The Misconception of Simple Groups and Their Abelianness

It's a common misconception that a group having no proper non-trivial normal subgroup necessarily makes it an abelian group. This article aims to clarify this misconception by presenting clear definitions, illustrative examples, and proofs. Specifically, we will prove that a group with no proper non-trivial normal subgroups is not always abelian, using the smallest counterexample: the alternating group ( A_5 ) (the group of even permutations on a set of 5 elements).

Key Definitions

To understand this topic, let's start by defining some key terms:

Abelian Group

A group ( (G, *) ) is called an abelian (or commutative) group if the binary operation ( * ) is commutative, i.e., for all ( a, b in G ), we have ( a * b b * a ).

Normal Subgroup

A subgroup ( H ) of a group ( G ) is called a normal subgroup (or invariant subgroup) if for every ( g in G ) and ( h in H ), we have ( g * h * g^{-1} in H ). This can be written as ( gHg^{-1} H ) for all ( g in G ).

Proper Non-Trivial Normal Subgroup

A proper non-trivial normal subgroup of a group ( G ) is a normal subgroup ( H ) such that ( H eq {e} ) and ( H eq G ), where ( e ) is the identity element of ( G ).

Counterexample: The Alternating Group ( A_5 )

The alternating group ( A_5 ) is the group of all even permutations of a set with 5 elements. It is a well-known simple group, meaning it has no proper non-trivial normal subgroups. We will now provide a proof to show that ( A_5 ) is not an abelian group.

Proof that ( A_5 ) is Not Abelian

To prove that ( A_5 ) is not an abelian group, we need to show that there exist at least two elements in ( A_5 ) such that their commutator is not equal to the identity element ( e ).

Consider two specific permutations in ( A_5 ): a 3-cycle and a product of two disjoint 2-cycles. Let's denote the 3-cycle by ( (123) ) and the 2-cycle product by ( (12)(34) ).

Step 1: Compute the Commutator

The commutator of two elements ( a ) and ( b ) in any group is defined as ( [a, b] a^{-1}b^{-1}ab ). We will compute the commutator of the 3-cycle ( (123) ) and the 2-cycle product ( (12)(34) ).

First, let's compute ( (12)(34)(123)(12)(34) ):

( (12)(34) ) swaps 1 and 2, and 3 and 4. ( (123) ) sends 1 to 2, 2 to 3, and 3 to 1. ( (12)(34) ) again swaps 1 and 2, and 3 and 4.

The result is a 3-cycle ( (13)(24) ).

Step 2: Compute the Inverse Commutator

Next, compute the inverse commutator ( (123)(12)(34)(123)^{-1}(12)(34) ):

( (123)^{-1} (132) ).

The result is a 3-cycle ( (14)(23) ).

Step 3: Compare the Results

We see that the commutator of ( (123) ) and ( (12)(34) ) is ( (13)(24) ), and the inverse commutator is ( (14)(23) ). Since these are not the same, the group ( A_5 ) is not abelian.

Conclusion

In conclusion, we have shown that the alternating group ( A_5 ) is a simple group with no proper non-trivial normal subgroups, but it is not an abelian group. This example disproves the notion that a group with no proper non-trivial normal subgroups must be abelian.

Frequently Asked Questions

What does it mean for a group to be non-abelian?

A group is non-abelian if there exist at least two elements in the group that do not commute, i.e., there exist ( a, b in G ) such that ( ab eq ba ).

Are there any simple abelian groups?

Yes, there are simple abelian groups. The cyclic groups of prime order are examples of simple abelian groups. For instance, ( mathbb{Z}_p ) (the integers modulo p) is a simple abelian group when p is a prime number.

Can a simple group ever be abelian?

A simple abelian group must be cyclic of prime order. If a group ( G ) is simple and abelian, it must have no proper non-trivial normal subgroups, and the only way this can happen is if ( G ) is isomorphic to ( mathbb{Z}_p ), where ( p ) is a prime number.