How to prove a group is abelian

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In summary, we use the fact that \sigma^2 is the identity map to obtain x=\sigma(x^{-1}), and then use this to show that the binary operation in G commutes, proving that G is abelian.
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Johnson04
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Homework Statement


Let [tex]G[/tex] be a finite group which possesses an automorphism [tex]\sigma[/tex] such that [tex]\sigma(g)=g[/tex] if and only if [tex]g=1[/tex]. If [tex]\sigma^2[/tex] is the identity map from [tex]G[/tex] to [tex]G[/tex], prove that [tex]G[/tex] is abelian.

Homework Equations


Show that every element of [tex]G[/tex] can be written in the form [tex]x^{-1}\sigma(x)[/tex] and apply [tex]\sigma[/tex] to such an expression.

The Attempt at a Solution


My first question is how to obtain [tex]x=x^{-1}\sigma(x)\mbox{, }\forall x\in G[/tex].

If [tex]x=x^{-1}\sigma(x)\mbox{, }\forall x\in G[/tex] is obtained, then by applying [tex]\sigma[/tex] to the identity, and denote [tex]\sigma(x)[/tex] by [tex]y[/tex], we get [tex]y=y^{-1}x[/tex]. Then take inverse on both sides to get [tex]y^{-1}=x^{-1}y\Rightarrow xy^{-1}=y[/tex]. So [tex]y^{-1}x=xy^{-1}[/tex]. It seems the binary operation in the group commutes, but since [tex]y[/tex] is actually the image of [tex]x[/tex] under the automorphism [tex]\sigma[/tex]. So my second question is how to reach the conclusion from this result.

Thanks in advance for any help!
 
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Hello,

To answer your first question, we can use the fact that \sigma^2 is the identity map to write x=\sigma(x^{-1}). Then we can substitute this into the expression x^{-1}\sigma(x) to get x^{-1}\sigma(x)=x^{-1}\sigma(\sigma(x^{-1}))=x^{-1}x^{-1}. Since \sigma^2 is the identity map, we can simplify this to x^{-1}\sigma(x)=x^{-2}.

To answer your second question, we can use the result we obtained above to show that xy=yx for any x,y\in G. We have xy=yx\Rightarrow x^{-1}xy=x^{-1}yx\Rightarrow y=x^{-1}yx. Since this holds for any x\in G, we can conclude that y=xyx^{-1}, or equivalently xy=yx, for all x,y\in G. This shows that the binary operation in G commutes, and therefore G is abelian.

Hope this helps!
 

1. What is an abelian group?

An abelian group is a type of mathematical group that follows the commutative property, meaning that the order in which you perform operations does not affect the outcome. In simpler terms, the elements of an abelian group can be rearranged without changing the result.

2. How do you prove that a group is abelian?

To prove that a group is abelian, you need to show that the group operation is commutative. This means that for any two elements a and b in the group, a * b = b * a. You can do this by showing that all possible combinations of elements in the group follow this property.

3. What are some common examples of abelian groups?

Some common examples of abelian groups include the integers under addition, the real numbers under addition, and the group of 2x2 matrices with real entries under matrix multiplication. Any group that follows the commutative property is considered an abelian group.

4. Can a subgroup of an abelian group also be abelian?

Yes, a subgroup of an abelian group will also be abelian. This is because a subgroup inherits the properties of the larger group, including the commutative property. So, if the larger group is abelian, then any subgroup will also be abelian.

5. Are all finite groups abelian?

No, not all finite groups are abelian. In fact, there are infinitely many finite non-abelian groups. However, there are some finite groups that are abelian, such as the cyclic groups, which are groups that can be generated by a single element.

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