Multipartite Ramsey number $m_{j}(K_{m},nK_{2})$

For $n=1$ it is clear that $m_{j}(K_{j},K_{2})=1$, hence assume that $n=2$. By considering $K_{j}=K_{3}\cup(K_{j-3}\oplus 3K_{1})$ it can be check that $m_{j}(K_{j},2K_{2})\geq 2$. As $n=2$ and any part of $K=K_{j\times 2}$ has two vertices, it is easy to say that in any two coloring of the edges of $K$ say $(G^{1},G^{2})$, either $K_{j}\subseteq G^{1}$ or $2K_{2}\subseteq G^{2}$. Hence $m_{j}(K_{j},nK_{2})=n$ for each $j\geq 2$ and $n=1,2$. ∎

For $n=1,2$, the proof is complete by Lemma 4, hence assume that $n\geq 3$. Let $K=K_{j\times(n-1)}$ and consider $2$-edge-coloring $(G^{1},G^{2})$ of $K$, where $G^{1}\cong K_{(j-1)\times(n-1)}$ and $\overline{G^{1}}=G^{2}$. It is easy to say that $G^{2}$ is a $2$-partite graph with $n-1$ vertices in the first part and $(j-1)(n-1)$ vertices in the second part. That is $G^{2}\cong K_{n-1,(j-1)(n-1)}$. By definition $G^{i}$, it can be check that $K_{j}\nsubseteq G^{1}$ and $nK_{2}\nsubseteq G^{2}$, that is $K$ is 2-colorable to $(K_{j},nK_{2})$, which means that $m_{j}(K_{j},nK_{2})\geq n$ for each $j\geq 2$ and each $n\geq 1$.

We prove $m_{j}(K_{j},nK_{2})\leq n$ by induction on $n$. For $n=1,2$ theorem holds by Lemma 4. Assume that for each, $n^{\prime}\leq n-1$, the theorem holds. By contrary suppose that $K=K_{j\times n}$ with partite sets $X_{i}=\{x_{1}^{i},x_{2}^{i},\ldots,x_{n}^{i}\}$ for each $i\in[j]$ is $2$-colorable to $(K_{j},nK_{2})$, that is there exist 2-edge-coloring $(G^{1},G^{2})$ of $K_{j\times n}$, where $K_{j}\nsubseteq G^{1}$ and $nK_{2}\nsubseteq G^{2}$. Without loss of generality (w.l.g) assume that $e=x_{1}^{1}x_{1}^{2}\in E(G^{2})$. Now, set $X^{\prime}_{i}=X_{i}\setminus\{x_{1}^{i}\}$ for each $i\in[j]$, and consider $K^{\prime}=K_{j\times(n-1)}=G[X^{\prime}_{1},\ldots,X^{\prime}_{j}]$. Consider a 2-edge-coloring $(G^{\prime 1},G^{\prime 2})$ of $K^{\prime}$. As $m_{j}(K_{j},(n-1)K_{2})\leq n-1$, $K_{j}\nsubseteq G^{\prime 1}\subseteq G^{1}$, one can check that $(n-1)K_{2}\subseteq G^{\prime 2}\subseteq G^{2}$. Suppose that $M^{\prime}$ be a M-M in $G^{\prime 2}$. Therefore, $M=M^{\prime}\cup\{e\}$ be a matching of size $n$ in $G^{2}$, a contradiction. Which means that $m_{j}(K_{j},nK_{2})\leq n$, for each $n\geq 2$. Therefore $m_{j}(K_{j},nK_{2})=n$ for each $j\geq 2$ and each $n\geq 1$, and the proof is complete. ∎

Consider a $2$-edge-coloring $(G^{1},G^{2})$ of $K=K_{j\times(n-2)}$, where $n\in[j]$, $G^{2}\cong K_{3\times(n-2)}$ and $\overline{G^{2}}=G^{1}$. It is easy to say that $G^{1}$ is a $(j-2)$-partite graph with $n-2$ vertices in $(j-3)$ parts and $3(n-2)$ vertices in one parts. That is $G^{1}\cong K_{n-2,\ldots,n-2,3(n-2)}$. By definition $G^{1}$, one can check that $K_{j-1}\nsubseteq G^{1}$, also as $|V(G^{2})|=3(n-2)\leq 2n-1$ for each $n\in\{3,4,5\}$, hence it can be say that $nK_{2}\nsubseteq G^{2}$, that is $K$ is 2-colorable to $(K_{j-1},nK_{2})$, which means that $m_{j}(K_{j-1},nK_{2})\geq n-1$ for each $n\in\{3,4,5\}$.

