Is there a property that is not preserved by isomorphism? [closed]

Solution 1:

Yes an no.

No: The point of introducing the notion of an isomorphism is exactly that it should preserve all the relevant structure. If, say, $G$ and $G'$ are isomorphic groups, then every sentence you can write in the language of group theory will be true for $G$ if and only if the same sentence is true for $G'$ (i.e. you cannot distinguish between $G$ and $G'$ if your vocabulary includes only the group operation, variables, the identity element and basic logic). You might want to look up and introduction to category theory for more details.

Yes: If you go outside the language relevant to the isomorphism you have in mind, then surely you can find lots of properties which will not be preserved. The groups $G = \mathbb{Z}/4\mathbb{Z}$ and $G' = (\{1,i,-1,-i\}, \cdot)$ are isomorphic, but one of then contains $i$ while the other one does not (it does contain the equivalence class $1 + 4 \mathbb{Z}$, but that's not quite the same thing). Of course, this is a trivial example, but in general isomorphism of groups will preserve the properties that apply to groups, but not any other type of properties.

Edit to add: It is worth also pointing out that there are various species of isomorphisms: group isomorphisms, isomorphisms of topological spaces (aka homeomorphisms), isomorphisms of topological groups, and so on.

Consider a triangle and a disc: they are isomorphic as topological spaces, but not as metric spaces (i.e. you can continuously deform a triangle into a circle, but you cannot do so while preserving distances between pairs of points).

Another example which I like is to take two different representations of the additive group $\mathbb{R}$: one is just the usual one, and the other one is $G = \{(t \bmod 1,\sqrt{2}t \bmod 1) \ : \ t \in \mathbb{R} \}$, equipped with addition modulo $1$. Then $(\mathbb{R},+)$ and $(G,+)$ are isomoprhic as groups. However, as topological spaces they look very different!

Solution 2:

If I understand it correctly, your question contains its own answer: consider the groups $(\Bbb R, +)$ and $\Big( (0, \infty), \cdot \Big)$, and the isomorphism $\exp : \Bbb R \to (0, \infty), \ \exp(x) = \textrm e ^x$. Choose now a property that has nothing to do with the algebraic operation, for instance choose some topological property: endow $\Bbb R$ with the trivial topology $(\emptyset, \Bbb R)$ and $(0, \infty)$ with the discrete topology. Then $\Bbb R$ with this topology will be compact and non-Hausdorff, while $(0, \infty)$ with the discrete topology will be non-compact and Hausdorff. Despite these tremendous topological differences, the two groups are algebraically isomorphic. So yes, there do exist properties that are not preserved by isomorphisms.

Solution 3:

This question is tautological, insofar that an isomorphism of a structure, by definition, preserves that structure.

That said, there are basically two potential issues here:

The first is that one can have an isomorphism between two structures on different underlying sets. A simple exactly is the permutation group $S_{\{A, B, C\}}$ on the letters $A, B, C$ and the group of symmetries $T$ of an equilateral triangle. One can define a group isomorphism $\phi : S_{\{A, B, C\}} \to T$ by labeling the corners of the triangle by $A, B, C$, and then declaring the image of a permutation to be the unique symmetry of the triangle that permutes the vertices according to the labels. Of course, the elements of $S_{\{A, B, C\}}$ are not those of $T$, so $\phi$ does not preserve the 'property' of membership in $S_{\{A, B, C\}}$.

Second, it is certainly possible to have an object with some structure and a map that preserves some of the structure but not all of it. A simple example: Define the partially ordered set $S := (\{0, 1\}, \prec )$, where declare $0 \prec 1$. Then, the map $\phi : S \to S$ defined by $\phi(0) = 1, \phi(1) = 0$ is an isomorphism of sets (this just means that $\phi$ is a set bijection), but not an isomorphism of partially ordered sets, as $0 \prec 1$ but $\phi(0) \not\prec \phi(1)$. (Of course, since $\phi$ is a map $S \to S$, it does preserve the 'property' of membership of $S$.)

The latter distinction can sometimes be subtle: For example, there is an isomorphism between $\Bbb R$ and $\Bbb R^2$ as vector spaces over $\Bbb Q$, but no such map is an isomorphism of real vector spaces, as $\Bbb R$ and $\Bbb R^2$ have different dimensions as real vector spaces.

Solution 4:

Names, basically. Otherwise an isomorphism indicates that the structures are identical (/compatible). If two (sets/groups/rings/...) are isomorphic, you can basically just think of them as being the same structure, with differently-named elements.