Cancellation Rule for Multiplication with 0 Preliminaries:
From basic group theory we have the following Definitions:
Definition (Group):
A group $(G,*)$ consists of a set $G$ together with an operation $*$ on $G$, defined as
\begin{align*}
*:G\times G\rightarrow G,\quad (a,b)\mapsto a*b
\end{align*}
for all $a,b\in G$, satisfying the following properties:

*

*Associativity: The operation $*$ is associative, i.e.,
\begin{align*}
(a*b)*c=a*(b*c),
\tag{G1}
\end{align*}
for all $a,b,c\in G$.

*Identity Element: There exists an element $e\in G$, such that
\begin{align*}
e*a=a*e=a,
\tag{G2}
\end{align*}
for all $a\in G$.

*Inverse Element: For each $a\in G$, there exists an element $a^{1}$ (called the inverse of $a$), such that
\begin{align*}
a^{-1}*a=a*a^{-1}=e.
\tag{G3}
\end{align*}
Definition (Abelian Group):
A group $(G,*)$ is called abelian if $*$ is commutative, i.e.,
\begin{align*}
a*b=b*a
\tag{G4}
\end{align*}
for all $a,b\in G$.
We then have the following Theorem (Cancellation Laws):
Theorem (Cancellation Laws):
Let $(G,*)$ be a group and let $a,b,c\in G$. Then the following cancellation rules hold:
\begin{align*}\tag{1}
a*c=b*c\Rightarrow a=b
\end{align*}
and
\begin{align*}\tag{2}
c*a=c*b\Rightarrow a=b
\end{align*}
I know we can prove that Theorem using the inverse element $a^{-1}\in G$.
Finally, we have the following Definition:
Definition (Field):
A field $(\mathbb{F},+,\cdot)$ consits of a set $\mathbb{F}$ together with two internal binary operations (addition $+$ and multiplication $\cdot$), satisfying the following properties:

*

*$(\mathbb{F},+)$ is an abelian group with identity element $0\in\mathbb{F}$

*$(\mathbb{F}\setminus\{0\},\cdot)$ is an abelian group with identity element $1\in\mathbb{F}$

*Distributivity: For all $a,b,c\in \mathbb{F}$, the following equations holds:
\begin{align*}
a\cdot (b+c)=a\cdot b+a\cdot c
\\
(b+c)\cdot a=b\cdot a + c\cdot a
\end{align*}
Question:
Now, let's assume * is the usual multiplication we know from the field $(\mathbb{R},+,\cdot)$ of real numbers (or any other field I guess). There are two cases:

*

*$c\neq 0$: In this case the Cancellation Law holds.

*$c=0$: The Cancellation Law obviously doesn't hold for $c=0$ since $a*0=0$ and $b*0=0$ (respectively $0*a=0$ and $0*b=0$) for all $a,b\in G$.

I conclude that the Cancellation Law holds for $G=\mathbb{R}\setminus\{0\}$ but not for $G=\mathbb{R}$.
Is the above Theorem not entirely accurate formulated or what did I miss? I guess, if the Cancellation Law holds for all groups, then $(\mathbb{R},\cdot)$ can't be a group, but why? From the definition of fields we know that $(\mathbb{R}\setminus\{0\},\cdot)$ is an abelian group, but how do we conclude that $(\mathbb{R},\cdot)$ isn't a group? Or am I guessing wrong and there is a completely different explanation for this?
Thanks in advance for trying to help me with this question.
 A: Here's an alternative approach.

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

Proof: Suppose $x\in G\setminus \{e\}$ is an idempotent. Then, by definition, $xx=x^2=x=ex.$ Now just apply the right cancellation law to get $x=e$, a contradiction. $\square$
Since $0\times 0=0$ and $1\times 1=1$, then, $(\Bbb R,\times)$ cannot be a group.
A: Right you are.  $\Bbb R$ is a field; but no one said that the real numbers under multiplication is a group.  They are different statements.
Just like $\Bbb Z$ isn't a group under multiplication, because there are not multiplicative inverses, and $\Bbb Q\,,\Bbb C$ and $\Bbb Z/(p\Bbb Z)$ are fields because they are commutative rings such that every nonzero element is invertible, just to take a  few examples.
Actually there is a finite field $\Bbb F_{p^n}$ of order $p^n$ for $p$ prime and each natural $n$, and $\Bbb F_{p^m}\subset \Bbb F_{p^n}$ is a subfield if and only if $m|n$.  And that is all the finite fields.  Also interestingly, the multiplicative group of a finite field is cyclic.
You have to be careful.
