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The following image shows a screenshot from a page of my lecture notes:

screen shot

As you can see, it keeps using the expression "up to isomorphism", but doesn't explain what this means. Can anyone tell me what this means in this context?

Thanks

M Smith
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    see: http://math.stackexchange.com/questions/1252081/what-does-it-mean-for-something-to-hold-up-to-isomorphism – Emilio Novati Dec 19 '15 at 20:55
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    It is about counting things. It means if two groups are isomorphic, we should only count this once (indeed we could label a group in infinitely many ways, and have "infinitely many" copies of essentially the same group, so we want to count them "up to isomorphism"). – hardmath Dec 19 '15 at 21:01
  • Can you post a link to the full lecture notes? – Olivier Dec 19 '15 at 21:02

4 Answers4

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Ok, let's take an example from those notes.

It says "Up to isomorphism, the groups of order $6$ are the cyclic group $C_{6}$ and the dihedral group $D_{3}$."

This means if $G$ is any group of order $6$, it must be the case that either $G$ is isomorphic to $C_{6}$ or to $D_{3}$. $G$ might be a completely different set living in a completely different universe than either $C_{6}$ or $D_{3}$, but we are assured by the theorem that regardless of this fact, the behavior of the elements of $G$ will either be exactly like those of $C_{6}$ or those of $D_{3}$.

That's what the word isomorphism means, anyway. If two groups $G$ and $H$ are isomorphic, they could live in completely different universes and have totally different looking elements and different operations. But we are assured that there is some way to identify the elements of $G$ and $H$ in a one-to-one correspondence so that they look and behave exactly like the same group, even though they superficially look different.

layman
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  • Thanks for the response. I still seem to be slightly confused by what you're saying though. You've said that "if $G$ is any group of order 6, it must be the case that either $G$ is isomorphic to $C_{4}$ or $D_{3}$". Did you mean to also say that the group $G$ may instead be the case that $G$ is not isomorphic? Also (from the notes), what does it mean to say that a group is "unique up to isomorphism"? – M Smith Dec 19 '15 at 23:40
  • @user2910074 I meant we cannot find a group $G$ of order $6$ satisfying the following: a) $G$ is not isomorphic to $C_{4}$ and b) $G$ is not isomorphic to $D_{3}$. Such a group cannot exist. Each group of order $6$ will either be isomorphic to $C_{4}$, or, if not that, then isomorphic to $D_{3}$ for sure. – layman Dec 19 '15 at 23:45
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    @user2910074 Also, "unique up to isomorphism" just means we can probably find examples of other groups that live on other universes that satisfy the same property, but if we are able to find such groups, they will be isomorphic to our original group. For example, if I said "there is only one group $G$ up to isomorphism satisfying property $p$", then that means if you find any other group $H$ satisfying property $p$, you will automatically know $H$ is isomorphic to $G$. We would never be able to find an $H$ satisfying $p$ that is not isomorphic to $G$. – layman Dec 19 '15 at 23:47
  • So, for a group of order 6, an isomorphism ALWAYS exists with one of those two groups? – M Smith Dec 19 '15 at 23:55
  • @user2910074 Yes, that's exactly what it means to say the words "up to isomorphism". And the proof should show that this is true. The proof should start with an arbitrary group $G$ of order $6$ and how that if it's not isomorphic to $C_{4}$, then it has to be isomorphic to $D_{3}$. – layman Dec 19 '15 at 23:57
  • @user2910074 Glad I could help! :) – layman Dec 19 '15 at 23:58
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This just means that the only groups you can have will be isomorphic to one given. Notice that we can define the "same" group simply be renaming its elements but in that case it is not the same group, but is isomorphic. This is all this accounts for and in algebra is what we think of as "equality".

Michael N
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Say you want to classify groups of order one. For any set with one element, there is a unique group structure on it. So every set with one element can be made into a group with one element in a unique way. In a manner of speaking, that means there is a group of order one for every possible "thing" that can be in a set. There is "too infinitely many things" to consider here - there is no set of all groups of order one, they instead form a "proper class." But intuitively, all of these groups of order one are "the same group," just as two giraffes are the same animal, and the psychological act of thinking of isomorphic groups as "the same group" is treating groups "up to isomorphism."

anon
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Here is an example. For every $a\ne b\in \mathbb R$, let $G_{(a,b)}$ denote the set $$\{a,b\}$$equipped with the binary operation $*_{(a,b)}$ given by$$a\ *_{(a,b)}\ a=a\\a\ *_{(a,b)}\ b=b\\b\ *_{(a,b)}\ a = b\\b\ *_{(a,b)}\ b=a$$

A simple check shows that this turns $(G_{(a,b)},*_{(a,b)})$ into a group of order $2$ with identity $a$.

But group theoretically, these groups are all exactly the same. They have the same order, their elements have the same structure - everything about their group structure is identical. Effectively these groups are all exactly the same, except we've called them and their elements different names.

This "sameness" is captured by the idea of isomorphism. If $a',b'\in \mathbb R$, then there is an isomorphism $$G_{(a,b)}\to G_{(a',b')}\\a\mapsto a'\\b\mapsto b'$$

Saying that we want to classify all groups "up to isomorphism" means, for example, that we consider all the $G_{(a,b)}$ to be the same group. Since it is the group structure that we're interested in, it makes sense to consider all the $G_{(a,b)}$ as the same group, because they are all isomorphic. It turns out there is exactly one group structure that can be given to a set with $2$ elements.

Mathmo123
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