Question

what is the order of

i) 14 in Z₂₄/ _

<8>?

ii) (Z₁₀⊕U(10)/<2,9>?

Answer 1

i)

(8) has order 3 in Z₂₄ , so Z₂₄/(8) is cyclic of order 8, generated by [14] +(8).

Now (8) = {[0], [8], [16]} ⊂ Z₂₄

[14] +(8)$\ne$ (8) (since 14 $\notin$ (8))

([14] + (8))² = [28] +(8)=[4] +(8) $\ne$ (8)

([14] + (8))³ = [42] +(8)=[18] +(8) $\ne$ (8)

([14] + (8))⁴ = [56] +(8)=[8] +(8) $=$ (8)

([14] + (8))⁵ = [70] +(8)=[22] +(8) $\ne$ (8)

([14] + (8))⁶ = [84] +(8)=[12] +(8) $\ne$ (8)

([14] + (8))⁷ = [98] +(8)=[2] +(8) $\ne$ (8)

([14] + (8))⁸ = [112] +(8)=[16] +(8) $=$ (8)

Therefore, [14] + (8) has order 6 in Z₂₄/(8)

ii)

$\Z^{\times}_{10}=\{1,3,7,9\}$

$\Z_{10}=\{0,1,2,3,4,5,6,7,8,9\}$

$<2,9> = (2,9),(4,1),(6,9),(8,1),(0,9),(2,1),(4,9),(6,1),(8,0),(0,1)$

The order of quotient group $\Z_{10}\times\Z_{10}^{\times}/<2,9>$ :

$\frac{|\Z_{10}\times\Z_{10}^{\times}|}{|<2,9>|}=\frac{10}{10}=1$