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I'm asked to verify which of these multiplication tables form a group. I'm having problems to see which of the axioms for a group are violated in each table. In (a), I couldn't find an element $e$ such that for all $x, ax = xa = e$. We have $aa = a, ba = b, ca = c, da = d$, so it seems like $a$ could be that identity element, but when I verify the converse, we can see that it doesn't work. So I would say (a) is not a table for a group.

In (b), the rule I searched in (a) Works, we have $a$ as the identity element, because $ax = x = xa$ for any $x$. Therefore, the axiom of the identity element of a group is satisfied.

Then, I tried to search for the inverse element. The axiom of invertabilitys states that each element $a$ must have an inverse $a^{-1}$ such that $aa^{-1} = e = a^{-1}a$. In this case, each element is its own inverse, so the rule $aa^{-1} = e = a^{-1}a$ works for every $a$.

Now I'm gonna try to verify the associativy axiom, that is: $(ab)c=a(bc)$ Do I have to verify all the possible multiplications? Is tghere any method to verify it faster? I didn't verify all them, but I think this table satisfies this property too. So I would say (b) is a group.

PS: i've read somewhere that if we have the same element appearing 2 times in a row, then it can't be a group. Is it because it it appears 2 times in a row, then, for example, in (d), we have: $da = db\implies a=b?$

What are the methods you guys utilize to verify a multiplication table? Am I doing this rigth?

I would love to read an article about recognizing multiplication tables, because I need to be fast in it for the exams. So please give me your tips <3

  • $\begingroup$ You were able to quickly discard table (a), based on lacking a (two-sided) identity element. Another quick "fail" is when a row (or column) contains the same entry more than once (equiv. does not contain each entry). This would eliminate table (d), as you've suggested. $\endgroup$ – hardmath May 25 '15 at 3:55
  • $\begingroup$ @hardmath and what about verifying associativity? $\endgroup$ – Guerlando OCs May 25 '15 at 3:56
  • 1
    $\begingroup$ Associativity can be tedious to check. I would fall back in this exercise on knowing something about groups of order 4. $\endgroup$ – hardmath May 25 '15 at 4:03

In practice, you should be able to tell that a table is a group table by matching it with a group that you already know. Assuming that you don't look at tables with more than six elements, here are the possible groups:

  • Cyclic groups: $C_2, C_4,\dots,C_6$
  • The Klein four group $V_4$
  • The symmetric group $S_3$.

So if you know what the multiplication tables of these groups look like, you'll be in good shape.

It's also good to know how to eliminate possibilities. Here are three necessary criteria for a table to be a group multiplication table. When I say necessary, I mean that these can only tell you if the table is not a group table. In other words, if these criteria are satisfied, it is nonetheless possible that the table is not a group table.

Say the elements to be multiplied are $a_1,a_2,\dots,a_n$ (in that order). Then

  1. There is a row $R$ that reads $a_1, a_2,\dots, a_n$ (in that order), and a column $C$ that reads $a_1,a_2,\dots,a_n$ (in that order). Edit: $R$ and $C$ must intersect on the diagonal.
  2. Every row contains each of $a_1,a_2,\dots,a_n$ exactly once.
  3. Every column contains each of $a_1,a_2,\dots,a_n$ exactly once.

Condition 1 comes from the existence of an identity element, and conditions 2 and 3 come from the existence of inverses. (Having inverses basically means that we can always undo multiplication, but if $ax=bx$ for $a\neq b$ -- i.e., if a row has duplicates -- then we can't undo multiplication by $x$. I can give more details if you're interested.)

The condition I haven't mentioned yet is associativity. If I add the condition

  1. The multiplication is associative.

then conditions 1 - 4 become necessary and sufficient, which means that conditions 1 - 4 are true if and only if the table is a multiplication table.

No reasonable professor will ever ask you to verify that a multiplication table is associative. There are algorithms to do so, which this MSE post mentions, but they're beyond the scope of an introductory algebra class. If conditions 1 - 3 hold and the table is small, it's easier to see if the table matches the multiplication table of one of the groups I listed above.

