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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?

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    Unfortunately, this argument does not work, since it is possible to have other elements act like an identity for a subset of the set of all matrices without actually being the identity matrix. This is for example the case here (try working out explicitly the $2\times 2$ case). – 2017-01-31
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    You must be knowing that the group $GL_n(\Bbb R)$ is totally a different group. So you change the name you have given to your group. – 2017-01-31

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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.

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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).

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    Forget the multiples; what is the _inverse_ of all-1 matrix? – 2017-01-31
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    1 . Ok so it need not have 0 in its identity – 2017-01-31
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    @IvanNeretin The OP question was about the identity element - that we do have. In fact, it turns out that $G$ is a group and there are inverses. Those are just not inverses within $GL_n(\Bbb R$), simply because we do not start from within GL_n$ in the first place (that was the "trivial reason" I alluded to) – 2017-01-31
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    Oh, I see. So it's a group, but not a subgroup of $GL_n$. Well, that should be expected: it is not even a _subset_ of $GL_n$. – 2017-01-31