# Is $(\mathbb{Z}, \times)$ also a group?

Addition $$+$$ is a closure operation for set of integers ($$\mathbb{Z}$$)

The identity element for set of integers is $$0$$

Definition of group:

Each group is a set of elements with one operation $$*$$ and is closed under $$*$$. Each element in the group has an inverse. Each element combine with its inverse gives the identity element $$e$$.

So, $$(\mathbb{Z}, +)$$ is a group

Is $$(\mathbb{Z}, \times)$$ also a group?

• Multiplication is a well-defined operation on $\mathbb{Z}$. While a single group has only one operation in its data, there is no restriction on how many operations can exist on a set. In other words, $(\mathbb{Z},+)$ and $(\mathbb{Z},\times)$ are both things that may or may not be groups, because $+$ and $\times$ are both operations $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$. It turns out that $(\mathbb{Z},+)$ is a group, but $(\mathbb{Z},\times)$ is not a group. Can you see why? Jul 22, 2020 at 0:29
• I don't think the term "closure operation" means anything. The integers are closed under both the operations addition and multiplication. The integers with the operation of addition is a group, with multiplication not. Jul 22, 2020 at 0:30
• @diracdeltafunk for (Z, x) identity element is 1 but each element does not have its inverse? this is why it is not group Jul 22, 2020 at 0:36
• Very nice! That's correct — because $1$ is the identity element under multiplication and there is no integer $x$ such that $2x = 1$, $(\mathbb{Z},\times)$ is not a group. Jul 22, 2020 at 0:39
• Thank you so much! I hope to teach as much as possible in my career :) Jul 22, 2020 at 19:58

Let's analyze the group, $$(\mathbb{Z}, \times)$$. First, we need an identity element. In this group, $$1$$ would be our identity element (there's your first condition). Now, it's also easy to see that $$\mathbb{Z}$$ is closed under multiplication. However, a problem arises with inverses. For any integer, $$a$$, $$a \times \frac{1}{a} = 1$$. However, for most integers, $$\frac{1}{a}$$ is not an element of $$\mathbb{Z}$$. For example, $$3 \times \frac{1}{3} = 1$$, but $$\frac{1}{3}$$ isn't an element of $$\mathbb{Z}$$

Definition of a Group:

1. (Closure) A set, $$G$$, is a group if it is closed under some binary operator, *
2. (Identity) There is an identity element, $$e$$, in G such that $$a * e = a$$ for all $$a$$ in $$G$$
3. (Inverse) For every $$a \in G$$, there exists an element, $$a^{-1}$$, such that $$a * a^{-1} = e$$
• What is the motivation behind Group definition? Jul 22, 2020 at 0:45
• @overexhange I included the definition of a group in my answer. The "group" you defined fails criteria $3$. Is this what you mean behind motivation? Jul 22, 2020 at 0:50
• I mean, where do we apply the concept of group definition? Solving equations Jul 22, 2020 at 1:01
• @N.Bar small typo, "$1 \times \frac{1}{3}$" should read $3 \times \frac{1}{3}$. Also on a very nitpicky level, the argument you use should include the point that inverses are unique in $\mathbb{Q}$ so the fact that $1/3$ is the inverse of $3$ in $\mathbb{Q}$ is enough to not check anything else in $\mathbb{Z}$. Jul 22, 2020 at 3:42
• @Mummytheturkey Thanks for catching that! And I was just trying to provide one simple counterexample for the OP, but you are certainly correct. Jul 22, 2020 at 15:23

The way you've stated is a bit too simplistic. We aren't interested in making extremely generic statements like "The set can have one and only one operation defined on it." when we're defining algebraic structures.

If we could PROVE that it can have one and only one operation defined on it, then that would be neat. But we don't say that a priori.

Here's the formal definition of a group.

Let $$G$$ be a set and $$\circ: G \times G \to G$$ be a function. Then, the pair $$(G, \circ)$$ is called a group iff the following statements hold:

1. $$\forall a,b,c \in G: a \circ (b \circ c) = (a \circ b) \circ c$$

2. $$\exists e \in G: \forall a \in G: a \circ e = a = e \circ a$$

3. $$\forall a \in G: \exists b \in G: a \circ b = e = b \circ a$$

That's it. So, for instance, $$(\mathbb{Z},+)$$ is a group, where we are careful in specifying that $$+$$ is the usual addition on the integers.

Now, this doesn't imply that a multiplication operation cannot be defined on $$\mathbb{Z}$$. You and I multiply integers on a daily basis and certainly, we get integers when we multiply integers with integers. In that sense, we say that $$\mathbb{Z}$$ is closed under multiplication. However, we note that $$(\mathbb{Z},\cdot)$$ is NOT a group.

We can see that not all elements of $$\mathbb{Z}$$ have a multiplicative inverse that is contained in $$\mathbb{Z}$$. For example, we note that $$1 \in \mathbb{Z}$$ is the identity element BUT:

$$2 \cdot \frac{1}{2} = 1 = \frac{1}{2} \cdot 1$$

so $$\frac{1}{2}$$ is an inverse of $$2$$ but it isn't actually an integer. So, $$(\mathbb{Z}, \cdot)$$ fails to satisfy the third condition and hence, it isn't a group.

A single set can have two different operations defined on it, both of which make it a group. And "$$\mathbb Z$$ with $$+$$" would be considered to be a different group from "$$\mathbb Z$$ with $$\times$$" (assuming both are groups).

As for "$$\mathbb Z$$ with $$\times$$", think about inverses.

Also, I don't know if there's some language barrier, but asking if something "is a closure operation" isn't how one talks about groups. I'm pretty sure you're asking "Does $$\mathbb Z$$ form a group under $$\times$$?".

In group theory, "closure" is a property of an operation on a set which means when you perform the operation on two members of the set you get back another element of the set. So, for example, the odd numbers are not closed under addition.

Definition: An idempotent with respect to an operation $$\ast:S\times S\to S$$ is an element $$e\in S$$ such that $$e\ast e=e$$.

Lemma: Each group has exactly one idempotent; namely, the identity.

Proof: Let $$(G,\circ)$$ be a group with identity $$e$$. Suppose $$g\in G$$ is an idempotent. Then $$g\circ g=g=g\circ e.\tag{1}$$ Multiply $$(1)$$ on the left by $$g^{-1}$$. Then

\begin{align} g^{-1}\circ(g\circ g)&=(g^{-1}\circ g)\circ g\\ &=e\circ g\\ &=g\\ &=g^{-1}\circ (g\circ e)\\ &=(g^{-1}\circ g)\circ e\\ &=g^{-1}\circ g\\ &=e. \end{align}

So, in particular, $$g=e$$. $$\square$$

But for $$0$$ and $$1$$ in $$\Bbb Z$$, $$0\times 0=0$$, $$1\times 1=1$$, and $$0\neq 1$$; thus $$(\Bbb Z,\times)$$ cannot be a group by the lemma above.

• This probably isn't appropriate to the O.P.'s level, but is a very slick way of doing it (plus it uses idempotents, which I have a personal fondness for). +1. Jul 22, 2020 at 14:58
• Thank you, @JonathanZsupportsMonicaC. Jul 22, 2020 at 15:01