4
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ You should always provide a link to the MSE post. $\endgroup$ Oct 23, 2015 at 12:28
  • $\begingroup$ My apologize! Link added $\endgroup$
    – Jimmy1990
    Oct 23, 2015 at 12:30
  • $\begingroup$ I didn't see your MSE post, because you didn't give it the group-theory tag. $\endgroup$
    – Derek Holt
    Oct 23, 2015 at 13:36
  • $\begingroup$ Thanks for the answer, it works! Yes sorry it was silly not to put the group theory tag. $\endgroup$
    – Jimmy1990
    Oct 23, 2015 at 13:43

1 Answer 1

8
$\begingroup$

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 $2$,

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.