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!

Let us leave out each Matrix in G and take only the Index of the Matrix.
That is $G=\{1,2,3,4,5,6,7,8,\cdots,k\}$
Let $B=(1,2,3,4,5,6,7,8,\cdots,k)$ which is a sequence of Integers.
Sum of elements of B is Constant.
Multiply by some element by B to get a new sequence of Integers which is a Permutation of the same Original Integers.
That is $C=AB=A(1,2,3,4,5,6,7,8,\cdots,k)=(\cdots)$
Sum of elements of C is same Constant.
Alternatively, using each Matrix instead of each Index, we will get same Sum.
This Constant or Sum is B itself.
