Let $G$ be a finite $p$-group. Suppose that $H,K$ are subgroups of $G$, such that $H \leq K$. If $\operatorname{Aut}(H)$ denotes the group of automorphism of $H$, then is there any relationship between $\operatorname{Aut}(H)$ and $\operatorname{Aut}(K)$? That is, may happen that $\operatorname{Aut}(H) \leq \operatorname{Aut}(K)?$ or $\operatorname{Aut}(K) \leq \operatorname{Aut}(H)?$ Or, in general, there is no relationship between these groups?
$\begingroup$
$\endgroup$
7
-
1$\begingroup$ What is the role of $ G $? $\endgroup$– Tobias KildetoftAug 8, 2013 at 14:24
-
$\begingroup$ $G$ is a any finite $p$-group $\endgroup$– user59969Aug 8, 2013 at 14:42
-
2$\begingroup$ Yes, but what role does it play. No part of the question seems to mention it. $\endgroup$– Tobias KildetoftAug 8, 2013 at 14:43
-
$\begingroup$ The coice of $G$ may play a role for the answer. For example, take $G=\mathbb{Z}/p\mathbb{Z}$. $\endgroup$– Dietrich BurdeAug 8, 2013 at 15:21
-
1$\begingroup$ If, for example, $H$ is a direct factor of $K$, then there is a natural embedding of ${\rm Aut}(H)$ in ${\rm Aut}(K)$, but there is not much more that you can say in general. $\endgroup$– Derek HoltAug 8, 2013 at 15:32
|
Show 2 more comments
1 Answer
$\begingroup$
$\endgroup$
The dihedral group $D_8$ of order $8$ contains a copy of the Klein $4$-group $V \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Now $\operatorname{Aut}(D_8) \cong D_8$ and $\operatorname{Aut}(V) \cong S_3$. So in this case $\operatorname{Aut}(D_8) \not\leq \operatorname{Aut}(V)$ and $\operatorname{Aut}(V) \not\leq \operatorname{Aut}(D_8)$ (by Lagrange's theorem), even though $V \leq D_8$.
answered Aug 8, 2013 at 15:46