[proofplan] We label the vertices of $K_{2n}$ by one distinguished point $\infty$ together with the cyclic group $\mathbb{Z}/(2n-1)\mathbb{Z}$. For each group element $t$, we construct a matching $F_t$ by pairing $\infty$ with $t$ and pairing the remaining finite vertices symmetrically around $t$. The odd order of the cyclic group makes division by $2$ possible, which lets us prove that every edge between two finite vertices appears in exactly one such matching, while edges incident to $\infty$ are accounted for directly. [/proofplan] [step:Label the vertices by an odd cyclic group plus one distinguished vertex] Fix an integer $n \geq 1$ and set $m := 2n-1$. Let \begin{align*} G := \mathbb{Z}/m\mathbb{Z} \end{align*} be the additive cyclic group of order $m$. Let $\infty$ be a symbol not belonging to $G$, and define the vertex set \begin{align*} V := G \cup \{\infty\}. \end{align*} Then $|V| = m+1 = 2n$, so the complete graph on $V$ is a copy of $K_{2n}$. Its edge set is \begin{align*} E := \bigl\{\{x,y\} : x,y \in V,\ x \neq y\bigr\}. \end{align*} For each $t \in G$, define a set of edges $F_t \subset E$ by \begin{align*} F_t := \bigl\{\{\infty,t\}\bigr\} \cup \bigl\{\{t+\overline{i},t-\overline{i}\} : i \in \{1,\dots,n-1\}\bigr\}, \end{align*} where $\overline{i}$ denotes the residue class of the integer $i$ in $G$. [/step] [step:Show that each constructed set is a one-factor] We prove that, for each $t \in G$, the set $F_t$ is a matching covering every vertex of $V$ exactly once. First, $\{\infty,t\}$ covers $\infty$ and $t$. For the finite vertices different from $t$, consider the subsets \begin{align*} S_i := \{\overline{i},-\overline{i}\} \subset G \end{align*} for $i \in \{1,\dots,n-1\}$. These subsets partition $G \setminus \{\overline{0}\}$. Indeed, every nonzero element of $G$ has a representative $r \in \{1,\dots,2n-2\}$; if $1 \leq r \leq n-1$, then $\overline{r} \in S_r$, while if $n \leq r \leq 2n-2$, then $j := m-r$ belongs to $\{1,\dots,n-1\}$ and $\overline{r} = -\overline{j} \in S_j$. The subsets are pairwise disjoint because $S_i \cap S_j \neq \varnothing$ implies either $\overline{i}=\overline{j}$ or $\overline{i}=-\overline{j}$; the first gives $i=j$, and the second would make $m$ divide $i+j$, impossible for distinct $i,j \in \{1,\dots,n-1\}$ since $2 \leq i+j \leq 2n-2=m-1$. Translation by $t$ is a bijection $G \to G$, so the pairs \begin{align*} \{t+\overline{i},t-\overline{i}\}, \qquad i \in \{1,\dots,n-1\}, \end{align*} partition $G \setminus \{t\}$. Hence the $n-1$ finite edges in $F_t$, together with $\{\infty,t\}$, are pairwise vertex-disjoint and cover all vertices in $V$. Therefore $F_t$ is a one-factor. [guided] Fix $t \in G$. The goal is to check that $F_t$ is not merely a collection of $n$ edges, but a one-factor: every vertex must occur in exactly one of those edges. The edge $\{\infty,t\}$ accounts for the distinguished vertex $\infty$ and the finite vertex $t$. It remains to show that the edges \begin{align*} \{t+\overline{i},t-\overline{i}\}, \qquad i \in \{1,\dots,n-1\}, \end{align*} cover each element of $G \setminus \{t\}$ exactly once. We first analyze the same question before translating by $t$. For each $i \in \{1,\dots,n-1\}$, define \begin{align*} S_i := \{\overline{i},-\overline{i}\} \subset G. \end{align*} These sets partition $G \setminus \{\overline{0}\}$. To see coverage, take any nonzero element $a \in G$. Choose its representative $r \in \{1,\dots,2n-2\}$. If $1 \leq r \leq n-1$, then $a=\overline{r}\in S_r$. If $n \leq r \leq 2n-2$, define $j:=m-r$. Since $m=2n-1$, this gives $1 \leq j \leq n-1$, and \begin{align*} a=\overline{r}=\overline{m-j}=-\overline{j}\in S_j. \end{align*} Thus every nonzero element appears in at least one $S_i$. Now we check uniqueness. Suppose $S_i \cap S_j \neq \varnothing$ for $i,j \in \{1,\dots,n-1\}$. Then one of the congruences \begin{align*} \overline{i}=\overline{j} \qquad\text{or}\qquad \overline{i}=-\overline{j} \end{align*} must hold. The first congruence gives $i=j$ because both integers lie between $1$ and $n-1$. The second congruence says that $m$ divides $i+j$. But \begin{align*} 2 \leq i+j \leq 2n-2=m-1, \end{align*} so $m$ cannot divide $i+j$. Therefore the subsets $S_i$ are pairwise disjoint. Finally, the map $G \to G$ given by $a \mapsto t+a$ is a bijection, so translating the partition \begin{align*} G \setminus \{\overline{0}\} = \bigsqcup_{i=1}^{n-1} S_i \end{align*} by $t$ gives the partition \begin{align*} G \setminus \{t\} = \bigsqcup_{i=1}^{n-1} \{t+\overline{i},t-\overline{i}\}. \end{align*} Hence the finite edges in $F_t$ cover every finite vertex except $t$ exactly once, while $\{\infty,t\}$ covers $\infty$ and $t$. Thus $F_t$ is a one-factor. [/guided] [/step] [step:Show that every edge incident to $\infty$ appears exactly once] Let $a \in G$. The edge $\{\infty,a\}$ belongs to $F_a$ by definition. If $\{\infty,a\} \in F_t$ for some $t \in G$, then the only edge of $F_t$ incident to $\infty$ is $\{\infty,t\}$, so $t=a$. Therefore every edge incident to $\infty$ appears in exactly one of the sets $F_t$. [/step] [step:Show that every edge between finite vertices appears exactly once] Let $a,b \in G$ with $a \neq b$. Since $m=2n-1$ is odd, $\gcd(2,m)=1$, so multiplication by $\overline{2}$ is a bijection $G \to G$. Hence there is a unique $t \in G$ satisfying \begin{align*} \overline{2}t=a+b. \end{align*} For this $t$, define \begin{align*} d := a-t \in G. \end{align*} Then \begin{align*} b-t = b-\frac{a+b}{2} = -\frac{a-b}{2} = -(a-t) = -d \end{align*} in $G$, where the notation $\frac{a+b}{2}$ denotes the unique element $t \in G$ with $\overline{2}t=a+b$. Also $d \neq \overline{0}$, because $d=\overline{0}$ would imply $a=t$ and then $b=t$ from $b-t=-d$, contradicting $a \neq b$. By the partition \begin{align*} G \setminus \{\overline{0}\} = \bigsqcup_{i=1}^{n-1} \{\overline{i},-\overline{i}\}, \end{align*} there is a unique $i \in \{1,\dots,n-1\}$ such that \begin{align*} \{d,-d\}=\{\overline{i},-\overline{i}\}. \end{align*} Therefore \begin{align*} \{a,b\} = \{t+d,t-d\} = \{t+\overline{i},t-\overline{i}\}, \end{align*} so $\{a,b\} \in F_t$. It remains to prove uniqueness of the factor. Suppose $\{a,b\} \in F_s$ for some $s \in G$. Then there exists $j \in \{1,\dots,n-1\}$ such that \begin{align*} \{a,b\}=\{s+\overline{j},s-\overline{j}\}. \end{align*} Adding the two endpoints in $G$ gives \begin{align*} a+b=(s+\overline{j})+(s-\overline{j})=\overline{2}s. \end{align*} Since multiplication by $\overline{2}$ is injective on $G$, this forces $s=t$. Thus every edge between two finite vertices appears in exactly one factor. [/step] [step:Conclude that the constructed one-factors partition the edge set] The preceding two steps show that every edge of the complete graph on $V$ lies in exactly one set $F_t$ with $t \in G$. Since each $F_t$ is a one-factor, the family \begin{align*} \{F_t : t \in G\} \end{align*} is a one-factorization of $K_{2n}$. Because $|G|=2n-1$, this one-factorization consists of exactly $2n-1$ one-factors, as required. [/step]