Now for each $n\in\{3,4,5\}$, consider $K=K_{j\times(n-1)}$ with partite sets $X_{i}=\{x_{1}^{i},x_{2}^{i},\ldots,x_{n-1}^{i}\}$ for each $i\in[j]$. Suppose that $(G^{1},G^{2})$ be a 2-edge-coloring of $K$, where $K_{j}\nsubseteq G^{1}$. Assume that $n=3$ and w.l.g let that $e=x_{1}^{1}x_{2}^{1}\in E(G^{2})$. Now, set $X^{\prime}_{i}=X_{i}\setminus\{x_{1}^{i}\}$ for $i=1,2$ and $X^{\prime}_{i}=X_{i}$ for each $i\geq 3$, and consider $K^{\prime}=G[X^{\prime}_{1},\ldots,X^{\prime}_{j}]$. It can be check that $K^{\prime}\cong K_{2}\oplus K_{(j-2)\times 2}$, therefore $K_{(j-1)\times 2}\subseteq K^{\prime}$. Consider a 2-edge-coloring $(G^{\prime 1},G^{\prime 2})$ of $K^{\prime}$. As by Theorem 5 $m_{j-1}(K_{j-1},2K_{2})=2$, $K_{{j-1}\times 2}\subseteq K^{\prime}$, and $K_{j-1}\nsubseteq G^{\prime 1}\subseteq G^{1}$, one can check that $2K_{2}\subseteq G^{\prime 2}\subseteq G^{2}$. Suppose that $M^{\prime}$ be a M-M in $G^{\prime 2}$. Therefore, $M=M^{\prime}\cup\{e\}$ is a matching of size $3$ in $G^{2}$, which means that $m_{j}(K_{j-1},3K_{2})=2$. Since by Theorem 5 $m_{j}(K_{j},nK_{2})=n$ for each $n\geq 3$, then for $n=4,5$ by the proof is same as the case that $n=3$, we have $m_{j}(K_{j-1},nK_{2})=n-1$, so the proof is complete. ∎

Consider a $2$-edge-coloring $(G^{1},G^{2})$ of $K=K_{j\times(t-1)}$, where $n\geq 3$, $t=\lceil\frac{2n}{j+2-m}\rceil$, $G^{2}\cong K_{(j+2-m)\times(t-1)}$ and $\overline{G^{2}}=G^{1}$. It is easy to say that $G^{1}$ is a $m-1$-partite graph with $t-1$ vertices in $(m-2)$ parts and $(j+2-m)(t-1)$ vertices in the last parts. That is $G^{1}\cong K_{t-1,\ldots,t-1,(j+2-m)(t-1)}$. By definition $G^{1}$, one can check that $K_{m}\nsubseteq G^{1}$, also as $|V(G^{2})|=(j+2-m)(t-1)$ and $t=\lceil\frac{2n}{j+2-m}\rceil$, then one can say that $|V(G^{2})|\leq 2n-1$ for each $n\geq 3$ and each $j\geq m+1$, hence $nK_{2}\nsubseteq G^{2}$, that is $K$ is 2-colorable to $(K_{m},nK_{2})$, which means that $m_{j}(K_{m},nK_{2})\geq t$ for each $j\geq m+1,m\geq 4$ and each $n\geq 3$. ∎

