Well, this is an exercise problem from

Hersteinwhich sounds difficult:

- How does one prove that if |G|>2, then G has non-trivial automorphism?
The only thing I know which connects a group with its automorphism is the theorem, G/Z(G) \cong \mathcal{I}(G) where \mathcal{I}(G) denotes the

Inner- Automorphism groupof G. So for a group with Z(G)=(e), we can conclude that it has a non-trivial automorphism, but what about groups with center?

**Answer**

As you note in the question, the group of inner automorphisms Inn(G) is isomorphic to G/Z(G). In particular, it’s trivial if and only if Z(G)=G. So there is a non-trivial (inner) automorphism unless G=Z(G).

Now, notice that, by definition, Z(G)=G if and only if G is abelian; so we have reduced to the abelian case.

If G is abelian then g\mapsto -g is an automorphism, and it is non-trivial unless g=-g for all g\in G. But g=-g if and only if the order of g divdes two. So we have now reduced to the case in which 2g=0 for all g\in G.

In this case, G is a vector space over the field \mathbb{Z}/2. As

|G| is equal to 2 raised to the power of the \mathbb{Z}/2-dimension of G,

the hypothesis that |G|>2 implies that \mathrm{dim}_{\mathbb{Z/2}} G>1. But now we can write down lots of linear automorphisms of G. For instance, you could fix any basis g_1,g_2,\ldots and take the automorphism g_1\mapsto g_2, g_2\mapsto g_1 and g_i\mapsto g_i for every i>2.

**Attribution***Source : Link , Question Author : Community , Answer Author : massy255*