Is $\langle \mathbb{R}^\ast, \ast \rangle$ Where $\mathbb{R}^\ast$ Are the Real Numbers Except $0$ and $a \ast b = |a|b$ a Group?

Question

Going by the definition of a group, that being that:

1. For all $a,b \in G$, $a \ast b \in G$.
2. For all $a,b,c \in G$, $a \ast (b \ast c) = (a \ast b) \ast c$.
3. There exists an element $e$ such that for all $a \in G$, $a \ast e = e \ast a = a$.
4. There exists an element $a^{-1} \in G$ for all $a \in G$ such that $a \ast a^{-1} = a^{-1} \ast a = e$.

$\langle \mathbb{R}^+, \ast \rangle$ obviously satisfies the first requirement since multiplication is closed under $\mathbb{R}$. As for the second condition, it is associative since, for $a \ast (b \ast c) = (a \ast b) \ast c$ that gives $|a||b|c = |ab|c$ which is true. The third requirement is where this gets iffy. There is only one number that satisfies the entire equation, but there are two that satisfy it if only the left-sided identity is considered. The number 1 is the multiplicative identity and it works as the identity for $\ast$ as well, however, the left-sided identity is both $1$ and $-1$. Then for the inverse, there is no solution for, $-3|x| = 1$, but if the identity were $-1$ instead, then there is a solution. I am pretty confused as to whether this is a group or not, but I imagine that it wouldn't be. I just want to know, would this be considered a group?

Answer 1

No, it would not be. As you said, if $e$ is such that $e*a=|e|a=a$ for all $a$, then $|e|=1$, or $e=\pm 1$.

If $e=1$, $(-1)*e=1\neq -1$, and if $e=-1$, $1*e=-1\neq 1$. Hence, there are no units in this system.

However, this system would be considered a semigroup, which only requires the first two axioms you list.

Answer 2

This operation has no right-identity, i.e. there isn't any $e$ such that, for all $x$, $x*e=x$. In fact, such an $e$ should satisfy $-1=(-1)*e=e$ and $1=1*e=e$.

Therefore, this isn't a group.

On the other hand there are two left-identities: $1$ and $-1$.

Answer 3

Take $a=2$ and $a=-2$ seperately and find the identity element in both the cases. You do not have unique identity element. It is not a group.