A question on automorphisms of finite abelian groups Which are the finite groups $(G,\cdot)$ with the following property: for every $f \in Aut(G)$, there are $g,h\in Aut(G)$ such that $f(x)=g(x)\cdot h(x), \forall x\in G?$   
I already have verified that:
 (i) such a group is abelian;
 (ii) all finite abelian groups of odd order have this property;
 (iii) a finite abelian 2-group satisfying this property must be of type $$(*)\hspace{2mm} \mathbb{Z}_{2^{\alpha_1}}\times \mathbb{Z}_{2^{\alpha_2}}\times \cdots\times \times \mathbb{Z}_{2^{\alpha_k}}$$where each $\alpha_i$ occurs at least twice.
In my opinion, the above class consists of all direct products $G_1\times G_2$, where $G_1$ is an abelian 2-group of type $(*)$ and $G_2$ is an abelian group of odd order, but I failed to prove it. 
 A: Here is an outline for odd order abelian $p$-groups:
The main point is that every non-zero element in a field with at least three elements is a sum of two non-zero elements. Using this, you can show that irreducible matrix in $GL_n(\mathbf Z/p\mathbf Z)$ can be written as a sum of two non-singular matrices, for such an irreducible matrix corresponds to multiplication by a primitive element of the field of order $p^n$.
Now, using the Jordan canonical form matrices over a finite field (see Theorem A.22 in my notes) you can show that every non-singular matrix over $\mathbf Z/p\mathbf Z$ is a sum of two non-singular matrices.
Finally, the endomorphism algebra of a finite abelian $p$-group modulo is radical is a product of matrix groups over $\mathbf Z/p\mathbf Z$ (see section 6 of this paper). So any invertible element of the endomorphism algebra can be written as a sum of two invertible elements modulo the radical of this ring. But adding something in the radical does not affect invertibility.
