Proving Isomorphism of a Group of Order 60 to the Alternating Group A5

In the study of group theory, one frequently encounters groups of various orders. One particularly challenging question is to prove the isomorphism of a specific group of order 60 to the alternating group (A_5). This problem is not a straightforward one, as not all groups of order 60 are isomorphic to (A_5). However, when considering simple groups of order 60, we can establish an isomorphism to (A_5). This article will outline a detailed proof using the Sylow theorems and concepts from finite group theory.

Introduction to the Problem

The problem can be approached by assuming that the group (G) is a simple group of order 60, which can be factored as (2^2 times 3 times 5). This means that (G) is a non-cyclic simple group of order 60, except possibly for the alternating group (A_5).

Termination of Simple Groups of Order Up to 100

A standard sequence of exercises in finite group theory shows that no non-cyclic simple groups exist through order 100, with the possible exception of order 60. This fact is crucial in understanding that if (G) is a simple group of order 60, it must be isomorphic to (A_5). This leaves open the question of whether there could be another simple group of order 60. The proof that there is no such simple group other than (A_5) requires a more detailed analysis.

Sylow Subgroups and Transitivity

Let (G) be a finite simple group of order 60. The number (t) of 2-Sylow subgroups can be 1, 3, 5, or 15. The first option, (t 1), is excluded as it would imply a normal 2-Sylow subgroup. The second option, (t 3), is also impossible because (G) acts transitively on the set of Sylow subgroups of any order, and a simple group of order 60 cannot act transitively or even non-trivially on a set of size 3.

Embedding in S5 and Isomorphism to A5

If (t 5), then (G) acts on a set of size 5, making it simple and thus the kernel of this action must be trivial. Therefore, (G) embeds as a subgroup of index 2 in (S_5), and the only such subgroup is (A_5). This establishes the isomorphism (G cong A_5).

Further Analysis for t 15

For the case where (t 15), we need to produce a subgroup of index 5. With this, (G) will embed in (S_5), and we will be done. Given that (G) has 15 2-Sylow subgroups each of order 4, if they all intersect trivially, that would result in 45 elements of order 2 or 4, which is too many. Specifically, we need at least 6 5-Sylow subgroups, so at least 24 elements of order 5. This is not possible.

Therefore, some 2-Sylow subgroups, say (A) and (B), must intersect non-trivially: (A cap B 2). Let (K A cap B). Since (A) and (B) are both abelian and all groups of order 6 are abelian, the centralizer (C C_G(K)) contains both (A) and (B). Therefore, (C) contains (C_4) and [G:C] 15). We certainly can't have (G C) because this would mean that (K) is in the center, making the center non-trivial, which is impossible as [G:C] cannot be 3 or 5.

Thus, we have established that there is no simple group of order 60 other than (A_5).