Question

Let G=(a) be a cyclic group of order n. If k is a positive divisor of n,prove that G has a unique subgroup of order k. [Hint: Consider the subgroups generated by a^n/k.]

Short answer

Answer

It is proved that G has a unique subgroup of order k.

Step-by-step solution

Step by Step Solution Step 1: Required Theorem

Theorem 7.17:

Every Subgroup of a cyclic group is itself cyclic.

Result from exercise 44

$〈a^m〉=〈a^{\left(m,n\right)}〉$

Prove that G has a unique subgroup of order k

Let us examine the subgroup generated by the element a^n/k.

Since the element a^n/k has order k, the subgroup also has order k.

Also, consider a subgroup H$\le$G.

As we know, G is a cyclic group. From theorem 7.17, we can deduce that H is also cyclic.

So, for subgroup H, there must be some generator a^m for $0\le m<n$.

This can be expressed as:

$$\begin{gathered} \begin{array}{l}H=〈a^{\left(m,n\right)}〉.....\left(1\right) \\ So, \\ k=\left|H\right|=\frac{n}{\left(m,n\right)}\\ k=\frac{n}{\left(m,n\right)}\end{array} \\ Thus,\left(m,n\right)=\frac{n}{k}.....\left(2\right) \end{gathered}$$

From the equation (1) and (2),

$H=〈a^{\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$k$}\right.}〉$

Hence, it is proved that G has a unique subgroup of order k.