The number of group homomorphisms from ℤ/150ℤ to ℤ/90ℤ is 

Concept:

If \(\mathbb{Z}/n\mathbb{Z} \) and \(\mathbb{Z}/m\mathbb{Z} \) are cyclic groups, the number of group homomorphisms from \( \mathbb{Z}/n\mathbb{Z} \) to \(\mathbb{Z}/m\mathbb{Z}\) is given by

\(\text{Number of homomorphisms} = \gcd(n, m),\) where \( \gcd(n, m) \) is the greatest common divisor of  n and m.

Explanation:

The number of group homomorphisms from \(\mathbb{Z}/150\mathbb{Z}\) to \(\mathbb{Z}/90\mathbb{Z}\) is given by the greatest common divisor (gcd)

of the orders of the two groups. That is  \(\text{Number of homomorphisms} = \gcd(150, 90)\)
  
Prime factorization of 150: \(150 = 2 \times 3 \times 5^2\)

Prime factorization of 90: \(90 = 2 \times 3^2 \times 5\)

Now, the gcd of 150 and 90 is the product of the lowest powers of the common prime factors:
   
\( \gcd(150, 90) = 2 \times 3 \times 5 = 30 \)

The number of group homomorphisms from \(\mathbb{Z}/150\mathbb{Z}\) to \(\mathbb{Z}/90\mathbb{Z}\) is 30.

Thus, option 1) is correct.