Q:

A logic statement. "or" in Abstract algebra - groups

Let H be the subset of $M_2(\mathbb{R})$ consisting of all matrices of the form 
$H^* = \left \{ \begin{pmatrix}
a &-b \\ 
 b&a 
\end{pmatrix} : a,b\in\mathbb{R} , a\neq 0 \; \text{or} \; b \neq 0\right\}$
Is (H*, .) a group under multiplication?

I said no because since $b \neq 0$, the identity matrix can't exist. But my prof briefly said "either a or b can't be 0, not a and b"
Am I right or his wrong?

A:

In mathematics the use of the word or is inclusive, not exclusive. It is possible that only $b\neq 0$, it is possible that only $a\neq 0$, and it is possible that both $a\neq0$ and $b\neq 0$.
It is only impossible that both $a=b=0$.
So the identity matrix is in $H^\ast$, since $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ has the wanted property.

A:

He’s right, and you’re wrong. The definition
$$H^* = \left \{ \begin{pmatrix}
a &-b \\ 
 b&a 
\end{pmatrix} : a,b\in\mathbb{R} , a\neq 0 \; \text{or} \; b \neq 0\right\}$$
makes $H^*$ the collection of all $2\times 2$ real matrices such that at least one of $a$ and $b$ is non-zero. The identity matrix is such a matrix: for it you have $a\ne 0$.
The condition $$a\ne 0\text{ or }b\ne 0$$ is equivalent to the condition $$(a\ne 0\text{ and }b\ne 0)\text{ or }(a\ne 0)\text{ and }b=0)\text{ or }(a=0\text{ and }b\ne 0)\;;$$ the only case that it excludes is $a=b=0$. In mathematics and logic, or is inclusive: ‘$a\text{ or }b$’ is ‘$a\text{ or }b\text{ or both}$’.

