Question

${\mathbf{\text{Letkbeapositivedivisorofthepositiveintegern.ProvethatH}}}_{\mathbf{\text{k}}}\mathbf{\text{=}}\left\{{\text{a}}_{\text{n}}{\text{∈U}}_{\text{n}}\text{|a}\equiv \text{1}\left(\text{modk}\right)\right\}{\mathbf{\text{isasubgroupofU}}}_{\mathbf{\text{n}}}\mathbf{\text{.}}$

Short answer

Answer

$Itisprovedthat{H}_k=\left\{a_n\in U_n|a=1\left(\mathrm{mod}k\right)\right\}isasubgroupof{U}_n.$

Step-by-step solution

Step by Step Solution Step 1: Required Theorem

A non-empty subset H of a group G is a subgroup of G provided that,

$$\begin{gathered} \text{(i)Ifa,b}\in \text{H,thenab}\in \text{Hand} \\ \left(ii\right)If\text{a}\in {\text{Hthena}}^{-\mathbf{1}}\in \text{H} \end{gathered}$$

Prove that Hk=an∈Un|a=1modk is a subgroup of Un

$$\begin{gathered} Leta,b\in H_k.Then{\left[ab\right]}_k={\left[a\right]}_k{\left[b\right]}_k \\ Sinceitisgiventhat{H}_{k}isasubsetof{U}_n,sotakea=b=1: \end{gathered}$$

$\begin{array}{c}{\left[ab\right]}_k={\left[a\right]}_k{\left[b\right]}_k\\ ={\left[1\right]}_k{\left[1\right]}_k\\ ={\left[1\right]}_k\end{array}$

So, we can say that H_k is closed under group operation

Also, we know that,

$\begin{array}{c}{\left[a^{-1}\right]}_k={{\left[a\right]}_k}^{-1}\\ ={\left[1\right]}_k^{-1}\\ ={\left[1\right]}_k\end{array}$

Thus, we see that H_kis also closed under inverse operation.

Since both the conditions for theorem 7.11 are satisfied above, we can conclude that $H_k=\left\{a_n\in U_n|a=1\left(\mathrm{mod}k\right)\right\}$is a subgroup of U_n.