Let X be the group generated by and , such that
X=\langle u, v \mid u^{46} = 1, v^{40,488,236} = 1, vu = uv^{27,636,633} \rangle $$. What is the order of the group X?The group (X) is defined by the presentation: [ X = \langle u, v \mid u^{46} = 1, v^{40488236} = 1, vu = uv^{27636633} \rangle. ]
From the relation (vu = uv^{27636633}), it follows that (u^{-1} v u = v^{27636633}). Conjugating (v) by (u) gives: [ u^{-i} v u^i = v^{(27636633)^i} \quad \text{for all integers } i. ] Setting (i = 46) and using (u^{46} = 1): [ v = u^{-46} v u^{46} = v^{(27636633)^{46}}, ] so: [ v^{(27636633)^{46} - 1} = 1. ] The order of (v) must divide both (40488236) and ((27636633)^{46} - 1). Computing the greatest common divisor: [ \gcd(40488236, (27636633)^{46} - 1) = 4. ] Thus, the order of (v) divides 4. Additionally, the relation (vu = uv^{27636633}) and the fact that (27636633 \equiv 1 \pmod{4}) imply that (v) commutes with (u), so ([v, u] = 1). This commutator relation, combined with the other relations, leads to (v^{27636632} = 1) (since (k - 1 = 27636632)), which is consistent with the order of (v) dividing 4.
Given that the order of (v) divides 4 and no relations force a smaller order, the order of (v) is exactly 4. The order of (u) is 46, and since (u) and (v) commute and their orders are coprime in the sense that the group is abelian with no additional identifications, (X) is isomorphic to (\mathbb{Z}/46\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}). The order of this group is: [ 46 \times 4 = 184. ]
All relations are satisfied:
- (u^{46} = 1),
- (v^{40488236} = (v^4)^{10122059} = 1^{10122059} = 1) (since (4 \mid 40488236)),
- (vu = uv) (since (v) and (u) commute), and (uv^{27636633} = uv^{27636632 + 1} = u v^{k-1} v = u \cdot 1 \cdot v = uv), so (vu = uv^{27636633}).
Thus, the order of (X) is 184.
\boxed{184}