unit of a ring distinct from neutral element in a subset-group If we examine the set $H$ of square matrices of order $2$ of the form $ M(x) := \begin{pmatrix}
x & x \\
x & x 
\end{pmatrix}$, we notice that the identity matrix isn't there. but still this set has a neutral element with respect to matrix multiplication, that is $M\left(\frac 12 \right)$. 
Now if we restrict ourselves to nonzero matrices, we see that this set is even a group under matrix multiplication though no matrix of $H$ is invertible in the usual sense.
I'd like to know how normal is this situation and look forward to get more information about that. Thanks.
 A: I don't know exactly what you mean by "how normal" it is (perhaps "how unusual"?) but I can assure you that this happens a lot more than you think.
The thing is that we are used to "nice" operations that we see every day. But really, if you had some group $G$, and a set $X$, you could take whatever arbitrary bijection you want $f:G\to X$ and make $X$ a group by declaring that
$x\cdot y:=f(f^{-1}(x) f^{-1}(y))$
(The dot is the new multiplication, juxtaposition denotes the old multiplication. The same thing works for rings, but of course you're transporting both operations.)  This is called transportation of structure from $G$ to $X$.
In particular, you could take the group of nonzero real numbers and the map $x\mapsto \frac{x}{2}$, and you get this perfectly defined group structure on the nonzero real numbers that has the peculiar identity:
$x\cdot\frac{1}{2}:=\frac{1}{2}(2x2\frac{1}{2})=x$.
Notice where the identity of the group of nonzero reals is being mapped: $1\mapsto \frac12$.
That's essentially what's happening here, except that it's disguised with matrix multiplication in $x\mapsto \begin{bmatrix}\frac{x}{2}&\frac{x}{2}\\ \frac{x}{2}&\frac{x}{2}\end{bmatrix}$.
The thing that makes this example somewhat striking is that the "weird" transported multiplication is linked with this ordinary multiplication (matrix multiplication.) 
Actually, if you leave the zero matrix in, you'll find that this is a subring of $M_2(\mathbb R)$ that is (ring) isomorphic to $\mathbb R$.
A: One can generate a large class of examples that includes this one.
First, a standard computation gives the Jordan form of the matrix $M(x)$ and the matrix via which it is similar to $M(x)$:
$$M(x) = \pmatrix{\frac{1}{2}&\frac{1}{2}\\-\frac{1}{2}&\frac{1}{2}}^{-1} \pmatrix{2x&0\\0&0}\underbrace{\pmatrix{\frac{1}{2}&\frac{1}{2}\\-\frac{1}{2}&\frac{1}{2}}}_P .$$
The Jordan normal form suggests writing $M(\frac{1}{2} x) = P^{-1} \pmatrix{x&0\\0&0} P$, so we see immediately that
\begin{align}
M\left(\tfrac{1}{2} x\right) + M\left(\tfrac{1}{2} y\right)
&= P^{-1} \pmatrix{x&0\\0&0} P + P^{-1} \pmatrix{y&0\\0&0} P\\
&= P^{-1} \pmatrix{x + y&0\\0&0} P\\
&= M\left(\tfrac{1}{2}(x + y)\right)
\end{align}
and
\begin{align}
M\left(\tfrac{1}{2} x\right) M\left(\tfrac{1}{2} y\right)
&= P^{-1} \pmatrix{x&0\\0&0} P \cdot P^{-1} \pmatrix{y&0\\0&0} P \\
&= P^{-1} \pmatrix{x&0\\0&0} \pmatrix{y&0\\0&0} P \\
&= P^{-1} \pmatrix{x y&0\\0&0} P \\
&= M\left(\tfrac{1}{2} x y\right) .
\end{align}
Thus
$$x \mapsto M\left(\tfrac{1}{2} x\right)$$
is a ring homomorphism $$(\Bbb F, +, \cdot) \to \left(\left\{\pmatrix{x&x\\x&x}\right\}, +, \cdot\right) ,$$ where $\Bbb F$ is the field underlying the matrix ring (necessarily not of characteristic $2$, so that the element $\frac{1}{2}$ is available). It's easy to check that it is surjective, so this is actually a ring isomorphism.
Now, $+$ and $\cdot$ denote (the restrictions of) the usual matrix multiplication and addition map, but $\left(\left\{\pmatrix{x&x\\x&x}\right\}, +, \cdot\right)$ is not a subring of $M(2, \Bbb F)$, because (as you observed) the neutral elements are not the same. By varying the coefficient of $x$ in the Jordan normal form and varying $P$ among the invertible elements, we get a large family of rings, all isomorphic to $\Bbb F$.
Much more generally, a bijection $\Phi: R \to S$ between the set underlying a ring $(R, +, \cdot)$ and a set $S$ induces a ring structure $(\oplus, \odot)$ on $S$ by regarding $\Phi$ as an identification, namely defining, $$s \oplus s' := \Phi(\Phi^{-1}(s) + \Phi^{-1}(s')), \qquad s \odot s' := \Phi(\Phi^{-1}(s) \cdot \Phi^{-1}(s')) .$$ (Similar constructions work for any admissible algebraic structure.)
In our case, $R = \Bbb F$, $\Phi = M$ and $S = \Phi(R)$. (It is an instructive exercise to work out the rather peculiar-looking ring operations for the choices $R = S = \Bbb R$ and $\Phi : x \mapsto x - 1$, for example.)
What is relatively special about the class of examples constructed above is that the induced maps $\oplus$ and $\odot$ are restrictions of familiar operations, namely, matrix addition and multiplication.
