Altough this sounds as a very basic question, I didn't receive any answer on stack exchange and by people more knowledgeable than me


Take $p$ a prime number and $P$ an abelian finite $p$-group. Let $A,A'$ be subgroup of $P$ such that $A \simeq A'$ and $P/A \simeq P/A'$ as groups. Can I conclude that there is $\phi \in Aut(P)$ such that $\phi(A)=A'$?


Thanks in advance!


p.s:<https://math.stackexchange.com/questions/1476641/obstruction-to-be-conjugated-by-an-automorphism-for-subgroups-of-an-abelian-grou>, here the MSE link



Obstruction for two subgroups to be conjugated by an automorphism
I will use additive notation, and let $Z\_k = \{0,1,\ldots,k-1\}$ with addition mod $k$.


Let $P = Z\_2 \oplus Z\_4 \oplus Z\_8$, and let $A$ and $B$ be the subgroups
$$A = \langle (1,1,0),(0,0,4) \rangle,$$ and $$B = \langle (1,0,0,), (0,2,2) \rangle.$$


Then $A \cong B \cong G/A \cong G/B \cong Z\_2 \oplus Z\_4$.


The element $(0,0,4)$ of $A$ is a multiple of $4$ in $G$, but is not a multiple of $2$ in $A$ (it generates the direct summand of $A$ of order $2$).


But the only nonzero element of $B$ that is a multiple of $4$ in $G$ is $(0,0,4)$, and this is a multiple of $2$ in $B$. So no automorphism of $G$ can map $A$ to $B$.


I expect the same construction will work with any prime $p$ in place of $2$,


