# Problem #92

This month we will venture into Group Theory. Recall that a group is a set G with a binary operation * satisfying the following properties:

For all a, b in G, a*b is also in G. (closure)
• For all a, b, c in G, (a*b)*c = a*(b*c). (associative)
• There is an element i in G such that for all a in G, i*a = a*i = a. (identity)
• For all a in G there is an element a-1 such that a*a-1 = a-1*a = i. (inverse)

Note that in general it is not the case that a*b = b*a. From now on, we will suppress the * and just juxtapose elements.

Consider the group generated by a,b,c, and d subject to the relations

ab = c, bc = d, cd = a, and da = b

Using the first relation, the second relations becomes bab = d. Using this expression and the first relation, we obtain

abbab = a and baba = b

Taking the second relation above and multiplying both sides on the left by a-1b-1 and on the right by a-1, we have b = a-2. Now first relation above becomes aa-2a-2aa-2 = a or a-4 = a, hence i = a5. Therefore our group consists of the five elements i, a, a2, a3, a4. The other elements can be expressed in terms of a as follows: b = a3, c = a4, and d = a2.

Finally, we get to his month's problems. How many elements are there in the groups given by the following generators and relations?

• Generators:a,b,c
Relations: ab = c, bc = a, ca = b

• Generators:a,b,c,d,e
Relations: ab = c, bc = d, cd = e, de = a, ea = b

• Generators:a,b,c,d,e,f
Relations: ab = c, bc = d, cd = e, de = f, ef = a, fa = b

Source: John H. Conway

No correct solutions have been submitted, so this problem is still open.

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