# Order of Automorphism and Abelian Groups

1. Nov 22, 2011

### 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

2. Nov 22, 2011

### 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.

3. Nov 23, 2011

### 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.

4. Nov 24, 2011

### 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.

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