I am currently writing a proof in which I need to use the fact that if a tree has no involutions, its automorphism group is trivial (ie, if a tree has any non-trivial automorphisms, then it has at least one automorphism of order 2). Equivalently, the order of the automorphism group of a tree is either equal to 1 or it is even.


I can prove this fact directly myself, but I assume I'm not the first person to have noticed it - does anyone know where I can find a published proof - I've tried a few graph theory text books, but I'm not really sure where to look in the literature.



Involution-free Trees are Asymmetric: Reference request
Look for "Automorphisms of Graphs" by P. J. Cameron (google finds the pdf), and look at the result of Polya on page 8 -- it states that the class of automorphism groups of trees is the smallest class containing the trivial group and closed under the operations of "direct product" and "wreath product with the symmetric group $S\_n$ for each $n>1.$" Since symmetric groups have even order, your result follows.


