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

Question

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

The identity element for set of integers is $0$

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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 $e$.

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

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

Answer 1

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 $a * a^{-1} = e$

Answer 2

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.

Answer 3

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.

Answer 4

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.