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 $H, K$)?

  • $\begingroup$ 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

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 $\phi \times id_3 \times \cdots \times id_n \in \text{Aut}(G)$.

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

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