9
$\begingroup$

In documents relating to group theory it seems common to use a multiplicative notation to represent the group operation. For example, I'm reading Herstein's "Topics in Algebra" and looking for some pointers about vector spaces in the section on groups. In the vector space section the group operation combining vectors is (quite logically) represented as addition ($v = v_1 + v_2$), but when I switch across to the section on groups it's multiplication ($c = ab$).

Besides finding the switch of notation unhelpful, I feel that the additive notation is a better analogy with real arithmetic: all group elements have an inverse as they do with addition, whereas the multiplicative notation carries an untrue suggestion that there may be a "0" which has no inverse.

So, have I missed something: is there some reason why the multiplicative notation is preferable ?

$\endgroup$
5
  • 4
    $\begingroup$ Just a guess: This is consistent with the conventions in ring theory: When considering rings, the 'addition' is always commutative, while the multiplication is not. So in group theory you usually do not assume commutativity, but if you use the additive notation, then it is clear that you are talking about a commutative group (and many groups can be considered as a ring but ignoring multiplication). So I think this convention makes perfectly sense. $\endgroup$
    – flawr
    Jan 29, 2015 at 14:30
  • $\begingroup$ I agree with @flawr. The additive notation suggests commutativity. If a group is commutative, then $+$ is used very frequently. But when we consider abstract groups $+$ could be misleading. $\endgroup$
    – Janko Bracic
    Jan 29, 2015 at 14:33
  • $\begingroup$ My guess is that group theory started with rotations and reflections, with reflections being seen as negative but the reflection of a reflection being seen as positive. $\endgroup$
    – Henry
    Jan 29, 2015 at 14:33
  • 2
    $\begingroup$ I'm not sure I'd call this "multiplicative" notation - we don't explicitly write out the multiplication sign. I think that this is a case of juxtaposition having two meanings, multiplication (when used for numeric variables) and an arbitrary group operation. $\endgroup$
    – Michael Lugo
    Jan 29, 2015 at 14:45
  • 2
    $\begingroup$ Before groups were known to be interesting on their own, permutation groups had been known for quite a while. As these are (invertible) functions on a set, the composition is written as $f\circ g$ or even $fg$ leading to a multiplicative notation. $\endgroup$
    – j.p.
    Jan 29, 2015 at 17:01

3 Answers 3

Reset to default
10
$\begingroup$

You surely know that many groups are not commutative. Because of this, an additive notation would bring confusion in non-abelian context.

Think about group of invertible matrices $\mathrm{GL}(\mathbf{K},n)$ over a field $\mathbf{K}$: this is a group under matrix multiplication: wouldn't be rather tricky to think $A+B$ as the result of $A\cdot B$, knowing that it can be very different from $B+A$?

Inverse elements are also another fact. In general, you can prove that for a group $G$ if $x,y\in G$ then $(xy)^{-1} = y^{-1} x^{-1}$. Imagine this in additive notation: $$ -(x+y)=-y - x$$ and that would be DIFFERENT from $$-x-y=-(y+x)$$

$\endgroup$
3
  • $\begingroup$ I'm not sure how this argument holds given that regular real multiplication which most people are familiar with is commutative $\endgroup$
    – Cubic
    Oct 9, 2019 at 12:12
  • $\begingroup$ @Cubic I don't get the point of your observation, but $\mathbf{R}$ is not a group under "ordinary multiplication", if this is what you mean. Instead, $\mathbf{R}^\times := \mathbf{R}\setminus \{0\}$ is a group and, despite being abelian, writing it additively would cause a great deal of confusion (for example, $1,-1\in \mathbf{R}^\times$ but then one has to write their product as $1+(-1)$, which I hope you'll agree is very misleading. $\endgroup$
    – Caligula
    Oct 13, 2019 at 10:34
  • $\begingroup$ If your argument is that additive notation is not good for beginners because it wrongly suggests that $-(A+B)=(-A)+(-B)$ from their experience in $\mathbb{R}$, then the same would be true for multiplicative notation. This is what Cubic was pointing out. $\endgroup$
    – Martin Frenzel
    Oct 15 at 21:38
5
$\begingroup$

Generally, multiplicative notation is used for the operation in an arbitrary group, and additive notation is reserved for the operation in Abelian (commutative groups). For some intuition into this, the general linear group $GL_n(\mathbb R)$ is the set of all invertible $n\times n$ matrices, with the operation of matrix multiplication. If $A,B\in GL_n(\mathbb R)$ then we would denote their product $AB$ or $BA$ (note that these are not equal in general!). If we consider $\mathbb R^n$ as a vector space, then it is an Abelian group under addition - so if $x,y\in\mathbb R^n$ then we write the sum as $x+y$ or $y+x$ (and these are indeed equal).

$\endgroup$
0
$\begingroup$

Let $(G,\star)$ be a group with $x\in G$. Suppose you want to write $x \star x \star \cdots \star x$ ($n$ times).

How would you write it with the multiplicative notation? $x^n$.

How would you write it with the additive notation? You can't say $nx$ or $xn$. You have to use an awkward construction like $\sum_{i = 1}^n x$.

Now lets assume we are working over a vector space with scalars in $\mathbb{R}$. How would we write it now? We would just use $nx$. In contexts where there is a reasonable definition to $nx$, usually the additive notation is used.

Add this to the fact that $+$ is used to mean an abelian groups.

$\endgroup$
2
  • 4
    $\begingroup$ What do you mean "you can't say $nx$"? For an abelian group, this is exactly the notation that is used. This is because abelian groups are best thought of as modules over $\mathbb{Z}$. $\endgroup$
    – Caleb Stanford
    Jan 29, 2015 at 16:24
  • $\begingroup$ In other words, everything you say earlier on in your answer really just ties back to whether or not the group is abelian. $\endgroup$
    – Caleb Stanford
    Jan 29, 2015 at 16:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .