2
$\begingroup$

The answer on this question states that $\text{Aut}(G_1 \oplus G_2)$ a subgroup of $\text{Aut}(G)$, where $G$ has been decomposed in the invariant factors $G_1 \oplus G_2 \oplus \ldots \oplus G_n$. I do not see why this is true. Any hints would be appreciated.

EDIT: Would the reason be that there is an injective group homomorphism from $\text{Aut}(H)\oplus \text{Aut}(K)$ to $\text{Aut}(H \oplus K)$ (for groups $H, K$)?

$\endgroup$
  • $\begingroup$ Because any automorphism of G that fixes $G_1 \oplus G_2$ must also fix $G$. $\endgroup$ – Kenny Lau Sep 18 '17 at 14:21
  • $\begingroup$ @KennyLau Thanks for your answer. I'll have to think about this one, because it is not immediately obvious to me. $\endgroup$ – Student Sep 18 '17 at 14:26
3
$\begingroup$

More precisely,

$\text{Aut}(G_1 \oplus G_2)$ can be embedded into $\text{Aut}(G)$

Indeed, if $G=G_1 \oplus G_2 \oplus \ldots \oplus G_n$ and $\phi \in \text{Aut}(G_1 \oplus G_2)$, then $\phi \times id_3 \times \cdots \times id_n \in \text{Aut}(G)$.

$\endgroup$
  • $\begingroup$ thank you very much, I did not thought about this embedding when I wrote my question. $\endgroup$ – Student Sep 18 '17 at 14:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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