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For example, to say that there are 2 such groups up to isomorphism such that the order of G is equal to $p^2$?

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    $\begingroup$ The phrase "up to isomorphism" means "any such things are isomorphic." For instance, any group of order $p^2$, where $p$ is prime, is isomorphic to one of two explicit examples (namely $\mathbf Z/(p^2)$ and $\mathbf Z/(p) \times \mathbf Z/(p)$. $\endgroup$ – KCd Apr 26 '15 at 2:02
  • $\begingroup$ "Isomorphism" means two groups are structurally the same and mathematicians "pretend" that they cannot distinguish between them at all, so "up to isomorphism" means we don't count the "same" group multiple times even if they are actually different objects. $\endgroup$ – MonkeyKing Apr 26 '15 at 2:07
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    $\begingroup$ Note that the title is not right: in the question it it is not the case that "something holds" up to isomorphism, but "things are counted" (namely groups are counted) up to isomorphism. I suggest you replace "for something to hold" in the title by "to count things". $\endgroup$ – Marc van Leeuwen Apr 27 '15 at 4:33
  • $\begingroup$ All the answers I've seen when looking around are regarding groups but what about in this context .. "A theory is called categorical if it determines a structure up to isomorphism" $\endgroup$ – Prince M May 22 '16 at 10:12
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A good example to add. "There is only one group of order 2 up to isomorphism". The group is $G=\{e,a\}$ with $ea=ae=a$ and $a^2=e$. Now you could argue "I have another group, $H=\{e,b\}$ with $eb=be=b$ and $b^2=e$". Your group IS different than mine, but your group is just the renamed version of mine. "Renaming" is isomorphism.

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On Isomorphism and Equality:

Intuitively, it means that there are two group structures of order $4$. But there's a problem: there are many non-equal groups of order $4$ (see below).

The proper way to formulate this idea is to say there are only two groups of order $4$ up to isomorphism (again, see below).


To illustrate, here are several groups of order $4$:

$\{0,1,2,3\}$ with the group operation addition modulo $4$

$\{(0,0),(0,1),(1,0),(1,1)\}$ with the group operation vector addition modulo $2$

$\{1, i, -1, -i\}$ with the group operation multiplication

$\{\textrm{leave as is}, \textrm{flip about the x axis}, \textrm{flip about the y axis}, \textrm{flip about the z axis}\}$ with group operation "composition of actions" (so e.g. flipping around the x axis and then flipping around the y axis is flipping around the z axis [try it!])

$\{alice, bob, carol, dave\}$ with group operation: alice is the identity, anyone plus themselves is alice, two non-alice people composed is the third non-alice person


The Upshot:

So, are any of these groups equal? No. They're all distinct groups (they're even distinct as sets), because their elements are not the same.

But the first and third are isomorphic, and the second, fourth, and fifth are isomorphic. The first and third work the same way ("have isomorphic group structures"), just with the elements renamed. Similarly for the second, fourth, and fifth.

But the first and the second are not isomorphic: the first one has an element which added to itself is not the identity. Not so for the second. So the first/third and the second/fourth/fifth are different isomorphism classes of groups.

Many mathematical statements of the sort "there are only two of these types of things" must be stated with "up to isomorphism" because equality is too strict: you're interested in how the group works, not what its elements actually are.

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More generally, "equal up to $\sim$", where $\sim$ is some first-order equivalence relation, means that we have temporarily redefined equality so that $x$ and $y$ are "equal" or "equivalent" if $x\sim y$. A property $P$ "holds up to $\sim$" or "holds modulo $\sim$" if $P$ depends only on the equivalence class.

(By first-order equivalence relation, I mean a formula at the first-order logic level which has the properties of an equivalence relation. If there is a set which contains all of the objects under study, then this technicality may be ignored, and a normal equivalence relation will do.)

Equivalences let us look at sets in a way which hides irrelevant detail. As Timothy Gowers wrote in Mathematics: A Very Short Introduction, "a mathematical object is what it does." In algebra we have isomorphisms, in topology we have homeomorphisms, in differential geometry we have diffeomorphisms, in algebraic topology we have homotopy equivalence, and in category theory we have natural isomorphisms. All of these preserve exactly the structure that can be measured when working with the set by some particular axioms. For instance, two groups are "the same" (equal up to isomorphism) if one cannot tell the difference between them when composing or inverting elements.

