Past exam of the short options course of the University of Oxford / 2015 / 2 / a

Solution

The defining relation

y x = x^{4} y

lets us move every occurrence of

y

to the right of every occurrence of

x

. More generally, repeated conjugation gives

y x^{r} = x^{4 r} y, y^{j} x^{r} = x^{4^{j} r} y^{j} .

Negative powers cause no problem because

x^{- 1} = x^{8}

and

y^{- 1} = y^{2}

. Thus every word in the group presentation can be put in the required form, reducing powers using

x^{9} = y^{3} = e

g = x^{i} y^{j}, 0 \leq i \leq 8, 0 \leq j \leq 2.

At this stage existence is proved; uniqueness will follow from the concrete semidirect product in part (b). Moving

x^{r}

past

y^{j}

(x^{i} y^{j}) (x^{r} y^{s}) = x^{i + 4^{j} r mod 9} y^{j + s mod 3} .

The action is consistent with reduction of

j

3

because

4^{3} \equiv 1 (mod 9)

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