I'm currently trying to learn abstract algebra myself, and the following is a quote from the book I am using, "A set of equations, involving only the generators and their inverses, is called a set of defining equations for $G$ if these equations completely determine the multiplication table of $G$."

Then the book proceeds to give an example: "Let $G$ be the group $\{e, a, b, b^{2}, ab, ab^{2} \}$ whose generators $a$ and $b$ satisfy the equations $a^{2} = e$, $b^{3} = e$, and $ba = ab^{2}$." And claims that the three equations determine the multiplication table of $G$.

So I worked out the multiplication table and displayed it below.

  • When they say, "completely determine the multiplication table of $G$," does that mean the product of two elements can be simplified to another element? For example, $(ab^{2})(ab^{2}) = ab(ba)bb = ab(ab^{2})b^{2} = abab(b^{3}) = a(ba)b = a(ab^{2})b = aab^{3} = e.$

  • I also don't see how inverses are used in determining the multiplication table in this case. I've only used substitution in this case. Can someone explain why inverses might be important?

  • How did the author know that only 3 equations were enough to determine the multiplication table? And why did he choose those equations?

  • Also what is the significance of determining a multiplication table for elements of a group?

Multiplication Table of G

          |  e      a      b      b^2    ab     ab^2  |
  |  e    |  e      a      b      b^2    ab     ab^2  |
  |  a    |  a      e      ab     ab^2   b      b^2   |
  |  b    |  b      ab^2   b^2    e      a      ab    |
  |  b^2  |  b^2    ab     e      b      ab^2   a     |
  |  ab   |  ab     b^2    ab^2   a      e      b     |
  |  ab^2 |  ab^2   b      a      ab     b^2    e     |
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    $\begingroup$ What text is this? $\endgroup$ Jul 30, 2011 at 14:35
  • $\begingroup$ @Pete: This is from the 2nd edition of Pinter's Abstract Algebra on pg 47-48. $\endgroup$
    – Student
    Jul 30, 2011 at 14:42
  • $\begingroup$ I am missing something but why is $b(ab^2) = a$? Should it not be $ab$? $\endgroup$
    – user38268
    Aug 1, 2011 at 12:24
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    $\begingroup$ @D Lim : Indeed, the table is wrong: look at the row for $b$ and the column for $ab^2$. A typo I suppose. $\endgroup$
    – yatima2975
    Aug 1, 2011 at 14:01

7 Answers 7


Fun fact to know: in 1992 Ales Drápal proved that if two finite groups agree on 89% on their multiplication tables, the groups must be isomorphic! He conjectured that the same holds true if the tables agree on 75% of their entries. The conjecture has not been proved yet. See also Groups St. Andrews 2001 at Oxford, featuring the paper of Drápal On the distance of 2-groups and 3-groups.

  • $\begingroup$ Interesting! I take it 75% is sharp because of $C_4$ versus $C_2 \times C_2$? Do you have an electronic version that's not behind a paywall, by any chance? $\endgroup$
    – yatima2975
    Aug 1, 2011 at 14:08
  • $\begingroup$ I have the book (two volumes). Maybe a math library near you has it too. $\endgroup$ Aug 1, 2011 at 21:10
  • $\begingroup$ Maybe I'll venture out to the campus one day, but then I'll have to remember to look for it... It may be easier to prove it myself - 89% is 8/9, I suppose? That should be a hint! $\endgroup$
    – yatima2975
    Aug 1, 2011 at 21:23
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    $\begingroup$ @yatima2975, I looked it up, the "seeding" paper is A. Drápal, How far apart can the group multiplication tables be?, European J. Combin. 13 (1992), no. 5, 335–343. If you Google on the title you will find a slew of other interesting papers. And yes, the 89% is 8/9 ... $\endgroup$ Aug 4, 2011 at 11:35
  • 1
    $\begingroup$ It has been proved for 78%, no? arxiv.org/pdf/1107.0133 $\endgroup$
    – user641
    Jan 6, 2012 at 9:44
  1. "Completely determine" means two things simultaneously. Firstly, "determine" here means that each entry in the multiplication table can be filled in by one of those six elements, which is basically what you said. Secondly, "completely" means that there is no ambiguity: some entries in the multiplication table can be worked out in different ways, and no matter how many different ways you try to work some given entry out, you will get the same thing.

  2. It's entirely conceivable that you might be faced with an equation like $ab^{-1} = cd^{-1}$. No matter how hard you try to get rid of the inverses, you can't.

  3. The author knew because this is a very standard group called $D_6$ (sometimes called $D_3$). You'll know it too once you meet the dihedral groups. But that's not a good answer. Another way to motivate this: notice that every element that he claims lives in that six-element set is of the form $a^ib^j$, where $i$ is 0 or 1 and $j$ is 0, 1 or 2. But what about $ba$, or other things that don't fit into this form? Our group axioms tell us that, since we have an element $b$ and an element $a$, we should be able to multiply them. So we need to know what to do when we have something of the 'wrong' form. And these wrong forms can be broken up into three types:

    • $a^i$ where $i > 1$ or $i < 0$,
    • $b^j$ where $j > 2$ or $j < 0$,
    • $ba$ (multiplied the 'wrong' way round).

      These three relations tell us exactly what to do in each of those situations.
  4. Exactly the same reason children write out their times tables. It shows you what's going on! It gives you much more of a clue of the structure of the group than you might expect at this point; on the other hand, you're right to be suspicious of them, because nobody uses them after a certain level. But pedagogically they're very important.

