# Group operation and a doubly stochastic matrix

If $$p(i)$$ is the probability of state $$i$$ with $$p(i)>0$$ and $$p_{ij}=p(j.i^{-1})$$ are transition probabilities where product and inverse refer to group operations, how can I show that the transition probability matrix is doubly stochastic?

Note that to show that the matrix $$P$$, with coordinates $$p_{ij}$$, is doubly stochastic, it is enough to show that the values of each rows and each column sum to $$1$$.
For this, observe that $$\sum\limits_i p(i) = 1$$.
Now consider the $$j'$$th column, the sum of of elements in this column is given by $$\sum\limits_i p_{ij} = \sum\limits_i p(j \cdot i^{-1}) = \sum\limits_{j^{-1}\cdot i^{-1}|i \in G} p(i) = 1,$$ where the second equality is just a change of variables.
• I'm not sure I understand the second equality. Could you please break it down or explain for me? Should the range of the sum be $j^{-1}\cdot i|i \in G$? How does it sum to 1? I think I'm quite naive! – Blain Waan Oct 17 '18 at 21:21
• The main idea is that to go over all elements in the group of the type $j \cdot i^{-1}$ (where $j$ is fixed and $i$ is every possible element in $G$), is exactly like simply going over all elements in the group. To see this, for $g \in G$, just take $i = g^{-1} \cdot j$, so that $j\cdot i^{-1} = g$. – Cain Oct 17 '18 at 21:29