If $X$ is a Markov chain then $X_{2m}$ is a Markov chain, I have the proof but don't understand one step, why is $\Pr(X_{2m+2}=k,X_{2m+1}=i_{2m+1},even,odd)=\Pr(X_{2m+2}=k,X_{2m+1}=i_{2m+1}|X_{2m}=i_{2m})\Pr(even, odd)$ with the notation given in the link
So one has to show that $X_{2m}$ is a Markov chain, but since X is, does that imply for example $\Pr(X_{2m+2}=k|even,odd)=\Pr(X_{2m+2}=k|X_{2m+1}=i_{2m+1},X_{2m}=i_{2m})$ and the proof shows, it is only dependent of $X_{2m+1}=i_{2m+1}$, am I correct ?
Your equation is incorrect: $$\Pr(X_{2m+2}=k\mid\mbox{even,odd})=\Pr(X_{2m+2}=k\mid X_{2m+1}=i_{2m+1},X_{2m}=i_{2m}).$$ One way to see this is that the state $i_{2m+1}$ appears on the right, but not at all on the left.
As for the question in the comment, \begin{eqnarray*} &&\Pr(X_{2m+2}=k, X_{2m+1}=i_{2m+1},\mbox{even,odd})\\[5pt] &=&\Pr(X_{2m+2}=k, X_{2m+1}=i_{2m+1}\mid\mbox{even,odd})\Pr(\mbox{even,odd})\\[5pt]&=&\Pr(X_{2m+2}=k, X_{2m+1}=i_{2m+1}\mid X_{2m}=i_{2m})\Pr(\mbox{even,odd}), \end{eqnarray*} where the last equation comes from the Markov property.