I have come across several groups, some of which have the same number of generating elements and of the same orders. Take, for instance, $D_{2n}$ and $S_n$. I have never seen it read explicitly, but it seems to me that many groups have some number of generating elements of a given order, and then also have some additional structure on top of this, such as the requirement that $sr=r^{-1}s$ for the Dihedral group, which allows the "object" to still satisfy the group axioms but it has a restricted set of allowed elements compared with the group generated by the elements with no restriction alone. Another example coming to mind is any Abelian group, which has this additional structure on it.

Are these properties (i.e. the number of generating elements, their orders, and any additional properties/constraints) necessary and sufficient to specify a group? I am trying to think in terms of isomorphisms; I would try to show two groups are 'the same' by showing that an isomorphism exists between them. onsidering this, it certainly seems sufficient that two groups are matched with these properties. The isomorphism can then simply map the corresponding elements on to each other. But is it necessary? Or, in other words, if two groups are not exactly matched in these properties, is there no isomorphism between them? Perhaps it also makes a difference if the groups are of infinite order. Then reerring to 'the number of generating elements' and 'their orders' seems a bit suspect...

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    $\begingroup$ We know $(\mathbb{R},+)\cong (\mathbb{C},+)$. We can't find any isomorphism between them. $\endgroup$ – 1ENİGMA1 Dec 13 '17 at 15:41
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    $\begingroup$ What you describe is that groups can be given py a presentation, e.g., $D_{2n}=\langle\, r,s\mid s^2=1, r^n=1, sr=r^{-1}s\,\rangle$. It is however not a trivial task to "see" an isomorphism between different presentations $\endgroup$ – Hagen von Eitzen Dec 13 '17 at 15:42
  • $\begingroup$ Also you can look there.math.stackexchange.com/questions/2247751/… $\endgroup$ – 1ENİGMA1 Dec 13 '17 at 15:43

Consider two groups $G_1$ and $G_2$, each of which is generated by two elements $a$ and $b$. In $G_1$, $a$ and $b$ satisfy these relations:

  • $a^2=e$;
  • $b^3=e$;
  • $aba^{-1}=b^{-1}$.

In $G_2$, $a$ and $b$ satisfy these relations:

  • $a^2=e$;
  • $b^2=e$
  • $(ab)^3=e$.

So, the two presentations are quite different. However, $G_1\simeq G_2$, because, in fact, both of them are isomorphic to $S_3$.

Here's another example. This time, I will provide two presentations of the group $\mathbb{Z}_6$:

  1. group generated by $a$ and the relation $a^6=e$;
  2. group generated by $a$ and $b$ and the relations $a^2=e$, $b^3=e$ and $ab=ba$.
  • $\begingroup$ @Arthur I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Dec 13 '17 at 16:02
  • $\begingroup$ Wow! But then what is the significance of an isomorphism? I agree that two groups being isomorphic still means there is a objective map between them, but I can't get my head around how this can be so without two groups being ' the same'. Actually, I read that mathematicians consider two groups to be the same if they are isomorphic. In what way are they the same? (This seems kind of like the donut/teacup concept in topology) $\endgroup$ – 21joanna12 Dec 13 '17 at 16:06
  • $\begingroup$ @21joanna12 I don't see in which way being generated by distinct sets of generators and relations makes them actually different. What we have here are two distinct descriptions of the same thing. It's like considering the sets$$\{\text{odd natural numbers}\}\text{ and }\{n\in\mathbb{N}\,|\,n\mid2^k-1\text{ for some }k\in\mathbb{N}\}.$$The descriptions are very different, but the sets are the same. $\endgroup$ – José Carlos Santos Dec 13 '17 at 16:14
  • $\begingroup$ Oh. I see now how the constraints on the elements given in the two examples are actually the same (of course, the labelling is different; b-> ab in the second case etc). In this case, the lements of the same orders still satisfy the same relations in both groups. I was wondering if two groups can be isomorphic without this being the case; i.e. the elements of corresponding orders do not satisfy the same relations between them, or indeed if two groups don't have the same number of elemets of the same orders (perhaps here with infinite groups?) $\endgroup$ – 21joanna12 Dec 13 '17 at 16:19
  • $\begingroup$ @21joanna12 Please take a look at my new example. $\endgroup$ – José Carlos Santos Dec 13 '17 at 16:38

Note that we can always add more (redundant) generators to the representation without changing the group, the easiest example is $\mathbb{Z}_{pq}$ for different primes. It can be defined as:

$$\mathbb{Z}_{pq}=\langle a \mid a^{pq}=e\rangle$$ $$\mathbb{Z}_{pq}=\langle a,b \mid a^{p}=b^{q}=e, ab=ba\rangle$$

I will now explain why this happens:

The problem with groups is that unlike vector spaces, it's not always possible to find a linearly independent set once you have a generating set. We have two important concepts in linear algebra:

  1. A generating/spanning set
  2. A linearly independent set

The first concept gives you the ability to write every element of your vector space as a linear combination of a few "generators". Notice that a vector space with its vector addition operation forms an Abelian group and therefore, $\vec{v} = \alpha_1 \vec{v_1} + \alpha\vec{v_2} + \cdots$ is in fact another way for writing $\vec{v}=\vec{v_1}^{\alpha_1}\cdot\vec{v_2}^{\alpha_2}\cdots$ in the additive notation instead of the "multiplicative notation". Well, we never use the multiplicative notation in this case but the point is that the concept of a generating set for a vector space is the same as a generating set for a group in essence.