One great benefit to using equivalence rather than equality is uniqueness. Otherwise it would be more difficult to express something such as "there is exactly one group with seven elements [up to isomorphism]." Of course there are many groups of seven elements, but all of them are the same as far as being a group is concerned.

A metamathematical aside: when we speak of the basic concept of equality of sets, two sets are considered equal if and only if they have the same elements, and so they are "equal up to extensionality" (though they might "really" be different, but not in any way we care about).

A practical aside: computer programming languages give an interface for determining equivalence between two objects, and there are many ways to approach this. Some options include: do they occupy the same memory location? do they have the same representation? what does their decision procedure for determining equality say? Whatever choice is made determines in which mathematical universe the language computes.

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    $\begingroup$ Best answer IMHO, but could be more explicit as to what it means to "hold up to". The point is that, when we look at a universe through the lense of an isomorphism, all isomorphic elements "look the same". If two things A and B are identical and you tell me A is P, but B is not P, then your property P makes no sense. Similarly, in the question's example, if you show me infinite identical copies of my two groups and tell me they're more than 2, you're making no sense. $\endgroup$ – Nemo Apr 26 '15 at 18:38
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A quick thought I'd like to add to the other answers that one needs to keep in mind.

Sometimes we get so caught up in the equivalence defined by isomorphism that we actually forget that two isomorphic groups can be different groups and sometimes this makes quite a difference to one's discussion. For example, a group can have two subgroups which are isomorphic copies of one another within the supergroup, but this is quite different from the groups being the same. In relation to the supergroup, two isomorphic copies of a subgroup can have quite different properties: consider $\mathbb{R}\times\mathbb{T}^2$, the product of the real line and the torus, where group entities can be represented as 3-tuples $(x,\,e^{i\,\phi},\,e^{i\,\theta})$ and the group operation is defined by:

$$(x_1,\,e^{i\,\phi_1},\,e^{i\,\theta_1})\,(x_2,\,e^{i\,\phi_2},\,e^{i\,\theta_2}) = (x_1+x_2,\,e^{i\,(\phi_1+\phi_2)},\,e^{i\,(\theta_1+\theta_2)})$$

The the subgroups:

$$G = \{(x,\,1,\,1)|\,x\in\mathbb{R}\}$$ $$H(\alpha) = \{(0,\,e^{i\,\phi},\,e^{i\,\alpha\,\phi})|\,x\in\mathbb{R}\}$$

where $\alpha$ is irrational are isomorphic copies of one another: they are both isomorphic to the real additive group $(\mathbb{R},\,+)$. But $G$ is closed in the obvious topology of $\mathbb{R}\times\mathbb{T}^2$ (i.e. that for which a base is all sets of the form $(\mathcal{I}_1,\,e^{i\,\mathcal{I}_2},\,e^{i\,\mathcal{I}_3})$ where the $\mathcal{I}_j$ are open intervals in $\mathbb{R}$), whereas $H(\alpha)$ is not: its closure is the whole torus $\mathbb{T}^2$.


(to prove that $H(\alpha)$ is isomorphic to $(\mathbb{R},\,+)$, assume that there is a member $(0,\,e^{i\,\phi},\,e^{i\,\alpha\,\phi})$ equal to the identity $(0,1,1)$ other than the identity itself, i.e. that the "thread" $H(\alpha)$ links up with itself: then $\phi=2\,m\,\alpha\,\phi = 2\,n\,\pi$ for $m,\,n\in\mathbb{Z}$, contradicting the assumption that $\alpha$ was irrational)

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  • $\begingroup$ Agreed, this is important. Galois theory went over my head the first time I saw it, essentially because of this issue. $\endgroup$ – Callus Apr 27 '15 at 1:22
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The following is a very general discussion of the idea of an isomorphism without going into detail about what types of structures we are talking about (e.g., groups, rings, vector spaces, etc.) since the general idea should be the same across each of these.

What does it mean for two structures to be isomorphic to each other? It means that there is a way to map the elements of one structure bijectively to the other such that under this map, which identifies elements with each other, the behavior of the identified elements is the same.

