Modern Algebra-Proving Elements Conjugate

[sephy]

1. If a₁,...,a_n is a list of (not necessarily distinct) elements of a group G, then, for all i, a_i...a_na₁...a_i-1 is conjugate to a₁,...,a_n.

Homework Equations

The Attempt at a Solution

I know that you have to prove the existence of an element g of the group G such that ga₁,...,a_ng^-1 = a_i...a_na₁...a_i-1, but I don't know how to find this element g, or how to define it or it's inverse. Very confused.

 

[mighty2000]

Firstly, let's define what it means for two elements in a group to be conjugate. Two elements a and b in a group G are said to be conjugate if there exists an element g in G such that g^-1ag = b. This essentially means that g "transforms" a into b.

Now, let's consider the elements ai...ana1...ai-1 and a1,...,an. We want to show that they are conjugate, which means we need to find an element g in G that transforms one into the other.

One way to approach this is to consider the element g = ai...an. This element is in G, since G is a group and therefore closed under multiplication. Now, let's see what happens when we apply g^-1 to ai...ana1...ai-1:

g^-1(ai...ana1...ai-1) = (ai...an)^-1(ai...ana1...ai-1) = (ai-1...a1)(ai...ana1...ai-1)

= ai-1...anai-1...a1ai...ana1...ai-1 (since G is associative)

= ai-1...a1ai...ana1...anai-1 (since G is commutative)

= a1,...,an

Therefore, we have shown that g^-1(ai...ana1...ai-1) = a1,...,an, which means that g^-1 transforms ai...ana1...ai-1 into a1,...,an. This also means that g transforms a1,...,an into ai...ana1...ai-1, since g^-1ag = a.

Hence, we have found the element g in G that conjugates ai...ana1...ai-1 to a1,...,an. This shows that for any i, ai...ana1...ai-1 is conjugate to a1,...,an.