# Does $(\{0,1\},*)$ form a group?

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?

Looking at the structure $S = (\{0, 1\}, \cdot)$, as you found, $\,0\,$ has no (multiplicative) inverse.

Therefore, $S$, under multiplication, cannot be a group, as it fails to have closure on taking inverses. It meets the other criteria of a group, but not the group axiom requiring that the inverse of every element of a group is contained in the group.

However, $S' = \{1\}$, under multiplication, is a group: it's called the trivial group, as it contains only one element: the (multiplicative) identity of the group itself.

If you know about the addition of integers modulo $n$, then you should know that $\,\mathbb Z_2 = \left(\{0, 1\}, +\right)\,$ is a group: it is the additive group of integers under addition modulo $2$.

• So is zero excluded from the set of reals, under multiplication, to define a group? If so, what does that imply when using reals? We can't divide by zero? Jun 16, 2013 at 2:24
• Exactly. Zero is excluded from the group of Reals under multiplication. But it is not excluded from the reals, or the real field. You'll learn more about fields, I suspect, down the road. Jun 16, 2013 at 2:26
• A group is a particular sort of structure. Zero is included in the group of reals, under addition. And Rings and Fields are other sorts of algebraic structures. What is true about a group isn't necessarily true about a field, and vice versa. Jun 16, 2013 at 2:29
• @amWhy: Hopefully this nice write up gets a nice badge! +1 Jun 17, 2013 at 0:08

Saying that a set is a group requires you specify the group operation. It is easy to miss this subtle point because it is commonly left out when the author assumes that operation is implicitly understood, e.g. you will see "$\mathbb{Z}$ is a group" everywhere, but it technically should be "$\mathbb{Z}$ is a group under addition". We just talk about $\mathbb{Z}$ often enough in group theory that we find it tedious to write "under addition" every time.

The answer is that $\{0,1\}$ is a group under addition modulo $2$, but not under multiplication.

It should be noted, however, that any finite$^\star$ set is a group under an induced action of the symmetric group of the order of that set. You probably won't understand what I mean by that until later (when you get to Cayley's theorem), but, in the meantime, keep in mind that any finite$^\star$ set can be made into a group.

EDIT: I must admit that in my first draft of this answer I forgot to include the word "finite." As Qiaochu points out, the statement is still true without "finite" (assuming choice), however it is important to note that this is not a result of Cayley's theorem. See his comment below for a link containing more details.

• "...any set is a group under the induced action of the symmetric group of the order of that set." What? The statement that any (non-empty) set can be made into a group is harder than this, and in fact is equivalent to the axiom of choice (mathoverflow.net/questions/12973/…). Jun 16, 2013 at 1:56
• @QiaochuYuan Maybe he meant finite set?
– Pedro
Jun 16, 2013 at 2:01
• Yes, Peter is right. I need to macro "group" to "finite group" on my keyboard. Jun 16, 2013 at 2:04
• @Alexander: I still don't understand. What does the symmetric group have to do with anything? Jun 16, 2013 at 4:27
• @Alexander: what? What do you mean by "an induced action of the symmetric group of the order of that set"? What do actions have to do with anything? The action of $S_n$ on an $n$-element set turns it into an $S_n$-set, not a group. Jun 16, 2013 at 7:05

Therefore it is not a group. If you do addition modulo 2, it is a group.

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.)

S is an idempotent commutative monoid, but not a group, since it has no inverse. So some other people have mentioned ({0, 1}, addition modulo 2) is a group here. Maybe the following will help you along. Suppose you consider all possible operations on {0, 1} defined purely by their tabular format like the following:

X  0  1
0  0  0
1  0  1

Does there exist another group? If any other group exists does there exist any relationship between those groups?