Why is $\text{Aut}(G_1 \oplus G_2)$ a subgroup of $\text{Aut}(G)$?

Question

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 $H, K$)?

Answer 1

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