Okay, i will create some cases to make you clear:
1)f(n) = O(g(n)) if and only if g(n) = Ω(f(n))
if g(n)=Ω(f(n)) is TRUE, means g(n) >= c1*(f(n)) for some value of c1>0
then there exists some value of c2>0 for which f(n) <= c2 * g(n)) which is the definition of : f(n)=O(g(n)).
example: 1 )f(n) = n2
g(n) = n2
g(n) = Ωf(n) means g(n) > = c1 * f(n) if c1 = 1 then the condition holds true i.e, n2 >= 1* n2 for every value of n.
Conversely,
f(n) = O g(n) means f(n) <= c2 * g(n) if c2 = 1 then the condition holds true i.e n2 <= 1 * n2 for every value of n.
2) f(n) = n
g(n) = n2
g(n) = Ωf(n) means g(n) > = c1 * f(n) if c1 = 1 then the condition holds true i.e, n2 >= 1* n for every value of n.
Conversely,
f(n) = O g(n) means f(n) <= c2 * g(n) if c2 = 1 then the condition holds true i.e n <= 1 * n2 for every value of n.
note: c1 and c2 values can be altered but the conditions should be satisfied.
2)f(n) = o(g(n)) if and only if g(n) = ω(f(n))
if g(n) = ω(f(n)) is TRUE, means ratio of g(n) to f(n) reaches to infinity when the value of n reaches infinity. that is possible only when g(n) >>> f(n) as n is increasing to infinity .
now think conversely, if very large value divided by very small value gives infinity as the n reaches infinity then what happens when very small value is divided by very large value as n reaches infinity, the ratio becomes ZERO. (read again). which is the definition of f(n) = o(g(n)) i.e, ratio of f(n) to g(n) becomes 0 when the value of n reaches infinity. that is possible only when f(n) >>> g(n) as n is increasing to infinity . You can try some examples by taking the limits and finding the ratio of both functions
1)f(n) = n
g(n) n3
2) f(n) = log n
g(n) = n
note : f(n) and g(n) cannot have same degree because it violates the '=' case in little oh and little omega
3)f(n) = Θ(g(n)) if and only if g(n) = Θ(f(n))
if g(n) = Θ(f(n)) is TRUE i.e. g(n)=O(f(n)) && g(n)= Ω(f(n)) .
⤶ ⤷
f(n)=Ω(g(n) (from 1) && g(n)=O(f(n)) (from 2) and these two conditions prove f(n) = Θ g(n).
You can try examples as well
1) f(n) = n5
g(n) = n5
2) f(n) = n10000
g(n) = n10000
Now, coming to your case f(n)=o(g(n)) if and only if g(n) = Ω f(n)
if g(n) = Ω f(n) is true, then g(n) >= f(n)
but f(n)=o(g(n)) says g(n) > f(n)
now if suppose, f(n) = n and g(n) = n ,
g(n) = Ω f(n) holds true but f(n)=o(g(n)) does not because when n reaches infinity ratio of f(n) to g(n) = 1 and not 0 because both functions are equal and are of same degree which disproves the condition.
pheww:D