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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.

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  • $\begingroup$ I'm not clear about what you can't prove? Is it whether $(*)$ is a sufficient condition for abelian 2-groups? In any case, for finite abelian groups I think the question reduces easily to the Sylow $p$-subgroups. $\endgroup$
    – Jeremy Rickard
    Commented Mar 16, 2013 at 13:40
  • $\begingroup$ Curious why you need this. $\endgroup$
    – Amritanshu Prasad
    Commented Mar 16, 2013 at 13:58
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    $\begingroup$ It's not true. You need to strengthen $(\ast)$ to say that each positive integer that occurs as an $\alpha_j$ occurs at least twice. $\endgroup$
    – Tom Goodwillie
    Commented Mar 16, 2013 at 15:44
  • $\begingroup$ Thank you very much. This is the strong form of a problem from a mathematics contest. I think that Tom is right: in a finite abelian 2-group $G$ satisfying the above property each $\alpha_i$ occurs at least twice. My problem is whether the converse is true (i.e. if a finite abelian 2-group $G$ such that each $\alpha_i$ occurs at least twice satisfies this property). $\endgroup$
    – Marius Tarnauceanu
    Commented Mar 18, 2013 at 6:21
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    $\begingroup$ Yes. It's enough if $G^k$ has the property when $G$ is cyclic. $G^k$ always has the property if $G$ is abelian, because there exists an invertible $k\times k$ matrix $M$ over $\mathbb Z$ such that $1-M$ is invertible. $\endgroup$
    – Tom Goodwillie
    Commented Mar 18, 2013 at 14:07

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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.

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  • $\begingroup$ Thank you very much. The real problem is for finite abelian 2-groups. $\endgroup$
    – Marius Tarnauceanu
    Commented Mar 18, 2013 at 6:34
  • $\begingroup$ The reduction-to-matrix-groups over $\mathbf Z/p\mathbf Z$ idea works for $2$-groups as well; so you need to figure out when an invertible matrix in $\mathbf Z/2\mathbf Z$ is a sum of two invertible matrices. Tom Goodwillie suggests that this holds for matrices that are at least $2\times 2$. $\endgroup$
    – Amritanshu Prasad
    Commented Mar 18, 2013 at 11:00

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