Are there any vertex-transitive digraphs with distinct in and out degree

Question

It seems to me like a question that must have been answered somewhere before, but I haven't been able to find anything on this.

Take a vertex-transitive digraph $\Gamma$, that is a directed graph, such that there exists a group $G$ acting on $\Gamma$ and with transitive induced action on vertices of $\Gamma$. If $\Gamma$ is undirected, it is clear it is regular: that is the degree of every vertex is the same. If $\Gamma$ is directed, it has a constant in degree and out degree, but these two degrees can still be distinct.

So my question is: Is there a vertex-transitive digraph, such that the in degree and the out degree of a vertex are distinct?

For Cayley digraphs, it is clear that they are the same, since to every edge $(g,gs)$ going out of $g$, we can associate the edge $(gs^{-1},g)$ going into $g$ and that gives us a bijection.

I have also shown that you can't have a situation where the out degree is 2, but the in degree is 1. Simply take $x$ a vertex, $(x,y)$ ,$(x,y')$ and $(y'',x)$ three edges. Then write $y=g_1\cdot x$ and $y'=g_2\cdot x$, with $g_1$ of order $n$ and $g_2$ of order $m$. We then end up with cycles $(x,g_1 \cdot x),(g_1 \cdot x,g_1^2 \cdot x) \cdots (g_1^{n-1} \cdot x,x)$ and $(x,g_2 \cdot x), \cdots (g_2^{m-1} \cdot x,x)$. Since the in degree of $x$ is $1$, then we get that $g_1^{n-1} \cdot x =g_2\cdot x=y''$. As $g_1\cdot x \neq g_2 \cdot x$, there has to be some vertex $x'$ where the two directed paths join. This vertex $x'$ however then has the in degree at least 2 and so $x$ has in degree at least 2 as well, which is a contradiction.

Unfortunately I don't think this approach can be generalized, since if we had three distinct cycles, they could end up joining at different vertices.

Answer 1

The answer turns out to be actually much simpler than I thought. If the out degree and in degree are both constant, they have to be equal.

The reason for it is the result known as the handshaking lemma. This lemma states that the sum of all out degrees is equal to the sum of all in degrees and equal to the total number of edges in the graph.