I need to show every automorphism of a tree fixes an edge or a vertex. I am aware of the classic inductive proofs of this fact, but I wanted to try this algebraically. I am fairly new to graph theory so I am not sure if this is a valid method.
Consider Tree $T = (V,E)$. Then an Automorphism of $T$ can be represented by permutation cycles of the vertices or a permutation cycle of edges. Since a tree has $v$ vertices and $v-1$ edges $\implies$ either the permutation of edges or the permutation of vertices must be even depending on $v$. But of course, every even permutation is a product of $3-cycles$.
It can easily be checked by hand or by a computer that any permutation of a $3-cycle$ of vertices of $T$ fixes a vertex or edge. Similarly, a $3-cycle$ of edges of $T$ fixes an edge or a vertex. So, since any automorphism of $T$ can be written as a product of $3-cycles$, then $T$ must fix an edge or vertex.
Any help showing if this is valid, or why it is invalid would be greatly appreciated.
It is not fair to conclude that because one of $|V|$ or $|E|$ is even, the permutation of either $V$ or $E$ must be even.
For example, consider the tree
and consider the automorphism that swaps $a,b$ (leaving $c,d,e$ fixed). This is an odd permutation of the vertices (because two vertices are swapped), and also an odd permutation of the edges (because the edges $ae, be$ are swapped, leaving the other edges alone).
Even a permutation of an even set with no fixed points can be odd. For example, we can permute $\{1,2,3,4\}$ by $(1\;2)\;(3\;4)$ (an even permutation) or by $(1\;2\;3\;4)$ (an odd permutation).