query on group theory Is the set (G)$G_n(\mathbb R) =\{a_{ij}=a\neq 0\,\forall\, i,j\}$ a group under matrix multiplication ? I think it is not as it's identity should be the identity matrix which has 0 and 0 can't be there in any element of G. Is is right?
 A: Let $$A = \begin{pmatrix} a & a \\ a & a\end{pmatrix}$$ and $$B = \begin{pmatrix} b & b \\ b & b\end{pmatrix}$$
Then, we have that:
$$AB = \begin{pmatrix} 2ab & 2ab \\ 2ab & 2ab\end{pmatrix}$$
So, the matrix:
$$A = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2\end{pmatrix}$$
clearly acts as the identity.
I've explicitly found the identity and inverse for the $n\times n$ case below:

 We can generalize this to show that in the $n\times n$ case that $[AB]_{ij} = \sum_{k = 1}^na_{ik}b_{kj} = na_{ij}b_{ij}$, so the "identity" matrix on this group is the matrix having $\frac{1}{n}$ as it's "value".

 Finally, given $B$ with "value" $b$, we have that $B^{-1}B$ has "value" $nb^{-1}b$, so for this to be equal to $\frac{1}{n}$ (the identity), we have that $b^{-1} = \frac{1}{bn^2}$.

 So, we've found the identity and inverse of this group, and the multiplication is clearly associative, so it's a group.

A: Let $A$ be the all-entries-1 matrix. Then $G$ contains all non-zero multiples $cA$, $c\in\Bbb R\setminus\{0\}$. See if you can figure out what $cA\cdot c'A$ is ...
(However, for $n>1$ this is not a subgroup of $GL_n(\Bbb R)$ for another trivial reason, if that’s your real question).
