Order of Automorphism and Abelian Groups

[RVP91]

If |Aut H| = 1 then how can I show H is Abelian? I've shown a mapping is an element of Aut H previously but didn't think that would help.

I have been looking through properties and theorems linked to Abelian groups but so far have had no luck finding anything that would help.

The closest I have is that it may have something to do with "any subgroup of an Abelian group is normal"

My line of argument was if |Aut G|=1 then Inn G = Aut G and so since Inn G is normal then G is normal?

However something tells me this is incorrect.

Thanks in advance for any help

 

[RVP91]

Here is my revised argument.
To show |Aut H| = 1 implies H is Abelian, we can show H is not Abelian implies |Aut H| does not equal 1.
If H is not Abelian then, we know firstly the identity map is always an automorphism of H. Furthermore we know that the mapping A: H ---> H given by A(x) = (h^-1)xh (where h,(h^-1),x are in H and h and (h^-1) are fixed). This tells us that we have |Aut H| does not equal to one. If H had been Abelian then A(x) = (h^-1)xh = (h^-1)x = x and this is just the identity map again.

 

[spamiam]

Your proof works, but try showing the result directly, i.e. without using contradiction.

Here's an even shorter proof. We know $G/Z(G) \cong \text{Inn}(G) \subseteq \text{Aut}(G)$. So if |Aut(G)|=1, then |G/Z(G)| = 1, implying Z(G)=G.

 

[Deveno]

along the same lines:

if |Aut(G)| = 1, then every inner automorphism is the identity.

this means that for every x and g in G:

gxg^-1 = x, so

gx = xg.

 

[CellCoree]

I understand your confusion and frustration in trying to find a connection between the order of automorphisms and the properties of Abelian groups. Let me provide some clarification and guidance to help you in your exploration.

Firstly, let's define what an automorphism is. An automorphism is a bijective mapping from a mathematical structure to itself that preserves the structure. In the case of groups, an automorphism is a mapping that preserves the group operation and the identity element. In simpler terms, it is a mapping that preserves the group's structure and properties.

Now, let's consider the statement |Aut H| = 1. This means that there is only one automorphism in the group H. This also means that this automorphism is the identity mapping, as it is the only one that preserves the group's structure and properties. Therefore, every element in H is mapped to itself. This leads us to the conclusion that H is a commutative group, or in other words, an Abelian group.

To understand why this is true, let's consider the definition of an Abelian group. An Abelian group is a group in which the group operation is commutative, meaning that the order in which the elements are multiplied does not matter. In other words, for any two elements a and b in the group, a*b = b*a.

Now, since we have established that the only automorphism in H is the identity mapping, this means that for any two elements a and b in H, their images under the automorphism will be a and b respectively. In other words, the automorphism does not change the order of the elements in the group operation. This is exactly the definition of a commutative group, thus proving that H is an Abelian group.

In conclusion, the statement |Aut H| = 1 is a strong indicator that the group H is an Abelian group. This is because the only automorphism in H is the identity mapping, which preserves the group's structure and properties, making H a commutative group. I hope this explanation helps in your understanding of the relationship between the order of automorphisms and Abelian groups.