  • $\begingroup$ Presumably you mean, $C_1, \ldots, C_6$, no? (Also, this is quite a nice answer, +1.) $\endgroup$ – Travis May 25 '15 at 5:42
  • $\begingroup$ Your condition 1 could even be stricter, in that the number of the row and column need to be the same (if your left identity and right identity are different, it can't be a group). $\endgroup$ – Paŭlo Ebermann May 25 '15 at 14:33
  • $\begingroup$ Paulo, you're right. At the time I thought that my condition took care of your objection, but I see now that it doesn't. $\endgroup$ – user134824 May 25 '15 at 17:32
  • $\begingroup$ But what about $\mathbb{Z}_2$ with the operation multiplication? Doesn't it has a row and a column of equal elements, the $0$ element? So it does not form a group, rigth? $\endgroup$ – Guerlando OCs May 26 '15 at 19:42
  • $\begingroup$ That's right, $\mathbb Z_2$ under multiplication does not form a group. This is because the element $0$ has no multiplicative inverse. But under addition, $\mathbb Z_2$ does form a group. $\endgroup$ – user134824 May 26 '15 at 22:18

Like you say, in order for a multiplication table for an operation to define a group structure, its rows and columns must each contain each element exactly once. (This is a consequence of divisibility: In a group $(G, \ast)$, if $a \ast b = a \ast c$ then $b = c$.) This property of an operation $\star: X \times X \to X$, called the Latin square property, because $\star$ has this property iff its multiplication table is a Latin square; in this case, we say that $(X, \star)$ is a quasigroup, but this guarantees neither existence of an identity element nor associativity. Like you say, the first row of (a) and the last column of (d) both have repeated elements, so neither of these are groups.

Also, it's easy to see that the operations in (b) and (c) both have identity elements ((a) in both cases). One can check associativity manually, but like you say, this is quite a lot of checking: Naively, for an operation on a set of $n$ elements, this is $n^3$ checks!

Here's a better method for this situation

Since $n := 4$ is a small enough order that one likely knows all of the groups up to isomorphism---just the cyclic group $\Bbb Z_4$ and the Klein $4$-group, $\Bbb Z_2 \times \Bbb Z_2$---one can try to find isomorphisms between those groups.

Table (b) We can see the element $b$ has order $4$ (in fact, provided that this table does define a group structure, to see this it's enough to check that $b^2$ is not the identity, as the order of any element of a group divides the order of the group). The elements of the group $\Bbb Z_2 \times \Bbb Z_2$ all have order $1$ or $2$, so if the table defines a group $(G, \ast)$, it must be $\Bbb Z$, and it must be generated by $b$. Any group isomorphism is determined by the images of its generators, so if the table defines a group, $b \mapsto [1]$ defines an isomorphism $\phi: G \to \Bbb Z$; in this case, we have $\phi(a) = [0]$, $\phi(c) = \phi(b \ast b) = \phi(b) + \phi(b) = [1] + [1] = [2]$ and similarly $\phi(d) = [3]$. In particular, this map is injective, so to check that this map is an isomorphism it's enough to check that $$\phantom{(ast)} \qquad \phi(x \ast y) = \phi(x) + \phi(y) \qquad (\ast)$$ for all $x, y \in G$.

In principle this is $n^2 = 16$ computations, but there are several simplifications we can make: First, since both $(G, \ast)$ and $(\Bbb z_4, +)$ are abelian (for the former, a group is abelian iff a group's multiplication table is symmetric), so we only need to check the ${{n + 1} \choose 2} = 10$ unordered pairs $(x, y)$. Next, we know that $\phi$ maps the identity $[a]$ to the identity $[0]$ of $\Bbb Z_4$, so the identity $(\ast)$ holds for pairs $(a, y)$, reducing to ${n \choose 2} = 6$ pairs. We implicitly checked the pairs $(b, y)$ when determining the map $\phi$ explicitly above, leaving just $3$ pairs to check, namely, $(c, c)$, $(c, d)$, and $(d, d)$.