It happens that in linear algebra when we are working over a field, every generating/spanning set contains a linearly independent subset. This gives us a unique representation for every element of the vector space. Failure of this property in an algebraic structure results in the existence of several different representations for the same element, which makes it tricky to define well-defined homomorphisms on it.

But if we indeed have a basis, i.e. a generating set that is linearly independent, we get a well-defined morphism just by determining where each generator should be mapped to. Because of the uniqueness of representation of an element, this gives us a well-defined morphism and we get to completely determine linear transformations (i.e. homomorphisms) easily.

What is missed in Group Theory is the concept of a basis. In fact, finite groups cannot have a basis over $\mathbb{Z}$. This leads to the concept of a free group which is way trickier than a vector space. Therefore, when you define $G$ by such a representation, you will not immediately have a way to see what other groups are isomorphic to it. However, I believe you can always add enough relations between generators to uniquely define a group, provided that one exists at all. For example, if $G_1$ and $G_2$ satisfy the same relations, but they are not isomorphic, there must be an equation in $G_1$ that does not hold in $G_2$ and adding that equation to our defining relations should separate these two.

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    $\begingroup$ @MorganRodgers: If you accept the axiom of choice or any equivalent form of it like Zorn's lemma, every vector space and in fact every module over a division ring has a basis. The proof is very simple in fact. You take an increasing chain of linearly independent vectors and at some point, this chain should stop. Then you'll show that it generates your module because otherwise, you'll get a contradiction by finding something that is not in this maximal linearly independent set. And indeed, $\mathbb{R}$ has a basis over $\mathbb{Q}$. See Hamel bases. $\endgroup$ – stressed out Dec 13 '17 at 17:03
  • $\begingroup$ "at some point, this chain should stop": well, your basis will be uncountably infinite, so it doesn't exactly stop, but I see what you mean. It also causes some issues with your implication that it is easy to "just" determine where each generator should be mapped to to get vector space isomorphisms. I don't know that this really ends up being any simpler (conceptually) than dealing with (minimal) generating sets for a group. $\endgroup$ – Morgan Rodgers Dec 13 '17 at 17:43
  • $\begingroup$ (but yes, I misspoke when I said it doesn't have a basis) $\endgroup$ – Morgan Rodgers Dec 13 '17 at 17:49
  • $\begingroup$ @MorganRodgers: Yes, that's indeed true that the basis will be uncountable, but if you aren't an intuitionist you won't be bothered by that. My point is that if we have linear independence, then because of the uniqueness of representing an element, we can define well-defined morphisms without going into the trouble of proving what we have defined is well-defined. Plus, isn't a minimal generating set conceptually the same as a maximal linearly independent set? $\endgroup$ – stressed out Dec 13 '17 at 18:00
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    $\begingroup$ I'm not an intuitionist, so I'm not bothered by it; and of course the morphisms will be well-defined. But it makes it a little complicated (not impossible) to determine whether two vector spaces with countably-infinite dimension are isomorphic. And yes, in some ways a minimal generating set is conceptually similar to a maximal LI set; your answer seemed to emphasize that things were different in groups because we don't have bases, so I'm trying to point out that there is a somewhat analogous object that we can work with. $\endgroup$ – Morgan Rodgers Dec 13 '17 at 18:13

You need to have the notion of isomorphism because two groups which are structurally "the same" can appear very different. But in the representations you have, you can have different numbers of generators with different orders. The point is that you can always find a generating set for each that "looks" the same.

As an example take the group $\mathbb{Z}_{6}$, the standard representation of this group is taking integers $\{0,1,2,3,4,5\}$ and addition is done mod 6.

Or you can take pairs $(a,b)$, where $a \in \mathbb{Z}_{2}$ and $b \in \mathbb{Z}_{3}$, so we are adding these like vectors but the first entry is then reduced mod 2, the second entry mod 3.

Or we can take the numbers $\{1,2,3,4,5,6\}$, using multiplication as the operation, and reducing the result mod 7.

We can write this group as $\{1, e^{2\pi \rm{i}/7}, e^{4\pi \rm{i}/7}, e^{6\pi \rm{i}/7}, e^{8\pi \rm{i}/7}, e^{10 \pi \rm{i}/7}, e^{12\pi \rm{i}/7} \}$ and use normal complex number multiplication as your operation.

You can even use powers of the matrix $$\begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 \end{bmatrix}$$ (this matrix will give the identity when you raise it to the 6th power)


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