Question

Let G=(a) be a cyclic group of order n. If H is subgroup of G, Show that $\left|H\right|$ is a divisor of n. (Hint: Exercise 44 and Theorem 7.17)

Short answer

Answer

Subgroup$\left|H\right|$is a divisor of n.

Step-by-step solution

Step by Step Solution Step 1: Theorem required

Theorem 7.17

Every Subgroup of a cyclic group is itself cyclic.

Result from exercise 44

Consider the given result from exercise 44,$〈a^m〉=〈a^{\left(m,n\right)}〉$ .

Prove that subgroup is a divisor of n

From Theorem 7.17, we know that every subgroup of a cyclic group is itself cyclic.

Therefore, H is itself cyclic. So, there must be some $0\le m\le n$, such that $H=〈a^m〉$.

From the result of exercise 44, we know that,

$〈a^m〉=〈a^{\left(m,n\right)}〉$

$Theorderof〈a^{\left(m,n\right)}〉isthesameastheorderof{a}^{\left(m,n\right)},whichis\frac{n}{m,n}.$

$Since\frac{n}{m,n}isadivisorofnthus,itisprovedthattheSubgroup\left|H\right|isadivisorofn.$