Determining the automorphism group of a disconnected graph There is this know formula for determining the automorphism group of a graph $G$: let the connected components of $G$ consist of $n_1$ copies of $G_1$, $\dots$, $n_r$ copies of $G_r$, where $G_1, \dots, G_r$ are pairwise non-isomorphic. Then $${\rm Aut}(G) = ({\rm Aut}(G_1) \wr S_{n_1}) \times \dots \times ({\rm Aut}(G_r) \wr S_{n_r}).$$
I know intuitively what the formula does, but I am not able to prove the formula formally, probably because I don't understand the definition of the wreath product properly.
I would also appreciate if someone would describe the automorphism group of a graph which is a disjoint union of three paths of length one (or any other simple example).
If you are aware of a text or a website, where this is explained, it would be nice if you gave me a link, I wasn't able to find anything.
Thank you for your help!
 A: Consider two simpler cases to understand first.  
If $G_1$ and $G_2$ are the two connected, nonisomorphic components of $G$, then every automorphisms of $G$ is obtained by effecting an automorphism of $G_1$ along with an automorphism of $G_2$.  It is thus clear that $Aut(G)$ is the direct product $Aut(G_1) \times Aut(G_2)$.  
Now suppose $G$ consists of $n_1$ copies of a connected graph $G_1$.  One kind of automorphism of $G$ is obtained by effecting automorphisms of the individual copies. Thus, $G$ has at least $|Aut(G_1)|^{n_1}$ automorphisms.  In addition, the copies can also be permuted amongst themselves.  Thus, the number of automorphisms of the graph $G$ is $|Aut(G_1)|^{n_1}  (n_1)!$.  This group is called the wreath product of $S_{n_1}$ with $Aut(G_1)$.  This result is in [Frucht, On the groups of repeated graphs. Bull. Amer. Math. Soc, 55 (1949), 418-420.]  As a special case, consider the octahedron graph which is formed from the convex hull of the six points $\{(\pm 1, 0, 0), (0, \pm 1, 0), (0,0,\pm 1)\}$. It's automorphism group is identical to that of its complement $3K_2$ (the graph you mention), and its automorphism group is $S_3[S_2]$ by Frucht's theorem.  This group has $2^3 \times 3!=48$ automorphisms, which is intuitively clear, since every automorphism is obtained by permuting the endpoints of an edge (giving $2^3$ automorphisms) as well as by permuting the 3 edges amongst themselves (this gives a factor of $3!$).
Putting together these two simpler cases, we obtain the more general result you mention where $G$ consists of $n_i$ copies of some connected graph $G_i$ and where the $G_i$'s need to be pairwise nonisomorphic.
The wreath product and this general result are described in Harary's Graph Theory text.