By Theorem 7 the lower bound holds. So we most show that $m_{5}(K_{4},nK_{2})\leq\lceil\frac{2n}{3}\rceil$. We prove $m_{5}(K_{4},nK_{2})\leq\lceil\frac{2n}{3}\rceil$ by induction on $n$. For $n=3,4,5$ theorem holds by Theorem 6. Assume that for each, $6\leq n^{\prime}\leq n-1$, the theorem holds. By contrary suppose that $K=K_{j\times t}$ with partite sets $X_{i}=\{x_{1}^{i},x_{2}^{i},\ldots,x_{t}^{i}\}$ for each $i\in[j]$ and $t=\lceil\frac{2n}{3}\rceil$ is $2$-colorable to $(K_{4},nK_{2})$, that is there exist 2-edge-coloring $(G^{1},G^{2})$ of $K_{5\times t}$, where $K_{4}\nsubseteq G^{1}$ and $nK_{2}\nsubseteq G^{2}$. For each $n\geq 6$, if $n\neq 3k$, then it can be check that $\lceil\frac{2(n-1)}{3}\rceil+1=\lceil\frac{2n}{3}\rceil$. Therefor assume that $n\neq 3k$ for each $k\geq 2$, and set $X^{\prime}_{i}=X_{i}\setminus\{x_{1}^{i}\}$ for each $i\in[j]$. Now consider $K^{\prime}=K_{j\times(t-1)}=G[X^{\prime}_{1},\ldots,X^{\prime}_{j}]$. Consider a 2-edge-coloring $(G^{\prime 1},G^{\prime 2})$ of $K^{\prime}$. As $m_{j}(K_{4},(n-1)K_{2})\leq t-1$, $K_{4}\nsubseteq G^{\prime 1}\subseteq G^{1}$, one can check that $(n-1)K_{2}\subseteq G^{\prime 2}\subseteq G^{2}$. Suppose that $M^{\prime}$ be a copy of $(n-1)K_{2}$ in $G^{\prime 2}$. Therefore, as $j=5$ and $K_{4}\nsubseteq G^{1}$, one can say that there is at least one edges of $E(G[V^{\prime}])$ say $e$, so that $e\in E(G^{2})$, where $V^{\prime}=\{x_{1}^{i},~{}i=1,2,\ldots,j\}$. So $M=M^{\prime}\cup\{e\}$ is a matching of size $n$ in $G^{2}$, a contradiction. Which means that $m_{j}(K_{4},nK_{2})\leq t$, for each $n\geq 3$ where $n\neq 3k$. Now assume that $n=3k$ for some $k\geq 2$. Hence, it can be check that $\lceil\frac{2(n-2)}{3}\rceil+1=\lceil\frac{2n}{3}\rceil=2k$. Set $X^{\prime}_{i}=X_{i}\setminus\{x_{1}^{i}\}$ for each $i\in[j]$. Consider $K^{\prime}=K_{j\times(t-1)}=G[X^{\prime}_{1},\ldots,X^{\prime}_{j}]$ and let $(G^{\prime 1},G^{\prime 2})$ be a 2-edge-coloring of $K^{\prime}$. As $m_{j}(K_{4},(n-2)K_{2})\leq k-1$, $K_{4}\nsubseteq G^{\prime 1}\subseteq G^{1}$, one can check that $mK_{2}\subseteq G^{\prime 2}\subseteq G^{2}$ where $m\geq n-2$. Suppose that $M^{\prime}$ be a M-M in $G^{\prime 2}$. If $|M^{\prime}|=n-1$, then, as $j=5$ and $K_{4}\nsubseteq G^{1}$, one can say that there is at least one edges of $E(G[V^{\prime}])$ say $e$, so that $e\in E(G^{2})$, where $V^{\prime}=\{x_{1}^{i},~{}i=1,2,\ldots,j\}$. So $M=M^{\prime}\cup\{e\}$ is a matching of size $n$ in $G^{2}$, a contradiction. Hence assume that $|M^{\prime}|=n-2$. Now, as $j=5$, $n=3k$ and $|M^{\prime}|=n-2$ it is easy to say that the following claim is true:

