TITLE: k-regular simple graph without 1-factor
QUESTION [8 upvotes]: Here's what I'm reading: every regular bipartite graph has a 1-factor.
But I understand that not every regular graph has a 1-factor.
So, I was thinking if it's possible to find a $k$-regular simple graph without 1-factor for any fixed $k$?

REPLY [13 votes]: If $k$ is even, then $K_{k+1}$ is $k$ regular with $k+1$ (odd) vertices, and hence has no $1$-factor (as a perfect matching requires an even number of vertices).
If $k$ is odd, consider the following graph. Start at an initial vertex, then branch out in $k$ directions. From each of those new vertices, branch out $k-1$many times. For each collection of $k-1$ vertices, draw another $k-1$ vertices opposite it, and connect each vertex on one side with every vertex on the other. Lastly, since $k-1$ is even, in the opposite $k-1$ collection, pair up each vertex with one other vertex. Note that: the initial vertex has $k$neighbors, each ensuing vertex has $k$ neighbors. Every vertex in the first $k-1$ collection is connected to the one that preceeded it and $k-1$ other ones in the matching $k-1$ set, so has $k$ neighbors. Lastly, each vertex in the $k-1$ matching set is connected to every vertex in the initial $k-1$ set, and one more. Thus our graph is $k$ regular. However, if we delete the initial vertex, each component has 
$$(k-1) + (k-1) + 1 = 2k - 1$$
 vertices, which is an odd number, and so we get $k$ odd components. By Tutte's theorem, this means that our graph could not have had a $1$-factor.
Here is an illustrative drawing in the $5$ case: