Is the multiplicative group $\{-1,1\}$ isomorphic with $\{0,1\}$?

Question

In my opinions of a math ignoramus, $\{0,1\}$ is not a multiplicative group since $0$ has no inverse.

The above was written by a Reviewer. If I am right, how to "prove" that $\{0,1\}$ is not a multiplicative group?

Answer 1

A set is not a group. A set with a closed binary operation is a group (or at the very least, it might be a group; there are some requirements).

The set $\{1,-1\}$ equipped with (standard) multiplication is a group. The set $\{0,1\}$ equipped with modulo-2 addition is a group. These two groups are isomorphic.

$\{0,1\}$ equipped with multiplication is not a group, as it doesn't fulfill the required properties. There are three of these properties, it's not difficult to find the one it violates.

Answer 2

I'm assuming you are considering $\{0,1\}$ as a group with the operation of addition mod $2$ (i.e. isomorphic to $\mathbb{Z}_2$).

In that case the answer is yes, check that $f : \{-1,1\} \to \{0,1\}$ given by $f(x)=\frac12(1-x)$ is an isomorphism.

Answer 3

As sets, $\{-1,1\}$ and $\{0,1\}$ are isomorphic to each other - that is, they have the same cardinality ($2$). In the category (or 'universe') of sets, isomorphisms take the form of bijective functions. In order for two groups to be isomorphic, the underlying sets must have the same cardinality - so there is at least a group $(G,*)$ with $G=\{0,1\}$ such that $(G,*)\cong (\{-1,1\},\times)$.

That said, you need to specify the operation '$*$' to determine whether or not a particular group $(\{0,1\},*)$ is isomorphic to the multiplicative group $(\{-1,1\},\times)$. As stated by others, $\{0,1\}$ does not form a group under the usual multiplication operation.

To prove that $(\{0,1\},\times)$ is not a [multiplicative] group, you need to show that at least one of the group axioms fails.

If, for some other operation $*$ (for example, addition modulo 2), $(\{-1,1\},*)$ is a group, then it suffices to show that there is a bijective homomorphism $f:\{-1,1\}\to\{0,1\}$ - i.e...

$$\forall a,b\in\{-1,1\}.f(a\times b)=f(a)*f(b)$$

...for this group to be isomorphic to $(\{-1,1\},\times)$.

Edit:

The group $(\{-1,1\},\times)$ is isomorphic to the group $(\{0,1\}, \oplus)$, where $a\oplus b = (a + b)\mod2$ (this is the same as a + b % 2). This means that there is an isomorphism $f$ between $(\{-1,1\},\times)$ and $(\{0,1\}, \oplus)$.

Check:

$$f(x)=\frac{1}{2}(1-x)\implies f^{-1}(x)=1-2x$$

$f:\{-1,1\}\to\{0,1\}$ is a bijection

$$f(-1\times-1)=f(-1)\oplus f(-1)=0$$

$$f(-1\times 1)=f(-1)\oplus f(1)=1$$

$$f(1\times 1)=f(1)\oplus f(1)=0$$

$f:\{-1,1\}\to\{0,1\}$ is a homomorphism

$f(x)$ is a bijective homomorphism, therefore $f$ is an isomorphism