Number of Mutually Non-Isomorphic Groups of Order 45

To determine the number of mutually non-isomorphic groups of order 45, we can use the concept of the isomorphism theorem and the prime factorization of the order.

Prime Factorization of 45

The prime factorization of 45 is 3 * 3 * 5, which means that the order 45 can be expressed as a product of three distinct prime numbers.

Existence of Groups of Order p^2 * q

According to the Sylow theorems, if a prime power p^k divides the order of a group G, then G must have a subgroup of order p^k. In this case, since the prime factorization of 45 is 3 * 3 * 5, we can deduce the following:

- There exists at least one subgroup of order 3^2 = 9.
- There exists at least one subgroup of order 5.

Number of Subgroups of Order p^2

To find the number of subgroups of order p^2, we can use the fact that the number of subgroups of order p^2 is congruent to 1 modulo p. Applying this to our case:

- The number of subgroups of order 3^2 = 9 is congruent to 1 modulo 3, which means there is 1 such subgroup.

Number of Subgroups of Order q

To find the number of subgroups of order q, we can use the same reasoning as before. In this case:

- The number of subgroups of order 5 is congruent to 1 modulo 5, which means there is 1 such subgroup.

Number of Mutually Non-Isomorphic Groups

Since the subgroups of order 9 and 5 are unique, we can conclude that there is only one mutually non-isomorphic group of order 45.

Summary

- The prime factorization of 45 is 3 * 3 * 5.
- There exists at least one subgroup of order 9 and one subgroup of order 5.
- The number of subgroups of order 9 is 1, and the number of subgroups of order 5 is also 1.
- Therefore, there is only one mutually non-isomorphic group of order 45.