Question about one of the definitions of an isomorphism between two binary algebraic strutures

Question

From Fraleigh's Introduction the Linear Algebra, there is one particular characterization of an isomorphism that I am confused on.

Have $\langle S, *\rangle$ and $\langle\dot S, \dot *\rangle $ be two binary structures. It makes intuitive sense to me that an isomorphism $\phi: S \rightarrow \dot S$ must be bijective (one-to-one and onto). However, there is one other property that I am confused about. Choose any $a,b \in S$ and any $\dot a, \dot b \in \dot S$. Then: $$ \phi(a *b) = \phi(a) \space\dot * \space \phi(b) = \dot a \space \dot * \space \dot b$$ While this property was very useful for solving some of the exercises in the book and proving whether functions are isomorphisms, it doesn't make intuitive sense to me why this property is mandatory for defining an isomorphism.

Any intuitive answer or proof would be greatly appreciated. Thank you in advance.

Answer 1

We can think of "the structure of a group $S$" as its image in $\operatorname{Sym}(S)$ via left multiplication, say $\epsilon(S)$. After all, this set encodes the whole information of $S$ as the group it is. So, if we want to compare the structures of two distinct groups, $\dot S$ and $S$, then we need "to transport" somehow $\dot\epsilon(\dot S)$ into $\operatorname{Sym}(S)$ (or $\epsilon(S)$ into $\operatorname{Sym}(\dot S)$), being $\operatorname{Sym}(\dot S)\cap\operatorname{Sym}(S)=\emptyset$. Intuition is quite personal, but I find "intuitive" to say that $\dot S$ and $S$ are isomorphic (=have the same structure) if and only if $$\forall a\in S,\space\epsilon_a=\text{ "transported }\dot\epsilon_{\phi(a)}"\tag 1$$ where $\phi\colon S\to \dot S$ is a bijection and the "transport" from $\operatorname{Sym}(\dot S)$ to $\operatorname{Sym}(S)$ is (non-canonically) provided by the map $\psi^{(\phi)}$, defined by $\sigma\mapsto\phi^{-1}\sigma\phi$: $$\require{AMScd} \begin{CD} S@>{\phi}>> \dot S\\ @VV{\epsilon}V @VV{\dot\epsilon}V \\ \operatorname{Sym}(S) @<{\psi^{(\phi)}}<< \operatorname{Sym}(\dot S) \end{CD}$$ Now, the "isomorphicity" condition $(1)$ is equivalent to requiring that the above diagram commutes: \begin{alignat}{1} \epsilon &= \psi^{(\phi)}\dot\epsilon\phi &&\iff \\ \forall a,b\in S, \space\epsilon_{a}(b) &=\phi^{-1}(\dot\epsilon_{\phi(a)}(\phi(b))) &&\iff \\ \forall a, b\in S, \space ab &=\phi^{-1}(\phi(a)\phi(b)) &&\iff \\ \forall a,b\in S, \space\phi(ab) &=\phi(a)\phi(b) \\ \end{alignat} Therefore, the isomorphicity between $S$ and $\dot S$ takes place precisely when (and only when) $\phi$ fulfils your "unintuitive" property, which hence rightly justifies the name "isomorphism" for $\phi$.

Answer 2

For the bijection $\phi:S\to\dot S$, that defining property of isomorphism can be rephrased more intuitively as:

For all elements $a,b,c$ in $S$: $a*b=c$ in $S$ if and only if $\phi(a)\mathop{\dot *}\phi(b)=\phi(c)$ in $\dot S$.

In other words, $a$ and $b$ multipy to $c$ in $S$ if and only if this is also true for the images under $\phi$ in $\dot S$. Intuitively, this captures the fact that $\phi$ respects or preserves the multiplicative structure between $S$ and $\dot S$.

Answer 3

Here is an example to illustrate the general idea.

Define a group $(G,*)$ with elements $G = \{1,x,y\}$ using the following multiplication table.

$$(G,*):\begin{array}{c|ccc} * & 1 & x & y \\ \hline 1 & 1 & x & y \\ x & x & y & 1 \\ y & y & 1 & x \end{array}$$

Then define another group $(G',*')$ with elements $G' = \{1',x',y'\}$ using the following multiplication table.

$$(G',*):\begin{array}{c|ccc} *' & 1' & x' & y' \\ \hline 1' & 1' & x' & y' \\ x' & x' & y' & 1' \\ y' & y' & 1' & x' \end{array}$$

Hopefully you will agree with me that these are basically the same group, just the labels of the elements are different. So you would expect that they are isomorphic, and indeed they are: the map $1 \mapsto 1'$, $x \mapsto x'$, $y \mapsto y'$ is an isomorphism.

Now suppose that you have another group $(H, \circ)$ which is isomorphic to $(G,*)$.

What does that mean? Well, you have an isomorphism $f: G \rightarrow H$. In particular $f$ is a bijection, so $H = \{f(1), f(x), f(y)\}$.

How do you multiply elements of $H$? Well, for example $f(x) \circ f(y) = f(x * y) = f(1)$ and $f(x) \circ f(x) = f(x * x) = f(y)$ since $f$ is an isomorphism. So the multiplication table is

$$(H,\circ):\begin{array}{c|ccc} \circ & f(1) & f(x) & f(y) \\ \hline f(1) & f(1) & f(x) & f(y) \\ f(x) & f(x) & f(y) & f(1) \\ f(y) & f(y) & f(1) & f(x) \end{array}$$

So as you see, $(H,\circ)$ is basically the same as $(G,*)$, just the labels of the elements are different. The new "labeling" of the elements is given by the isomorphism $f$.

All of this generalizes to isomorphisms between binary structures. If two binary structures are isomorphic, they literally have the same binary operation, just with different names for the elements.