### 38th Putnam 1977

Problem B6

G is a group. H is a finite subgroup with n elements. For some element g ∈ G, (gh)3 = 1 for all elements h ∈ H. Show that there are at most 3n2 distinct elements which can be written as a product of a finite number of elements of the coset Hg.

Solution

It is essential to notice first that g3 = 1 (since 1 ∈ H).

A little messing around should now convince us that we can simplify finite products of elements of the form hg. In fact, we show that they can always be written as (1) h1gh2, (2) h1g2h2, or (3) h1g2h2g.

Clearly hg is of the form (1), so it is sufficient to show that given an element k of form (1), (2) or (3), then k (hg) is also of one of these forms.

It is convenient to note that: ghg = h-1g2h-1 (*) (post-multiply ghghgh = 1 successively by h-1, g2, h-1); and g2hg2 = h-1gh-1 (**) (pre-multiply gh-1gh-1gh-1 = 1 successively by g2, h, g2).

So dealing with the three cases in turn: (h1gh2) h3g = h1h3-1h2-1g2h3-1h2-1g, which is of form (2).

(h1g2h2) h3g is obviously of form (3).

(h1g2h2g) h3g = h1g2h2h3-1g2h3-1 = h1h3h2-1gh3h2-1h3-1, which is of form (1).