#Elements that are automorphic images in finite abelian groups
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rrc08
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Twenty
Sorry I think this is false (this is only a necessary condition)
I think it's good, maybe it works by induction on card(G)
I will take take a look later
I think i found something :
If $G = \langle a_1, ..., a_n \rangle$ then $\forall x\in G, \exists i_k\in\mathbb{N}, x = a_1^{i_1}...a_n^{i_n}$ because G is abelian.
from that we can deduce the following condition :
Let $x, y\in G$ so $x = a_1^{i_1}...a_n^{i_n}$ and $y = a_1^{j_1}...a_n^{j_n}$,
$$\exists \phi\in Aut(G), \phi(x)=y \Longleftrightarrow \exists \sigma \in S_n, \forall k \in \left{1, ..., n\right}, \left{
\begin{array}{ll}
od(a_k) = od(a_{\sigma(k)})\
i_k = j_{\sigma(k)}
\end{array}
\right.$$
Ævan
where $$od(x)= min\left{p\in\mathbb{N}^*, x^p = e\right}$$
Ævan
I can write the proof if you want