5
$\begingroup$

I was wondering about the ways we say something is a group. For example, the set $$G = \{z \in \mathbb{C} \mid z^n = 1 \text{ for some }n \in \mathbb{Z}^+\}$$ is a group under multiplication. But if I were to just say "Assume that $G = \{1,a,b,c\}$ is a group of order $4$," how can we not specify the binary operation? Is this just saying there exists a binary operation for which it is a group?

Edit: Here is a definition of a group that seems to show what I am saying:

enter image description here

$\endgroup$
  • 10
    $\begingroup$ The way that definition in the image is phrased is simply horrid. A group is a pair (G,*) of a set and an operation satisfying certain conditions. It is difficult not to suggest you to pick a different textbook. $\endgroup$ – Mariano Suárez-Álvarez Oct 29 '16 at 7:26
  • 1
    $\begingroup$ Completely agree with Mariano. Please quit that book and get a good one. $\endgroup$ – Martin Argerami Oct 29 '16 at 13:39
  • 1
    $\begingroup$ According to that definition, every set is a group. $\endgroup$ – Federico Poloni Oct 29 '16 at 20:00
9
$\begingroup$

In general, you are right. A group always has to be specified along with a binary operation, e.g. $(G,\times)$. But in many situations, the binary operations are understood. I'll give a few examples:

  • $\mathbb{C}\backslash \{0\}$ would be a group with multiplication as its binary operation, because if it were addition, $0$ would be in the group.
  • Group of 2x2 invertible matrices. In this case, we have either matrix multiplication or matrix addition as a choice for binary operation. But matrix addition would mean that the identity element is the zero matrix, which is not invertible.

Some others are just by convention, such as $\mathbb{Z}/n\mathbb{Z}$, the group of integer modulo n, which has the binary operation $+$. As compared to the multiplicative group $(\mathbb{Z}/n\mathbb{Z})^*$, which has the binary operation $\times$.

That said, if there is any ambiguity at all, it is still necessary to state the binary operation.

$\endgroup$
5
$\begingroup$

Yes. If you are assuming $G$ is a group, then it certainly must have the properties of a group by assumption. So, in particular, you are indeed assuming that $G$ must have such a binary operation, even if you do not explicitly state what it is.

$\endgroup$
1
$\begingroup$

When an author talks of 4-element groups without specifying group operation, the statements made he is going to make expected to be valid for all groups of order 4 (there are just two actually). Then there is no need to go into specifics.

This not exactly about groups. When we make similar statements like "Let $X$ be topological space". The topology is not mentioned. So we do not know if it is a compact space or not. Do not know if connected or not. But the type of statement that author makes in that context is valid for all those cases. So that information is not needed.

Of course in your case there is one specific issue involved. Is there a group of order 4? Or else the whole discussion would be pointless. This, the author is expected to justify. But again in this case possibly the existence of finite groups of any given order was already established.

$\endgroup$
1
$\begingroup$

When you tell me "$G = \{1,a,b,c\}$", you have told me the operation is (if it is written at all, to be written as) multiplication (and that I should not assume the group is Abelian). You have told me this because you have written the identity element of the group as "$1$".

If instead you had told me "$G = \{0,a,b,c\}$", you would have told me that the group operation is (to be written as) addition (and that this group is likely Abelian). You have told me this because you have written the identity as "$0$".

Of course, one could write an abelian group multiplicatively and a nonAbelian group additively. One can do any number of things that are terrible ideas. Since I write so as not to confuse my reader, I wouldn't do either of those things.

I say "(to be written as)" in both cases, because there is nothing intrinsic about the operation that is used here. We could use any consistent symbol to write the group operation.
$$ a \smile b, a \ast b, a \star b, a \oplus b, a \wr b, a \wedge b, \dots $$ In a group with operation written as multiplication, it is standard to not even write the operation; the group elements $\beta^{2}\alpha^{-1}$ and $\delta\beta\alpha$ recently showed up in my work. (These are written multiplicatively. The group is very non-Abelian (trivial center).)

$\endgroup$
-2
$\begingroup$

"What is a group? Algebraists teach that this is supposedly a set with two operations that satisfy a load of easily-forgettable axioms. This definition provokes a natural protest: why would any sensible person need such pairs of operations? "Oh, curse this maths" - concludes the student (who, possibly, becomes the Minister for Science in the future).

We get a totally different situation if we start off not with the group but with the concept of a transformation (a one-to-one mapping of a set onto itself) as it was historically. A collection of transformations of a set is called a group if along with any two transformations it contains the result of their consecutive application and an inverse transformation along with every transformation.

This is all the definition there is. The so-called "axioms" are in fact just (obvious) properties of groups of transformations."

From V. Arnol'd's article "On teaching mathematics", http://pauli.uni-muenster.de/~munsteg/arnold.html

$\endgroup$
  • 5
    $\begingroup$ You should not treat everything what Arnold ever said as the ultimate truth, including this case. There are "naturally occurring" groups which do not immediately present themselves as transformation groups, and Arnold was well aware of this. Few examples (from topology): Homology groups, homotopy groups, K-theory, etc. $\endgroup$ – Moishe Kohan Oct 29 '16 at 3:32
  • 1
    $\begingroup$ @MoisheCohen: Arnold makes this remarks on "Teaching of Mathematics". Typically a student who starts groups/abstract algebra with definition of groups is mostly going to study $S_n$, finite groups$, subgroups of real numbers, groups in modular arithmetic, and groups of matrices, groups of symmetries/rigid motions etc. So it is a valid viewpoint. Even group actions make sense if understood as groups of transformations. $\endgroup$ – P Vanchinathan Oct 29 '16 at 7:11
  • 4
    $\begingroup$ This isn't an answer to the question at all... $\endgroup$ – Eric Wofsey Oct 29 '16 at 7:46
  • 1
    $\begingroup$ If your algebraists are teaching that groups must have two operations, you're not in a good place to be discussing mathematics. And this is general discussion that doesn't address the specific question, by copy-pasting with no further explanation at that. $\endgroup$ – Nij Oct 29 '16 at 8:29
  • $\begingroup$ @MoisheCohen, while a homology group is, formally speaking, a group, its key useful properties (as I understand them) are more the properties of a module over the integers. Grothendieck groups are also abelian groups over a commutative monoid; i.e., the module-like properties are of key interest. With homotopy groups, how are the not immediately transformation groups? If a group is abelian, it seems interesting only as a module over a ring. Otherwise, the central object of interest is the noncommutativity of transformations. $\endgroup$ – avs Oct 29 '16 at 21:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.