## Structures #2

For an introduction, see my previous post, “Structures #1”.

Throughout, I will denote the set of real numbers $R$, the set of complex numbers $C$, the set of rationals $Q$. Furthermore, $R^*$, $C^*$, and $Q^*$ will represent the set of nonzero elements of those sets. (Zero causes problems when it comes to multiplicative groups, because it has no inverse.)

So I’m going to throw on some more “morphism” terminology just for fun.

• An automorphism is an isomorphism from a structure onto itself.
• An epimorphism is a surjective (“onto”) homomorphism.
• A monomorphism is an injective homomorphism.

So we can see that isomorphisms are precisely the monomorphisms which are also epimorphisms, or the epimorphisms which are also monomorphisms. (At this point, I’m more or less trolling. No one would ever refer to them like that.)

Now, I’ll show you why isomorphisms are important.

• Isomorphisms map identity elements to identity elements.
• They also map inverses to inverses.
• They also preserve structural properties like associativity and commutativity.

For general interest, I’ll prove the last statement. Suppose $(E, \oplus)$ is an algebraic structure satisfying associativity (these are called semigroups). Suppose also that it is isomorphic to $(F, \otimes)$. Then I claim that $(F, \otimes)$ is also a semigroup.

To see this, let $x,y,z \in F$. Since the structures are isomorphic, there exists some isomorphism $\psi$. Note that by the surjectivity of $\psi$, there exist elements $u, v, w \in E$ so that $\psi(u) = x$, $\psi(v) = y$, $\psi(w) = z$. Then we note that

$x \otimes (y \otimes z) = \psi(u) \otimes (\psi(v) \otimes \psi(w))$

by our remark. OK, but now since $\psi$ is a homomorphism this becomes

$\psi(u) \otimes \psi(v \oplus w) = \psi(u \oplus (v \oplus w))$

Hehe. Now we can just apply the associativity of $E$, and rip it apart again:

$\psi((u \oplus v) \oplus w) = (\psi(u) \otimes \psi(v)) \otimes \psi(w)$

Which of course is equal to

$(x \otimes y) \otimes z$.

Isn’t that beautiful? You are probably beginning to see now why the concept of isomorphism is really what we need to gauge whether two things are structurally identical. In the proof, we not only used the fact that $\psi$ was a homomorphism, but also that it was surjective. Injectivity, too, is important, as we are about to see.

Question. Is $(C^*, \cdot)$ isomorphic to $(R^*, \cdot)$?
Idea. After thinking about it for a bit, you’ll realize that in the former group, there are actually four fourth-roots of 1, whereas in the latter, there are only two fourth-roots of 1. Namely, in $(C^*, \cdot)$ we have that $i^4 = (-i)^4 = 1^4 = (-1)^4 = 1$. But in the latter, only $1^4 = (-1)^4 = 1$. This is a structural difference between these two groups, and so it seems intuitive that they should not be isomorphic. This is indeed the case. Suppose $\psi$ is an isomorphism. Then $\psi(i)^4 = \psi(-i)^4 = \psi(-1)^4 = \psi(1)^4 = \psi(1^4) = \psi(1) = 1$. So by the injectivity of $\psi$ we would be seeing four distinct fourth-roots of 1, all lying in $R^*$. That’s impossible; there are only two. This is a contradiction. Hence no such isomorphism could possibly exist, and we conclude that the two groups are not isomorphic.

Question. Is $(R^*, \cdot)$ isomorphic to $(R, +)$?
Idea. Again I claim they are not isomorphic. This time we consider the set of solutions to the equation $x \oplus x = e$. In the setting of the first group this translates to $x \cdot x = 1$, or in other words $x^2 = 1$. In the setting of the second group this would be written $x + x = 0$, or in other words $2x = 0$. (You can see that multiplication is to an additive group as exponentiation is to a multiplicative group. In the abstract, the same general thing is going on: the binary operation is being iterated on a single element.) Certainly the equation $x^2 = 1$ is solved both by $1$ and $-1$. Suppose we had some isomorphism. Then by its injectivity we would obtain two distinct solutions to $x + x = 0$, which is absurd; 0 is the only solution to this equation. This completes the proof.

Exercise for you. Prove that $(Q, +)$ is not isomorphic to $(Q^*_+, \cdot)$ where $Q^*_+$ are the strictly positive rationals. [Hint: $\sqrt{2}$ is irrational.]

For a nice overview of abstract algebra, check out the book Modern Algebra. It has been criticized since it is quite dense and substantial, but it’s what I’m currently reading. If you’re really bent on going intensely into one branch of abstract algebra such as group theory, you might want to look elsewhere, but it gives a surprisingly appreciable depth of many different branches.