Existence of a grothendieck group

follows any \(t\), e.g. \(t = 0\) and

\begin{align*} m_1 + n_2 + 0 = m_1 + n_2 + 0 \end{align*}

follows from the symmetry of \(=\)

Let \((m_1, m_2) \sim (n_1,n_2) \sim (l_1,l_2)\) with

\begin{align*} m_1 + n_2 + t =& n_1 + m_2 + t \\ n_1 + l_2 + s =& l_2 + n_2 + s \end{align*}

for suitable \(l,s \in M\) Thus

\begin{align*} m_1 + n_2 + t + n_1 + l_2 + s =& n_1 + m_2 + t + l_1 + n_2 + s \\ m_1 + l_2 + (n_1 + n_2 + t + s) =& l_1 + m_2 + (n_1 + n_2 + t + s) \end{align*}

and hence

\begin{align*} (m_1,m_2) \sim (l_1,l_2) \end{align*}

Let

\begin{align*} +: G(M) \times G(M) \rightarrow& G(M) \\ ([x,y], [x',y']) \mapsto& [x + x', y + y'] \end{align*}

Suppose \([x_1,x_2] \sim [x_1',x_2']\) and \([y_1,y_2] \sim [y_1',y_2']\) Then there exist \(s,t \in M\) such that

\begin{align*} x_1 + x_2' + t =& x_1' + x_2 + t \\ y_1 + y_2' + s =& y_1' + y_2 + s \end{align*}

Therefore,

\begin{align*} (x_1 + y_1) + (x_2' + y_2') + (s + t) =& (x_1' + y_1') + (x_2 + y_2) + (s + t) \end{align*}

or equivalently

\begin{align*} [x_1 + y_1, x_2 + y_2] =& [x_1' + y_1', x_2' + y_2'] \\ [x_1,x_2] + [y_1,y_2] =& [x_1',x_2'] + [y_1',y_2'] \end{align*}

follows from

\begin{align*} [x,y] + [x',y'] =& [x + x', y + y'] \\ =& [x' + x, y' + y] \\ =& [ \end{align*}

follows from

\begin{align*} [e,e] + [x,y] =& [x + e, y + e] \\ =& [x, y] \end{align*}

and the commutativity

for \([x,y]\), the inverse element is \([y,x]\), as

\begin{align*} [x,y] + [y,x] =& [x + y, y + x]\\ =& [x + y, x + y] \end{align*}

and furthermore

\begin{align*} (x + y) + 0 + 0 =& 0 + (x + y) + 0 \end{align*}

thus

\begin{align*} [x + y,x + y] = [e,e] \end{align*}

Let \(\varphi: M \rightarrow H\) be a monoid-homomorphism for an abelian group \(H\). Then let

\begin{align*} \psi: G(M) \rightarrow& H \\ [x,y] \mapsto& \varphi(x) - \varphi(y) \end{align*}