Multiplicative group $(\mathbb R^*, ×)$ is group but $(\mathbb R, ×)$ is not group, why?

$\mathbb R^*$ refers to $\mathbb R$ without zero.

Please explain the statement "$\mathbb R$ is not a group under multiplication, it is a group under addition." in the comment. Basically:

1. Why is multiplicative group $(\mathbb R^*, ×)$ a group?

2. Why is $(\mathbb R, ×)$ not group?

3. Why are $(\mathbb R^*, +)$ and $(\mathbb R, +)$ groups?

Solution 1:

$(\mathbb{R}^*, \times)$ is a group, since

• It has a neutral element $1$
• It has inverses $1\over x$ for all $x \in \mathbb{R}^*$
• The operation is associative

$(\mathbb{R}, \times)$ is not a group, because $0$ has no multiplicative inverse.

$(\mathbb{R}, +)$ is a group, since

• It has a neutral element $0$
• It has inverses $-x$ for all $x \in \mathbb{R}$
• The operation is associative

$(\mathbb{R}^*, +)$ is not a group, since it is not closed under operation: $x+(-x)=0 \not \in \mathbb{R}^*$

Solution 2:

Something is a group if it satisfies the group axioms (https://en.wikipedia.org/wiki/Group_(mathematics)#Definition). So, for instance, $(\mathbb{R}, *)$ is not a group because one of the group axioms is the existence of inverses, and $0$ has no multiplicative inverse. Similarly, your (3) is partly wrong: $(\mathbb{R}^*, +)$ is not a group because (a) it is not closed under the operation $+$, and (b) does not have an identity for the operation $+$.

Showing that something is a group is slightly more complicated than showing that something isn't a group, since in the latter case you just have to exhibit a single counterexample; but looking at the group axioms, do you see how to verify them for the remaining examples?

Solution 3:

$(\mathbb{R},*)$ is not a group because there does not exist a multiplicative inverse for $0$. If you remove $0$, the problem is solved. Thus, $(\mathbb{R}^*,*)$ is a group. (I do not want to show the group axioms here). With addition, we do not have this problem since $0$ is inverse to itself under addition.