Does having an isomorphism between two could-be groups imply that they are groups?

Question

I know that the title is a bit of a word salad, but I couldn't find a better way to ask it concisely. Here is the question:

Let's say I have an operation $*$ such that for any $a,b \in \mathbb R$,

$$x * y = \sqrt[3]{(x^3 + y^3)}.$$

I also have an application $f: x \mapsto x^3$ that is an isomorphism of $(\mathbb R, *) \to (\mathbb R, +)$. How can I deduce that $(\mathbb R, *)$ is a commutative group from the fact that f is an isomorphism from $(\mathbb R, *) \to (\mathbb R, +)$?

I'm not really sure how to even go about this. I also assumed that the the source and destination of isomorphisms were groups by definitions, but I guess that's wrong? A point towards the right direction would be greatly appreciated, thanks!

Answer 1

I think the main point of confusion is that "isomorphism" here is being used in a broader context than you are used to. In this context we have, on the face of it, a magma: this is a fancy name for a set together with a binary operation. Note that every group is a magma, but not conversely: the set $\{0,1\}$ equipped with the binary operation $*$ which always outputs "$1$" is a magma but not a group (there is no identity element, since $x*0=1\not=0$ regardless of what $x$ is).

We now define an isomorphism of magmas. If $\mathcal{A}_1=(A_1, *_1)$ and $\mathcal{A}_2=(A_2, *_2)$ are magmas, then a magma isomorphism from $\mathcal{A}_1$ to $\mathcal{A}_2$ is a function $i: A_1\rightarrow A_2$ such that:

• $i$ is a bijection.

• For all $a, b\in A_1$, we have $i(a*_1b)=i(a)*_2i(b)$.

Note that this looks exactly like the definition of an isomorphism for groups! In general, every kind of mathematical structure has an associated notion of "isomorphism." (This theme is taken up in more detail in abstract algebra, and then developed even further in category theory.)

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Now let's look at the actual question. The point is that if we have two things (in this case, magmas) which are isomorphic, then they are "algebraically identical" - any algebraic property which one has, the other also has. In particular, if one is a group so is the other.

Here's a concrete example:

Suppose $(A_1, *_1)\cong (A_2, *_2)$ are isomorphic magmas and $(A_1, *_1)$ has an identity element. Then $(A_2, *_2)$ has an identity element.

Proof. Let $e$ be the identity element of $(A, *_1)$ and $i$ be some isomorphism from $(A_1, *_1)$ to $(A_2, *_2)$. I claim $i(e)$ is the identity element of $(A_2, *_2)$.

To see this, pick $b\in A_2$; we want to show $i(e)*_2b=b*_2i(e)=b$. Since $i$ is a bijection, $b$ "comes from" some element of $A_1$; let $a=i^{-1}(b)$. We now argue as follows:

• Since $e$ is the identity in $(A_1, *_1)$, we have $e*_1a=a$. Since $i$ is an isomorphism, we have $$i(a)=i(e*_1a)=i(e)*_2i(a).$$ But $i(a)=b$, so this tells us that $$b=i(e)*_2b.$$

• Similarly, since $e$ is the identity in $(A_1, *_1)$, we have $a*_1e=a$. Since $i$ is an isomorphism, we have $$i(a)=i(a*_1e)=i(a)*_2i(e).$$ But $i(a)=b$, so this tells us that $$b=b*_2i(e).$$

Now you want to show:

If $\mathcal{A},\mathcal{B}$ are isomorphic magmas and $\mathcal{A}$ is a group, then so is $\mathcal{B}$.

By the above argument, we've shown that the existence of identities "ports over" from $\mathcal{A}$ to $\mathcal{B}$. We now just need to check the other group axioms!

• Show that the associativity of $\mathcal{A}$ implies that $\mathcal{B}$ is associative.

• Show that the existence of inverses in $\mathcal{A}$ implies the existence of inverses in $\mathcal{B}$.

Once you've done this, you'll be finished.