Let $G(V,E)$ be a graph. I am searching for graphs with only **disjoint** [perfect matchings](https://en.wikipedia.org/wiki/Matching_(graph_theory)) (i.e. every edge only appears in at most one of the perfect matchings).


Examples:


* [Cyclic graph](https://en.wikipedia.org/wiki/Cycle_graph) $C\_n$ with even $n$, with $m=2$ disjoint perfect matchings.
* [Complete graph](https://en.wikipedia.org/wiki/Complete_graph) $K\_4$, with $m=3$ disjoint perfect matchings.


I have three questions:


1. How are such graphs called?
2. Are there other examples than $C\_n$ and $K\_4$?
3. What is the maximum number $m$ of perfect matchings, if the graph has only completly disjoint perfect matchings?


For question 3, it seems to me that $K\_4$ with $m=3$ different, disjoint perfect matchings is the optimum, but I have no proof for that.


Every hint to an answer or to relevant literature would be very much appreciated!


**Edit:** I am interested in undirected graphs only for the moment.


**Edit2:** The answer to this question I have used in a recent article in [Physical Review Letters](https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.119.240403), where I cite this MO question as reference [24]. See Figure 2 for a detailed variant of the application of Ilya's answer. Thanks Ilya!



Graphs with only disjoint perfect matchings
$m=3$ is indeed the maximum, and $K\_4$ is the only example for this value of $m$.


Two perfect matchings form a disjoint union of cycles. If there is more than one cycle, then you may swap one of them, obtaining a third matching on the same edges. So any two of the $m$ matchings form a Hamiltonian cycle.


Assume that $m\geq3$; consider a Hamiltonian cycle $v\_1,\dots,v\_{2n}$ formed by the first two matchings, and check how the third one looks like.


If some its edge $(v\_i,v\_j)$ subtends an arc of odd length (i.e. if $i-j$ is odd), then we may split the vertices outside this arc into pairs, and split the cycle $v\_i,\dots,v\_j$ into edges including $(v\_i,v\_j)$, obtaining a matching intersecting the third one but not coinciding with it. This should not be possible; thus each subtended arc is even.


Now take an edge $(v\_i,v\_j)$ subtending minimal such arc, and consider an edge $(v\_{i+1},v\_k)$ of the third matching (going otside this arc). Now split the cycle $v\_i,v\_j,v\_{j+1},\dots,v\_k,v\_{i+1}$ into edges containing $(v\_i,v\_j)$, and split all the remaining vertices into edges of the Hamiltonian cycle (it is possible, according to the parity). If $2n>4$, you again get a fourth matching sharing edges with the third ones bot diferent from it. 


Thus $2n\leq 4$ and $m\leq 3$.


