In Dihedral group $D_{2n'}=\langle r,\Delta\mid r^{n^{'}}=\Delta^2=1,\Delta r \Delta^{-1}=r^{-1}\rangle $. Now compute $m,n$ in terms of $i,j,k,l$ from $(r^i \Delta^j) (r^k \Delta^l)=r^m \Delta^n$
now $(r^i \Delta^j) (r^k \Delta^l)=(r^i \Delta \Delta^{j-1})( r^k \Delta^l)=(r^i \Delta^{-1} )(\Delta^{j-1} r^k \Delta^l)=(\Delta^{-1}(\Delta r^i \Delta^{-1} ))(\Delta^{j-1} r^k \Delta^l)=(\Delta^{-1}(\Delta r \Delta^{-1} \Delta r \Delta^{-1}...\Delta r \Delta^{-1} ))(\Delta^{j-1} r^k \Delta^l)=\Delta^{-1} r^{-i}(\Delta^{j-1} r^k \Delta^l)$.
now I can't understand what to do, how to get the form $r^m \Delta^n$.I think I am making this more complicated. any help is appreciated. thanks
Define $s_1=\Delta$ and $s_2=\Delta r$. You can check that $\{s_1,s_2\}$ generate $D_{2n}$ with presentation $$\langle s_1,s_2\mid s_1^2=s_2^2=(s_1s_2)^n=1\rangle.$$ This is the Coxeter presentation of $D_{2n}$. Note that $r=s_1s_2$. Given $i,j,k,l$ we can compute \begin{equation}r^i\Delta^j r^k\Delta^l=(s_1s_2)^is_1^j(s_1s_2)^ks_1^l.\tag{1}\end{equation}
If $j$ is even, $s_1^j=1$, so $(1)$ becomes $$(s_1s_2)^i(s_1s_2)^ks_1^l=(s_1s_2)^{i+k}s_1^l=r^{i+k}\Delta^l.$$
If $j$ is odd, $s_1^j=s_1$, so $(1)$ becomes $$(s_1s_2)^is_1s_1(s_2s_1)^{k}s_1^{l-1}=(s_1s_2)^{i-k}s_1^{l-1}=r^{i-k}\Delta^{l-1},$$ using the fact that $(s_1s_2)^ks_1=s_1s_2s_1\cdots s_1=s_1(s_2s_1)^k$.
Thus $r^i\Delta^j r^k\Delta^l=r^a\Delta^b$ where $$a=\begin{cases} i+k\;(\textrm{mod}\;n) &\textrm{if}\;j\;\textrm{is even}\\ i-k\;(\textrm{mod}\;n) &\textrm{if}\;j\;\textrm{is odd} \end{cases}$$ and $$b=\begin{cases} l\;(\textrm{mod}\;2) &\textrm{if}\;j\;\textrm{is even}\\ l-1\;(\textrm{mod}\;2) &\textrm{if}\;j\;\textrm{is odd} \end{cases}$$