I am stuck on the following two questions on graph automorphisms?
My questions are the following :
1.How is $\def\Aut{\operatorname{Aut}} \Aut(G_1\cup G_2)$ related to $\Aut(G_1), \Aut(G_2)$?
2.How is $\Aut(G_1+ G_2)$ related to $\Aut(G_1), \Aut(G_2)$?
Here $G_1\cup G_2$ is the disjoint union of $G_1,G_2$ and $G_1+G_2$ is the join of $G_1,G_2$.
If someone can help, I will be grateful.
My try:
For the 1st question, since $G_1\cup G_2$ has two disjoint components so I guess $\Aut(G_1\cup G_2)=\Aut(G_1)\times \Aut(G_2)$, where $\times$ denotes the direct product of two groups.
For the 2nd question, $\Aut(G_1+ G_2)=\Aut(\overline{G_1+ G_2})=\Aut(\overline{G_1}\cup \overline{G_2})$
So if I can answer the 1st question, then the 2nd question will be answered.
Can someone please help me verify the 1st question?
Very brief hints.
If graphs $G_1$ and $G_2$ are connected and non-isomorphic, then indeed $$\operatorname{Aut}(G_1\cup G_2)\cong\operatorname{Aut}(G_1)\times \operatorname{Aut}(G_2).$$ If connected graphs $G_1$ and $G_2$ are isomorphic, then $$\operatorname{Aut}(G_1\cup G_2)\cong(\operatorname{Aut}(G_1)\times \operatorname{Aut}(G_2))\rtimes C_2.$$ I think that if graphs $G_1$ and $G_2$ are disconnected, then there is no simple formula.