Let $A,B$ be matrices under certain conditions, then $AB=B$ Let $k, n$be positive integers. Let $G= \{ A_1, A_2, \cdots, A_k\}$ be a set of real $n\times n$ matrices such that $G$ is a group under the usual matrix multiplication. Let $B=A_1 + A_2 + \cdots +A_k$. Prove that $A_i B=B$ for every $i\in \{ 1,2, \cdots, k\}$.
At first I tried writing out the matrices. Then it became a mess of notations. To make the notations a bit nicer, I used $A^i$ instead of $A_i$. Then I got $(A^iB)_{pq} = \sum_{x=1}^n A^i_{xp}(\sum_{i=1}^kA^{i}_{px})$..
Then I tried something different. I got $I= (B^{-1}B) = (A_iB)^{-1}B = B^{-1} A_i^{-1}B$. So $B=A_i^{-1}B$. But then I realized that $B$ might not be invertible.
Thanks in advance for any help!

Notice that the map $f_{i}:G\to G$ defined by $f_i(g)=A_ig$ is a bijection for every $i$.
