MHB Proving matrix group under addition for associative axiom

[cbarker1]

Dear Everyone,

I have some feeling some uncertainty proving one of the axioms for a group. Here is the proof to show this is a group:
Let the set T be defined as a set of 2x2 square matrices with coefficients of integral values and all the entries are the same.
We want to show that T is an abelian group under addition. Let's denote that $(x)=\left[\begin{array}{cc}x&x\\x&x\end{array}\right]$.
Let $(a)$ and $(b)$ $\in T$ where (a),(b) are a arbitrary elements in the set T. Then by the definition of (a) and (b) as well the definition of matrix addition, So, \begin{equation}
(a)+(b)=\left[\begin{array}{cc}
a & a \\
a & a
\end{array}\right]+\left[\begin{array}{cc}
b & b \\
b & b
\end{array}\right]=\left[\begin{array}{cc}
a+b & a+b \\
a+b & a+b
\end{array}\right]= (a+b)\in T. \end{equation}
Let $(a)$ be an arbitrary element in the set $T$. Let $(e)=(0)$. Then, by the definition of the binary operation,
\begin{equation}
(a)+(0)=\left[\begin{array}{cc}
a & a \\
a & a
\end{array}\right]+\left[\begin{array}{cc}
0 & 0 \\
0 & 0
\end{array}\right]=\left[\begin{array}{cc}
a+0 & a+0 \\
a+0 & a+0\end{array}\right]=\left[\begin{array}{cc}
a & a \\
a & a
\end{array}\right]=(a).\end{equation} And \begin{equation}
(0)+(a)=\left[\begin{array}{cc}
0 & 0 \\
0 & 0
\end{array}\right]+\left[\begin{array}{cc}
a & a \\
a & a
\end{array}\right]=\left[\begin{array}{cc}
0+a & 0+a \\
0+a & 0+a\end{array}\right]=\left[\begin{array}{cc}
a & a \\
a & a
\end{array}\right]=(a)
\end{equation}
Let $(a)$ be an arbitrary element in the set $T$. Let $(a')=(-a)$. Since the coefficients of the matrix is the set of integers, so the additive inverse for the integers is the negative element of the set of integers. Then, by the definition of the binary operation, \begin{equation}
(a)+(-a)=\left[\begin{array}{cc}
a & a \\
a & a
\end{array}\right]+\left[\begin{array}{cc}
-a & -a \\
-a & -a
\end{array}\right]=\left[\begin{array}{cc}
a-a & a-a \\
a-a & a-a\end{array}\right]=\left[\begin{array}{cc}
0 & 0 \\
0 & 0
\end{array}\right]=(0).
\end{equation}
And \begin{equation}
(-a)+(a)=\left[\begin{array}{cc}
-a & -a \\
-a & -a
\end{array}\right]+\left[\begin{array}{cc}
a & a \\
a & a
\end{array}\right]=\left[\begin{array}{cc}
-a+a & -a+a \\
-a+a & -a+a\end{array}\right]=\left[\begin{array}{cc}
0 & 0 \\
0 & 0
\end{array}\right]=(0).
\end{equation}
Let $(a)$,$(b)$,$(c)$ $\in T$ where $(a),(b),(c)$ are a arbitrary elements in the set T. Then by the definition of (a),(b),(b) as well the definition of matrix addition, So, \begin{equation}
((a)+(b))+(c)=(\left[\begin{array}{cc}
a & a \\
a & a
\end{array}\right]+\left[\begin{array}{cc}
b & b \\
b & b
\end{array}\right])+\left[\begin{array}{cc}
c & c \\
c & c
\end{array}\right]=\left[\begin{array}{cc}
a+b & a+b \\
a+b & a+b
\end{array}\right]+\left[\begin{array}{cc}
c & c \\
c & c
\end{array}\right]=\left[\begin{array}{cc}
(a+b)+c & (a+b)+c \\
(a+b)+c & (a+b)+c
\end{array}\right]=\left[\begin{array}{cc}
(a+0+b)+c & (a+0+b)+c \\
(a+0+b)+c & (a+0+b)+c
\end{array}\right] =\left[\begin{array}{cc}
(a+0)+(b+c) & (a+0)+(b+c) \\
(a+0)+(b+c )& (a+0)+(b+c)
\end{array}\right]=\left[\begin{array}{cc}
a+(b+c) & a+(b+c) \\
a+(b+c )& a+(b+c)
\end{array}\right]=(a)+((b)+(c))\end{equation}.

Did I correctly prove the associative axiom correctly?

Thanks
Cbarker1

 

[Olinguito]

Yes, your steps are correct (although there are a couple of redundant steps you could have skipped in the proof of associativity). Another way to prove the result is to show that the obvious mapping $f:\mathbb Z\to T;x\mapsto(x)$ is an isomorphism.