Markov chain doesn't sum up to 1 
Let $\{X_n\}$ be a Markov chain on $S=\{1,2,3,4,5,6\}$ with the matrix 
suppose we define a new sequence $\{Y_n\}$ by $$Y_n=\cases{1\quad X_n=1\vee X_n=2\\2\quad X_n=3\vee X_n=4\\3\quad X_n=5\vee X_n=6}$$does for $a=1$ this is a Markov chain?

I thought summing up and computing for example the first row of the stochastic matrix to better understand $\{Y_n\}$ so I did the following $$P(Y_n=1\mid Y_n=1)=P(X_n=1\vee X_n=2\mid X_n=1\vee x_n=2)=\frac{a+7}{10}\\P(Y_n=2\mid Y_n=1)=P(X_n=3\vee X_n=4\mid X_n=1\vee x_n=2)=\frac{3-a}{10}\\P(Y_n=3\mid Y_n=1)=P(X_n=5\vee X_n=6\mid X_n=1\vee x_n=2)=1$$but doesn't matter the value of a, they sum up to $2$ and not to $1$ means $$P_{Y_n}=\left(\begin{array}{ccc}
\frac{a+7}{10} & \frac{3-a}{10} & 1\\
\dots & \dots & \dots\\
\dots & \dots & \dots
\end{array}\right)
 $$ Do I need to normalize the results? what am I doing wrong? Is checking that it's Markov chain is as simple as summing up each row or I need to show a specific path which doesn't meet the property of Markov chain?

Let's write out the transition matrix of $X$ when $a = 1$:  $$\mathcal P = \frac{1}{10} \begin{bmatrix} 1 & 3 & 0 & 1 & 5 & 0 \\ 2 & 2 & 1 & 0 & 1 & 4 \\ 6 & 0 & 1 & 1 & 2 & 0 \\ 6 & 0 & 1 & 1 & 0 & 2 \\ 3 & 0 & 4 & 1 & 0 & 2 \\ 3 & 0 & 1 & 4 & 2 & 0 \end{bmatrix}$$  Now split it up into $2 \times 2$ blocks, and observe for example that $$\begin{align*} \Pr[Y_2 = 1 \mid X_1 = 1] &= 0.4, \\ \Pr[Y_2 = 2 \mid X_1 = 1] &= 0.1, \\ \Pr[Y_2 = 3 \mid X_1 = 1] &= 0.5. \end{align*}$$  But in fact, the same thing is true if $X_1 = 2$; thus $Y_2 \mid Y_1 = 1$ follows the above distribution.  And we can also see that the same is true for the other two pairs of rows.  Therefore, $Y$ is also a Markov chain, with transition matrix $$\mathcal P^* = \frac{1}{10} \begin{bmatrix} 4 & 1 & 5 \\ 6 & 2 & 2 \\ 3 & 5 & 2 \end{bmatrix}.$$