  1. Yes, the author means that the product of any two elements can always be determined by referencing and manipulating the facts from defining set of equations.
  2. Inverses aren't relevant in the given example because they aren't present. If the author replaced the first equation with the equivalent form $a = a^{-1}$, then you could say they're relevant. It's all just a matter of acceptable presentation.
  3. I might not be familiar enough with abstract algebra to know this, but my guess is that the author actually started with the defining set of equations and then swiftly established a multiplication table. It's also possible they just tinkered with the algebra a bit until they found an acceptable example for the text - this sort of thing is absolutely trivial for people who know this stuff inside and out.
  4. It presents the group's entire multiplicative structure for easy reference.
  • 4
    $\begingroup$ "this sort of thing is absolutely trivial for people who know this stuff inside and out" does not seem particularly useful :-) $\endgroup$
    – Srivatsan
    Jul 30, 2011 at 15:02

The use of inverses (or more precisely, the use of negative exponents) is superfluous in the case of finite groups. In a finite group, every element has finite order. In other words, for every element $a$, there is a positive integer $k$ such that $a^k=e$. Hence instead of writing $a^{-1}$, one can write $a^{k-1}$, which is the same thing. In your example $b^{-1}=b^2$.

On the other hand, as soon as your group has elements of infinite order, it is no longer possible to write inverses as positive powers of the elements, and so the "$-1$" notation may become necessary. This can only happen in groups of infinite size.

The three relations in your example tell you

  1. what the order of $a$ is,
  2. what the order of $b$ is,
  3. how $a$ and $b$ move past each other.

With two generators, this is enough to determine the results of arbitrarily complicated multiplications, so you won't need additional relations.

In groups of large size, it may be impractical to write out the multiplication table. What's important is being able to produce the multiplication table in principle. The group is only defined once you've specified how to multiply arbitrary elements of the group. The multiplication table is one way to do that, but if you can prove that your relations allow you to compute arbitrary products, that's fine too.


A group multiplication table is always a latin square. A latin square is a group multiplication table if it satisfies associativity. So technically you can find all groups of order $n$ by finding all associative $n\times n$ latin squares.


I am sure nicer answers will be posted, but I will post my attempt at explanation, though I am a group theory amateur myself. (Sorry for the extremely long answer!) Also check the wikipedia page for more details and for precise definitions/theorems.

This is an example of a "presentation" of a group. A few definitions will be useful for the discussion. The elements $a$ and $b$ are called the "generators" of the group, and the "equations" $a^2 = e$, $b^3 = e$ and $ba = ab^2$ are called "relations". A "word" is a string involving the generators and their inverses (e.g.: $a^{-2} b^{-4} bab^2$). Note that it is sometimes straightforward to rewrite a word in terms of just the generators. For instance, in this example, we can rewrite $a^{-1}$ as $a$ and $b^{-1}$ as $b^2$; when we do this $bab^2 a^{-2} b^{-6}$ becomes $b a b^2a^{2} b^{12}$.

Completely determining the multiplication table

Yes, it means that the product of two elements can be expressed as one of these elements themselves. This can be done by simplifying the product by repeatedly using the relations $a^2 = e, b^3 = e, ba = ab^2$. (In simple examples such as this, you can usually do routing group operations just by inspection.)

Where are the inverses? Inverses are automatically and implicitly defined by the relations that you have prescribed. For instance, $a^{-1} = a$ and $b^{-1} = b^2$. Why? Because the relation $b^3 = e$ implies that multiplying $b$ by $b^2$ (to the right or to the left) gives you $e$. So, $b^2$ must be the inverse of $b$. You can similarly check that $a^{-1}$ is $a$. (@WillO's comment that you can always write $b^{-1}$ as a power of $b$ in a finite group is relevant here.)

To compute the inverse of words (like $bab^2 a^{-2} b^{-6}$), you can use the identity $(xy)^{-1} = y^{-1} x^{-1}$. For example, we have $(bab^2 a^{-2} b^{-6})^{-1} = b^{6} a^2 b^{-2} a^{-1} b^{-1}$. This can, of course, be simplified further using the usual $a^{-1}=a$ and $b^{-1} = b^2$ trick.

Why 3 equations? In one sense, $3$ equations is not sacred. You can take any set of generators and any set of relations between the generators, and you will end up with some group. For instance, taking the generators to be $a$ and $b$, and the set of relations to be the empty set, you will get what is called the free group on $\{a, b\}$.

But not all examples that we obtain this way are nice; for instance, the group may turn out to be infinite or even uncountable. If you want to express your finite group in terms of a presentation, that is also quite easy. We will come to this point later.

Why is a multiplication table significant?

Pedagogical value. A multiplication table is a proof that the group that the author defines is in fact a group. I remember that when I first read about representing groups using relations, I used to hand-compute the whole table, and check the group axioms. Though tedious even for medium-sized groups, the exercise helped me intuitively understand the idea of group presentations and also taught me some simple tricks in simplifying the products of elements of a finitely generated group.

As a finite presentation. The multiplication table of a finite group $G$ also automatically gives you a finite presentation of the group. (I found this point in the Wikipedia page while researching for writing this answer.) Simply define the elements of $G$ as the generators, and for each entry $g_i g_j = g_k$ in the multiplication table, add a relation $g_i g_j g_k^{-1} = e$. (In this sense, finite presentation is a generalization of finite number of elements.) But of course, out of the $|G| \times |G|$ relations that we can write this way, only a few are useful; the remaining can be deduced from other relations. For example, you show how to deduce the relation $(ab^2)(ab^2)e^{-1} = e$ from the usual relations $a^2 = e$, $b^3=e$ and $ba=ab^2$.


To amplify what others have said about the multiplication table, by inspection we can see this group doesn't satisfy commutation, it isn't an idempotent group, it has neutral (or identity element) of e, and the inverse of each element can also get determined readily.


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