TITLE: Does $(\{0,1\},*)$ form a group?
QUESTION [8 upvotes]: I am reading my first book on abstract algebra. I am not enrolled in a class on the subject.
Given $S = \{0,1\}$. Is $(S,\cdot)$ a group?
$S$ is closed under multiplication. $$0\cdot1=0,\,1\cdot0=0,\,0\cdot0=0,\,1\cdot1=1.$$ 
$S$ has an identity, $1$, I think.
$$0\cdot1=0,\,1\cdot1=1.$$
I don't believe $S$ satisfies $a\cdot a^{-1}=\operatorname{id}.$
However, zero is excluded when stating that $\mathbb R$ satisfies $a\cdot a^{-1} = \operatorname{id}$ under multiplication.
$S$ would be a group if zero is excluded. $1\cdot1=1.$
So is $S$ a group or not?

REPLY [2 votes]: Everyone else has correctly pointed out that the set {0, 1} under multiplication does not form a group. This seems to be the source of your confusion, so I shall only answer this:

However, zero is excluded when stating that $\mathbb{R}$ satisfies $a\cdot a^{−1}=id$ under multiplication.

Let's be clear: $\mathbb{R}$ is not a group under multiplication. (Why? Because there's no $0^{-1}$.) The "exclusion" of zero that you're talking about requires you to remove it from the set entirely. That is, the set $(\mathbb{R} - \{0\})$ is a group under multiplication, but when you put 0 back in, it fails to be a group any more.
This is a common feature of objects like this (I won't make that precise, but you might like to look up the word nilpotent):

if you want to modify $M_4(\mathbb{C})$ ($4\times 4$ matrices with complex entries) so that it forms a group under multiplication, you must first throw away all of the non-invertible matrices, e.g. $\left(\begin{smallmatrix} 1&1&1&1\\1&1&1&1\\1&2&3&4\\5&6&7&8\end{smallmatrix}\right)$. That is, what you are left with at the end may or may not be a group under multiplication, but this won't be $M_4(\mathbb{C})$, because $M_4(\mathbb{C})$ definitely isn't.
if you want to modify $\mathbb{Z}/10\mathbb{Z}$ (the set containing the numbers 0, 1, 2, ..., 9) to form a group under multiplication (mod 10), you must first throw away all of the non-invertible numbers, e.g. $4$ and $5$. (You end up throwing away 6 of the 10 elements.)