Group of Order 39 Classification Homework

[cuses]

Homework Statement

Classify all groups of order 39

Homework Equations

The Attempt at a Solution

I can show that if the group of abelian, it is isomorphic to Z/3Z*Z/13Z.
If the group is not abelian, there are two types of groups, they either follow the rule : x^3=e, y^13=e, and xy=y^3x; or x^3=e, y^13=e, and xy=y^9x. But I want to show that they are the same thing if I rename x^2=z.
But I want to prove that there is a homomorphic between these two groups. Could some one help me?

 

[mighty2000]

I understand the importance of classifying and organizing information in a systematic way. In this case, we are looking at groups of order 39, which is a specific number that can help us narrow down the potential groups.

First, we can start by listing out the prime factors of 39: 3 and 13. This gives us a hint that the group may be isomorphic to Z/3Z*Z/13Z, as mentioned in the attempt at a solution. This is because the order of an abelian group is equal to the product of the orders of its cyclic subgroups.

Next, we can consider the non-abelian groups of order 39. As mentioned, there are two types of groups that can potentially fit this criteria. One way to approach this problem is to use the concept of group presentations. We can define a group presentation as <x, y | x^3=e, y^13=e, xy=y^3x> or <x, y | x^3=e, y^13=e, xy=y^9x>, depending on which type of group we are considering.

To show that these two groups are isomorphic, we can use the concept of homomorphisms. A homomorphism is a function between two groups that preserves the group operation. In this case, we can define a function f: G1 -> G2, where G1 and G2 are the two groups defined by the group presentations above. We can show that this function is an isomorphism by proving that it is both injective and surjective.

To prove injectivity, we can show that the kernel of f, defined as Ker(f) = {g in G1 | f(g) = e}, is equal to the identity element e. This means that the only element that maps to the identity in G2 is the identity element in G1.

To prove surjectivity, we can show that every element in G2 has a preimage in G1. This means that for every element in G2, there is an element in G1 that maps to it under f.

By proving both injectivity and surjectivity, we can conclude that f is an isomorphism between the two groups, and thus they are the same thing under a different naming convention.

In conclusion, as a scientist, I would approach this problem by using the concepts of group presentations