclude that the number of such isomorphisms is 120 <---where does this number come from?

one specific example worked out:

f(35) = f(lcm(5,7)) = f(5+7) = f(5) + f(7) = f(5) U f(7) = {c} U {d} = {c,d}.

lets see
f(35), f(110), f(210), f(330)

f(110) = f(lcm(11,2,5)) = f(11+2+5) = f(11) + f(2) + f(5) = f(11) U f(2) U f(5) = {e} U {a} U {c} = {a,c,e}

f(210) = f(lcm(7,5,3,2)) = f(7+5+3+2) = f(7) + f(5) + f(3) +f(2) = f(7) U f(5) U f(3) U f(2) = {d} U {c} U {b} U {a} = {a,b,c,d}

f(330) = f(lcm(11,5,3,2)) = f(11+5+3+2) = f(11) + f(5) + f(3) +f(2) = f(11) U f(5) U f(3) U f(2) = {e} U {c} U {b} U {a} = {a,b,c,e}

f((30 + 5*7)^c) = 2310 / f( f(lmc(2,3,5)) + f(gcd(5,7))) = {a,b,c,d,e} / {{a,b,c} U Ø} = {d,e}

B₁ have 5 elements and every element is an atom

there is 5! = 120 (permutation) different isomorphism between B₁ and B₂ because we have 2 sets with 5 elements.
For the first element we have 5 combinations (f(2)={a}, f(2)={b}, f(2)={c}, f(2)={d}, f(2)={e}), second element have 5-1 combinations (because one element applies to f(2)), third 5-2 etc.
so came to the result of 120 different isomorphisms