$X, X'$ be two connected graphs and let $Z$ be there $Disjoint$ union. $D=Aut(X) \times Aut(X')$ . How does $|D|< |Aut(Z)|$ if and only if $X \simeq X'$ ? An example would be helpful.
In Non-Isomorphic case, should $D$ be lesser ?
Context:-
See the red-boxed line.
--From "Lecture Notes in Computer Science" by Christoph M. Hoffmann.
It is easy to see that $D \leq Aut(Z)$. But, I could not comprehend the statement, " $D$ is a proper subgroup iff $X$ and $X'$ are isomorphic." Why $D$ is a proper subgroup ?
Thanks.
The real key is that, since $X$ and $X^{\prime}$ are disjoint connected components of $Z$, any automorphism of $Z$ either stabilizes both $X$ and $X^{\prime}$ (in which case it is in $\mathrm{Aut}(X)\times \mathrm{Aut}(X^{\prime})$), or else it switches them (which can only happen when $X\simeq X^{\prime}$).
If $X\simeq X^{\prime}$ then there is some $\tau \in \mathrm{Aut}(Z)$ sending the vertices of $X$ to the vertices of $X^{\prime}$, that means $\tau \not\in D$ and so $D$ is a proper subgroup of $\mathrm{Aut}(Z)$. If $X\not\simeq X^{\prime}$ then there is no such $\tau$ and so $D=\mathrm{Aut}(Z)$ (i.e. $D$ is not a proper subgroup).