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A finite abelian $p$-group $H$ is homogenous when it is the direct sum of cyclic groups of the same order $p^r$, i.e. $H \cong \big(\mathbb{Z}/p^{r}\mathbb{Z}\big)^{e}$. Every finite abelian $p$-group $G$ can be decomposed into homogenous factors $$ G = H_{d_1} \oplus H_{d_2} \oplus \cdots \oplus H_{d_k} $$ where $H_{d_i}$ has exponent $p^{d_i}$ and $d_1 > d_2 > \cdots > d_k > 0$. While this decomposition is unique up to isomorphism, unless $k = 1$ there are many distinct choices for the homogenous subgroups themselves.

Now suppose $\varphi$ is an automorphism of $G$ of order coprime to $p$. Must there be a homogenous decomposition that is invariant under $\varphi$?

If it helps, you can assume that the order of $\varphi$ is prime. For the intended application, all I really need is an invariant homogenous subgroup of the largest possible exponent.

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It is true that if $\phi$ is automorphism of order prime to $p$ of a finite Abelian $p$-group $A$, then $A$ is a direct sum of indecomposable $\phi$-invariant subgroups ( that is, no summand is a direct sum of two non-zero $\phi$-invariant subgroups). It is also true that each of these indecomposable summands is homocyclic, that is,a direct sum of cyclic subgroups of the same order, ( what you call homogeneous). Hence the answer to your question is yes. This can be found in D.Gorenstein's book "Finite Groups" (the original results may be due to P. Hall, I am not sure).

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    $\begingroup$ Thanks! The precise reference in Gorenstein is Theorem 5.2.2 on page 176. $\endgroup$ Commented Jun 19 at 20:01
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    $\begingroup$ Glad you found the precise reference. $\endgroup$ Commented Jun 19 at 20:22
  • $\begingroup$ I guess it's true for any action of a finite group of order coprime to $p$? $\endgroup$
    – YCor
    Commented Jun 20 at 7:00
  • $\begingroup$ @YCor: Indeed that is the case( and is done in Gorenstein). $\endgroup$ Commented Jun 20 at 7:34

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