Automorphism of order 2 fixing just identity. Prove that G is abelian.

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In summary, the given conversation discusses a problem involving a finite group $G$ and an automorphism $T$ with specific properties. It is proven that $G$ must be abelian by using the permutation representation of $T$ on $G$ and showing that $T$ has a homomorphism property. This is done by defining a function $f$ and showing its injectivity, which leads to the conclusion that $G$ is abelian.
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Let $G$ be a finite group, $T$ an automorphism of $G$ with the property that $T(x)=x$ if and only if $x=e$. Suppose further that $T^2=I$, that is, $T(T(x))=x$ for all $x\in G$. Show that $G$ is abelian.

I approached this problem using the permutation representation afforded by $T$ on $G$. Its easy to deduce that the cycle representation of the permutation of $G$ caused by $T$ has $(n-1)/2$ disjoint transpositions, where $n=|G|$. We know, from this, that $n$ is odd but so what? I am not able to exploit the homomorphism property of $T$.
 
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hint: (i don't want to spoil your fun because this is a beautiful theorem)

define $f:G \to G$ by:

$f(g) = g^{-1}T(g)$

show $f$ is injective (and thus bijective).

now...what is $T\circ f(g)$?
 

FAQ: Automorphism of order 2 fixing just identity. Prove that G is abelian.

What is an automorphism of order 2 fixing just identity?

An automorphism is a type of function that maps a mathematical structure to itself while preserving its structure. An automorphism of order 2 means that when the function is applied twice, it results in the identity element. In this case, the automorphism only fixes the identity element and no other elements in the structure.

What does it mean for a group to be abelian?

A group is said to be abelian if its group operation is commutative, meaning that the order in which elements are multiplied does not affect the final result. In other words, if a and b are elements of an abelian group G, then a * b = b * a for any combination of a and b.

Why does an automorphism of order 2 fixing just identity prove that G is abelian?

If G is an abelian group, then for any two elements a and b in G, the order in which they are multiplied does not matter. This means that the function that fixes just the identity element must also fix all other elements in the group, as swapping their order would result in a different element being fixed. Therefore, the group G must be abelian.

Can you give an example of an automorphism of order 2 fixing just identity?

One example of an automorphism of order 2 fixing just identity is the function that maps every element in the group to itself except for the identity element, which is mapped to its inverse. This function satisfies the criteria of an automorphism of order 2 fixing just identity, as applying it twice results in the identity element being fixed.

How does this proof relate to the properties of an abelian group?

This proof shows that the defining property of an abelian group, commutativity, is preserved by the automorphism of order 2 fixing just identity. This is because the function only fixes the identity element, leaving all other elements to be freely multiplied in any order. Therefore, the proof strengthens the argument that G must be abelian.

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