Therefore, since $j=5$, by Claim 9, and by pigeon-hole principle, there exist three partition of $K$, say $X_{i_{1}},X_{i_{2}},X_{i_{3}}$, so that for each $r\in[3]$ there exist a vertex of $X_{i_{r}}$ say $v_{r}$ so that $v_{r}$ not belongs to $M$. W.l.g assume that $X_{i_{1}}=X_{i}$ and $v_{i}=x_{2}^{i}$ for each $i\in[3]$. Set $V^{\prime\prime}=\{x_{2}^{1},x_{2}^{2},x_{2}^{3}\}$. Therefore, it can be check that $K^{\prime\prime}=G[V^{\prime}\cup V^{\prime\prime}]\cong K_{2}\oplus K_{3\times 2}\subseteq G^{1}$. Therefore $K_{4\times 2}\subseteq K^{\prime\prime}$. Consider a 2-edge-coloring $(G^{\prime\prime 1},G^{\prime\prime 2})$ of $K^{\prime\prime}$. As by Lemma 4, $m_{4}(K_{4},2K_{2})=2$, $K_{4\times 2}\subseteq K^{\prime\prime}$, and $K_{4}\nsubseteq G^{\prime\prime 1}\subseteq G^{1}$, one can check that $2K_{2}\subseteq G^{\prime\prime 2}\subseteq G^{2}$. Suppose that $M^{\prime\prime}$ be a M-M in $G^{\prime\prime 2}$. Therefore, $M=M^{\prime}\cup M^{\prime\prime}$ is a matching of size $n$ in $G^{2}$, which means that $m_{5}(K_{4},nK_{2})\leq 2k$ where $n=3k$. Therefore for each $n\geq 3$ we have $m_{5}(K_{4},nK_{2})\leq\lceil\frac{2n}{3}\rceil$. Which means that for each $n\geq 3$, $m_{5}(K_{4},nK_{2})=\lceil\frac{2n}{3}\rceil$ and the proof is complete. ∎

To prove $m_{j}(K_{4},nK_{2})\leq\lceil\frac{2n}{j-2}\rceil$ consider $K=K_{j\times t}$ with partite sets $X_{i}=\{x_{1}^{i},x_{2}^{i},\ldots,x_{t}^{i}\}$ for each $i\in[j]$ and $t=\lceil\frac{2n}{j-2}\rceil$. Suppose that $K$ is $2$-colorable to $(K_{4},nK_{2})$, that is there exist 2-edge-coloring $(G^{1},G^{2})$ of $K_{j\times t}$, where $K_{4}\nsubseteq G^{1}$ and $nK_{2}\nsubseteq G^{2}$. Thus $K_{4}\nsubseteq G^{1}[V(K^{\prime})]$ where $K^{\prime}=K_{j\times t^{\prime}}\subseteq K$, $t^{\prime}=\lceil\frac{2n-2}{j-2}\rceil$, so it can be check that $(n-1)K_{2}\subseteq G^{2}[V(K^{\prime})]$. As $t=\lceil\frac{2n}{j-2}\rceil$ and $|M|=n-1$, where $M$ ba a M-M in $G^{2}$, then one can show that the following claim is true:

If either $Y\cap X_{i}\neq\emptyset$ for at least four parts of $K$ or $t=1$, then one can check that $K_{4}\subseteq G^{1}[Y]$, a contradiction. Hence, let $t\geq 2$, as $|Y|=2t+2$ and $|X_{i}|=t$, we can assume that $Y\cap X_{i}\neq\emptyset$ for three $i,i\in[j]$. W.l.g let $Y\cap X_{i}\neq\emptyset$ for each $i\in[3]$. As $t\geq 2$, it can be check that $K^{\prime\prime}=K_{3\times 2}\subseteq G^{1}[Y]$. Also as $j\geq 6$ and $|M|=n-1$, it is easy to say that there exist at least one edges of $M$ say $e=v_{1}v_{2}$, so that $v_{1},v_{2}\notin X_{i}$ for $i=1,2,3$. Therefore w.l.g let $v_{1}\in X_{4}$ and $v_{2}\in X_{5}$. Now, by considering $v_{1},v_{2}$ and $Y$, where $Y$ is the vertices of $K$ not belongs to $M$, we have the following claim:

Therefore by Claim 12 it can be check that for at least one $i\geq 4$, there exist at least one vertex of $X_{i}$, say $w$, so that $|N_{G^{2}}(w)\cap Y|\leq 1$. Hence $|N_{G^{1}}(w)\cap Y|\geq|Y|-1$. So, one can check that $K_{4}\subseteq G^{1}[V(K^{\prime\prime})\cup\{w\}]$, a contradiction again. So, we have $m_{j}(K_{4},nK_{2})\leq\lceil\frac{2n}{j-2}\rceil$ for each $j\geq 6$ and $n\geq 3$. Therefore by Theorem 7, we have $m_{j}(K_{4},nK_{2})=\lceil\frac{2n}{j-2}\rceil$ for each $j\geq 6$ and $n\geq 3$, which means that the proof is complete.