Prove that Z is not a group under multiplication

Prove that \mathbb{Z} is not a group under multiplication .

I do expect that when you are learning group theory and you search for an answer to the question above, you want to know how to tackle such a question. While this can be answered in one sentence, I will take you through the whole process of what you need to do.

Proof. Let \mathbb{Z} be the set of integers. To be a group under multiplication, i.e., (\mathbb{Z},\times), we must verify three things: associativity, identity, and inverse elements. Let a,b,c \in \mathbb{Z}.

It is straightforward by the definition of integers that (a \times b)\times c = a \times (b \times c). So for all elements in \mathbb{Z} we know that the operation multiplication is associative.

Next, we need to find the identity element of \mathbb{Z} under multiplication. This is easy since 1 is identity element (note that a \times 1 = 1 \times a = a).

Finally, the last step is to find the inverse elements of \mathbb{Z} under multiplication. Now we take the element a \in \mathbb{Z}. Does this mean that there is an inverse element of a under multiplication? In other words, is there an element a^{-1} \in \mathbb{Z} such that:
a \times a^{-1} = a^{-1} \times a = 1?
Assume it is true with a \neq 1. Then it means that a^{-1} = \frac{1}{a} \in \mathbb{Z}. But then \frac{1}{a} is a rational number, which isn’t contained in the set of integers. So \mathbb{Z} can’t be a group under multiplication since for all integers (expect for the identity element 1) there is no inverse element.

A short counterexample you could give: 2 has no inverse element since \frac{1}{2} \not \in \mathbb{Z}.

Leave a Reply