A similar argument for (c) shows that the table there defines a group isomorphic to $\Bbb Z_2 \times \Bbb Z_2$.

Remark A quasigroup with an identity element is called a loop, and it turns out that all loops with $\leq 4$ elements are automatically groups, and so with this fact in hand, our first observations that (b) and (c) both have the Latin square elements (i.e., no repeated entries in rows or columns) and both have an identity is enough to conclude that they both define groups. This fact takes some work to show, though, and is anyway surely outside the scope of your course, so I would recommend against citing it as a justification for a solution! Furthermore, this fact does not hold for operations on sets with $> 4$ elements, see this counterexample for a $5$-element loop that is not a group.


One easy way to check for associativity is to see if we can regard $\{a,b,c,d\}$ as functions on some set, and regard our "multiplication" as functional composition, which you should already know is associative. We already have a set at hand, $\{a,b,c,d\}$ itself (it may seems strange to use the same letters for functions and their arguments, so I will call the functions by $a',b',c',d'$).

Let's look at example (c): Suppose $a'$ is the identity function:

$a'(a) = a, a'(b) = b, a'(c) = c, d'(d) = d$.

It should be clear that no matter what functions we define $b',c'$ and $d'$ to be, the first row and column (except for using primed letters for the entries) will look like our example (c) table.

Now suppose we take $b'$ to be the function:

$b'(a) = b,b'(b) = c,b'(c) = d, b'(d) = a$.

So far, the only "new" entry we can make is $b' \circ b'$, which by the rules of composition must be:

$(b'\circ b')(a) = b'(b'(a)) = b'(b) = c\\ (b'\circ b')(b) = b'(b'(b)) = b'(c) = d\\ (b'\circ b')(c) = b'(b'(c)) = b'(d) = a\\ (b'\circ b')(d) = b'(b'(d)) = b'(a) = b.$

Since in our (c) table, we have $b\circ b = c$, our hope is we will mimic this table by defining $c'$ to be:

$c'(a) = c, c'(b) = d, c'(c) = a, c'(d) = b$.

Finally, we will take $d' = b'\circ c' = b'\circ b'\circ b'$ (leaving off parentheses because functional compostion is associative). That is:

$d'(a) = b'(b'(b'(a))) = b'(b'(b)) = b'(c) = d\\ d'(b) = b'(b'(b'(b))) = b'(b'(c)) = b'(d) = a\\ d'(c) = b'(b'(b'(c))) = b'(b'(d)) = b'(a) = b\\ d'(d) = b'(b'(b'(d))) = b'(b'(a)) = b'(b) = c.$

Now we need to check all possible products by comparing then with our table in example (c). We have already checked the ones involving $a'$, no confusion there. We also know that $b'\circ b' = c'$ (just as in our table, without the primes). So we need to check the following compositions:

$b'\circ c' = c' \circ b' = d'$ (but this follows from the way we defined $d'$, and the associativity of composition).

$b' \circ d' = d' \circ b' = a'$ (this is two equations in one). So:

$(b'\circ d')(a) = b'(d'(a)) = b'(d) = a\\ (b'\circ d')(b) = b'(d'(b)) = b'(a) = b\\ (b'\circ d')(c) = b'(d'(c)) = b'(b) = c\\ (b'\circ d')(d) = b'(d'(d)) = b'(c) = d$

so, evidently, $b' \circ d'$ is the identity function, which is $a'$.

The other compositions are checked similarly, it's now just a matter of computing them, and seeing you get an identical table. You still have to check:

$d'\circ b', c' \circ c',c'\circ d', d'\circ c'$ and $d'\circ d'$.

Although this is a fair amount of work, it's slightly less work than checking all possible triple products (done both ways), which is $64$ verifications, instead of $16$.

My hope is, when doing this, is that you will start to see a pattern emerging, and realize halfway-through you don't need to keep going to be sure it works.


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