Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 4 / 5 / ii

Table of contents
- Solution ii

Solution

Yes, for $p = 2$ . The displayed group presentation is that of the infinite dihedral group. Put

u = a t

and

v = t

. Then

v^{2} = 1

and

u^{2} = a t a t = a (t a t^{- 1}) t^{2} = a a^{- 1} = 1.

Conversely, from

u^{2} = v^{2} = 1

a = uv

and

t = v

; then

t a t^{- 1} = vu = a^{- 1}

. These inverse substitutions give

G_{2} ≅ ⟨ u, v ∣ u^{2}, v^{2} ⟩ ≅ C_{2} * C_{2} .

Each relator has free-group

2

-root exponent one, so

def_{2} (⟨ u, v ∣ u^{2}, v^{2} ⟩) = 2 - \frac{1}{2} - \frac{1}{2} = 1.

The change of presentation matters: the original presentation's mixed conjugation relator has weight one and would give a smaller p-deficiency.

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version