I am working through a group theory textbook and I am having trouble with this question.

Find a binary operation on a set of $5$ elements such that there is an identity element, each element is its own inverse, and the cancellation laws hold (if $ax=ay$ then $x=y$, and if $xa=ya$ then $x=y$). Show that the set is not a group under this operation.

Here is what I have done so far: If there is an identity ($e$) and each element is its own inverse then I have a set that looks like $\{e, a, b, c, d\}$ where $ee = aa = . . . = dd = e$. I have noticed that the question mentions nothing about associativity, and I do not know what exactly I get from having the cancellation law in terms of group axioms.

From what I can tell, this is a group, but clearly from the question is isn't, so any hints?

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    $\begingroup$ if there is no assumption of associativity, then you are free to fill in the multiplication table as you wish, subject only to the stated requirements. thus multiplication on the left by $a$ must permute the three elements $b,c,d$, and similarly for the other non-identity elements. these permutations must be selected so that every row or column of the multiplication table contains each element exactly once. $\endgroup$ – David Holden Apr 2 '15 at 1:19
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    $\begingroup$ The cancellation rule is telling you that you should have a latin square: en.wikipedia.org/wiki/Latin_square $\endgroup$ – Alex Zorn Apr 2 '15 at 1:26
  • $\begingroup$ I've asked a follow-up question: math.stackexchange.com/questions/1216907/… . $\endgroup$ – Travis Willse Apr 2 '15 at 7:16

If one has any binary operation $\ast$ on the given set $S$ with an identity $1$ for which any nonidentity element $a$ is its own inverse, then $\ast$ cannot define a group structure on $S$: If it did, the subgroup $\langle a \rangle = \{1, a\} < S$ generated by $a$ would have $2$ elements, but any subgroup of a group structure defined on $S$ has order dividing $5$. (By the same reasoning, the same is true for any set with an odd number of elements.)

Now, we're asked to exhibit an explicit product $\ast$ satisfying the given hypotheses. The products $1x$, $x1$, $x^2$, $x \in S$, are already determined by the hypotheses, leaving $12$ entries unspecified in the multiplication table of $\ast$. The hypotheses together only impose one condition on the remaining entries: Since the cancellation rule holds, each element in $S$ must appear exactly once in each row and each column of the table. Now, we must have $ab = c$ or $ab = d$, and by relabeling if necessary, we can assume the former, which forces $ac = d$, $ad = b$, $cb = d$, and $db = a$.

These in turn force other relations, and after a few iterations this determines the entire table, so in fact there are only two products $\ast$ that satisfy the conditions (determined by the assignment of $c$ or $d$ to $ab$), and these are isomorphic. As you say, this multiplication must be nonassociative; indeed we have $(ab)d = cd = a$ but $a(bd) = ac = d$.

Since this example is both small and unique, it would be interesting to know if this structure could be described in some way more suggestive than a multiplication table. I've posed this as a new question.

Remark The identity and cancellation rules mean we are in particular looking for an operation $\ast$ that defines a loop structure on $S$. This example is minimal in the sense that all loops with $< 5$ elements are actually groups. Up to isomorphism there are exactly $6$ loop structures on a set of $5$ elements (of course, only one of them, $(\mathbb{Z}_5, +)$, is a group structure).

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  • $\begingroup$ Thank you everybody for comments and hints. The answer is pretty straightforward when you do a Caley table. I actually feel like an idiot. $\endgroup$ – ocyeung Apr 2 '15 at 2:29
  • $\begingroup$ You're welcome, I'm glad you found it useful. $\endgroup$ – Travis Willse Apr 2 '15 at 2:31

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