For example, suppose I have two structures $G$ and $H$ (you can consider them groups if you like), and I say that $G$ and $H$ are isomorphic. Then that means I can find a bijection between the sets $G$ and $H$, call it $\phi : G \to H$, such that if $a, b \in G$, and we look at $\phi(a)$ and $\phi(b)$ in $H$, then via the isomorphism we know that if we look at the interaction of $a$ and $b$ in $G$, and call this interaction some new element $c$, then $\phi(a)$ and $\phi(b)$ will interact in $H$ exactly as $\phi(c)$. For example, let's consider groups: Suppose $\cdot$ is the operation of $G$ and $*$ is the operation of $H$. Then $a \cdot b = c$ implies $\phi(a) * \phi(b) = \phi(c)$, that is, if $a$ and $b$ interact in a way called $c$ in $G$, then $\phi(a)$ and $\phi(b)$ will interact exactly in the way $\phi(c)$ in $H$.

So, for all intents and purposes, two structures that are isomorphic are the same structure, even though they might live in different universes. For example, take the group of integers $\Bbb Z$ under addition, and consider the set $A = \left \{ \begin{bmatrix} c & 0 \\ 0 & 0 \end{bmatrix} \mid c \in \Bbb Z \right \}$. Under matrix addition, $A$ forms a group and it's actually isomorphic to $\Bbb Z$. To see this, let $\phi : \Bbb Z \to A$ be the map sending $c \to \begin{bmatrix} c & 0 \\ 0 & 0 \end{bmatrix}$.

Then if $a, b \in \Bbb Z$, we have $a + b$ is sent to $\phi(a + b) = \begin{bmatrix} a + b & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} a & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} b & 0 \\ 0 & 0 \end{bmatrix} = \phi(a) + \phi(b)$.

In other words, if $a$ and $b$ interact as $a + b$, then $\phi(a)$ and $\phi(b)$ will interact as $\phi(a + b)$ so that the structure is preserved. If you look far away, the two groups look the same. The only problem is one is a set of matrices and the other is a set of integers, so we can't strictly say they are equal. (Remember that to be isomorphic, you also need the map to be a bijection -- so check for yourself that $\phi$ is a bijection; it's a very easy exercise).

Now, when we say something like "there are only two structures up to isomorphism that do blah blah blah" what we mean is that if you look at any structure that does "blah blah blah", it is guaranteed that this structure will look exactly like one of the two structures listed in the statement. So saying "there are only two groups up to isomorphism of order BLAH" means if you look at any group of order BLAH, it can only have one of two possible structures. There isn't a third distinct structure we can find. So if we think somehow about all groups of order BLAH, we can divide them up into two distinct sets, where each set is the set of groups with the same structure.

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Say $\varphi: G\rightarrow G'$ is an group isomorphism. If $|G|=p^2$, then clearly $|G'|=p^2$. In particular, each element of $G$ corresponds to exactly one element in $G'$ [which exists by definition], and those two elements preserve the exact same properties in both groups. That is to say, if two groups are isomorphic under some operation, they are (basically) the same group [despite having different-looking elements] -- 'the groups structure is the same'. So you can't quite say $G$ and $G'$ are the exact same group. To get around this, you say '$G$ is $G'$ up-to-isomorphism.'

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When proving something like uniqueness, for example, by saying that something holds up to isomorphism, you mean that no mathematical entity which is not isomorphic to the entity that is unique can satisfy the given property. However, since isomorphic entities are essentially the same, they can be clubbed together and said to have that particular property.

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For any group with order $p^2$, there are two groups up to isomorphism: \begin{align} 1.) \ \ & \mathbb Z_{p^2}\\ 2.) \ \ & \mathbb Z_p \times \mathbb Z_p \end{align}

This means that any other group with order $p^2$ has the same structure as one these. If the group is cyclic, then it has the same structure as $\mathbb Z_{p^2}$. If it is not cyclic, then it has the same structure as $\mathbb Z_p \times \mathbb Z_p$.

In general, when two groups are isomorphic, this means that they behave exactly the same way. Isomorphism is transitive as well. If one group $G$ is isomorphic to another group $H$, and $H$ is isomorphic to $K$, then $G$ is also isomorphic to $K$.

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