Matrices in group theory 
Let
  $$ S = \left\{ \left(\begin{matrix}
        a & a \\
        a & a  \\
        \end{matrix}\right) \ \middle| \ a \in \mathbb{R} \right\}. $$
  and $\cdot$ be a usual matrix multiplication operation. 
Then is $(S,\cdot)$ a group or semigroup or monoid ?


We know that matrix multiplication is associative. We need to check that $S$ is also closed under this operation. Indeed,
$$
\begin{pmatrix}
a & a \\ a & a
\end{pmatrix}
\begin{pmatrix}
b & b \\ b & b
\end{pmatrix}
=
\begin{pmatrix}
2ab & 2ab \\ 2ab & 2ab
\end{pmatrix}
$$
shows that it is closed under multiplication. Therefore $S$ is at least a semigroup.
To check whether $S$ is a monoid, we need to see if there is an identity element. If
$$
\begin{pmatrix}
e & e \\ e & e
\end{pmatrix}
$$
is to be an identity element, then we would need that
$$
\begin{pmatrix}
e & e \\ e & e
\end{pmatrix}
\begin{pmatrix}
a & a \\ a & a
\end{pmatrix}
=
\begin{pmatrix}
a & a \\ a & a
\end{pmatrix}
=
\begin{pmatrix}
a & a \\ a & a
\end{pmatrix}
\begin{pmatrix}
e & e \\ e & e
\end{pmatrix}
$$
holds for every $a$. This is equivalent to requiring that
$$
\begin{pmatrix}
2ea & 2ea \\ 2ea & 2ea
\end{pmatrix}
=
\begin{pmatrix}
a & a \\ a & a
\end{pmatrix}
=
\begin{pmatrix}
2ae & 2ae \\ 2ae & 2ae
\end{pmatrix}
$$
holds for every $a$, which occurs iff $a = 2ae$ and thus iff $e=1/2$. Thus we have an identity element
$$
\begin{pmatrix}
1/2 & 1/2 \\ 1/2 & 1/2
\end{pmatrix},
$$
so $S$ is at least a monoid.
Finally, to check whether $S$ is a group, we need to see if there are inverses. Given a matrix
$$\begin{pmatrix}
a & a \\ a & a
\end{pmatrix}$$
in $S$, its inverse 
$$\begin{pmatrix}
i & i \\ i & i
\end{pmatrix}$$
must satisfy
$$
\begin{pmatrix}
a & a \\ a & a
\end{pmatrix}
\begin{pmatrix}
i & i \\ i & i
\end{pmatrix}
=
\begin{pmatrix}
1/2 & 1/2 \\ 1/2 & 1/2
\end{pmatrix}
=
\begin{pmatrix}
i & i \\ i & i
\end{pmatrix}
\begin{pmatrix}
a & a \\ a & a
\end{pmatrix}.
$$
Such an inverse exists iff $2ai=1/2$, which occurs iff $a\ne0$ and $i=1/(4a)$. This reveals that the element
$$
\begin{pmatrix}
0 & 0 \\ 0 & 0
\end{pmatrix}
$$
does not have an inverse, so $S$ is not a group.
In conclusion $S$ is a semigroup and a monoid, but not a group.
