Let $G$ be an abelian (not elementary) finite $p$-group. In what conditions the automorphism group of $G$ is solvable?
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
|
0
1
|
|
||||||||
|
|
1
|
I believe that for $p\ge5$, these are the direct products of cyclic groups of pairwise distinct orders, for if $C_{p^e}\times C_{p^e}$ is a direct factor of $G$, then $GL(2,p)$ is a homomorphic image of a subgroup of $A=\text{Aut}(G)$. Similarly for $p=2$ or $3$, I expect the groups $G$ you are looking for are those where no order in the direct product appears more than $2$ times. In order to show that these groups indeed have a solvable automorphism group, you probably may apply induction: Let $N$ be the subgroup of $G$ generated by the $p$-th powers in $G$. Then $N$ has the same shape as above (with the order of each direct factor divided by $p$). So $\text{Aut}(N)$ is solvable by induction. As $C_A(N)$ is the kernel of the restriction of $A$ to $N$, all what remain to show is that $C_A(N)$ is solvable too. Alternatively, one could try to use the induction hypothesis for $G/K$, where $K$ is generated by the elements of order $p$. Note that $N$ is (non-canonically) isomorphic to $G/K$. |
|||
|
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
|
1
|
The Frattini quotient G/Phi(G) is the maximal elementary abelian quotient of G; the group of automorphisms of G which act trivially on G/Phi(G) is a p-group, and so Aut(G) is solvable if and only if its image in the linear group Aut(G/Phi(G)) is; that's at least a partial answer. Also, this holds for general pro-p groups, not only for finite abelian p-groups. |
|||
|
