If $G = \langle a\rangle$ and $b$ $\in$ $G$, the order of $b$ is a factor of the order of $a$

Question

Prof. Pinter's "A Book of Abstract Algebra" presents the following exercise from the "Cyclic Groups" chapter:

If $G = \langle a\rangle$ is finite and $b$ $\in$ $G$, the order of $b$ is a factor of the order of $a$

I believe that the proof is given by Theorem $2$ of this chapter:

Every subgroup of a cyclic group is a cyclic.

Does this theorem provide an adequate proof to this exercise?

Answer 1

The theorem you have indicated argues that every subgroup is cyclic. The question requires something slightly different.

Assuming $G$ is a finite group, what can you say about the order of the subgroup $H=\langle b \rangle$? Try and use Lagrange's theorem after that point.

Answer 2

Hint: Consider $m$ the least integer such that $a^m \in H$, where $H = \langle b \rangle \leq G$. Clearly $\langle a^m \rangle \subseteq H$. Take any $a^u \in H$ and divide $u$ by $m$ use the minimality of $m$.

Finally $H = \langle a^m \rangle$ where $H$ has order $n/m$ where $|G| = n$.

Answer 3

Consider the subgroup $H = \langle b \rangle$. The order of $b$ will be the order of $H$. But we know (Lagrange's theorem, for instance) that $|H| = {\rm ord} b $ divides $|G| = {\rm ord} a$, which is your conclusion.

Answer 4

I would say no, in itself, regarding purely logically, the cited statement would allow that the order of the (cyclic) subgroup is not a factor of the order of the group.

However, together with Lagrange's theorem ($|H|$ divides $|G|$ for all $H\le G$), it is enough.

Answer 5

In the previous chapter, chapter 10, exercise 10.D.2 says:

Let a be any element of finite order of a group G. Prove the following:

The order of a^k is a divisor (factor) of the order of a.

That exercise has an answer in the back of the book.

Answer 6

Here's an answer more in line with the concepts learned in Pinter's book up to that point (ie. no Lagrange's theorem etc).

Let ord$(a) = n$. Therefore, there are $n$ elements in $G$.

Then since $b \in G$, we have $b = a^i$ for some $1 \leq i \leq n$. This is because since $G = \langle a \rangle$ is cyclic, $G$ contains all the powers of $a$ and nothing else.

So we have $b^n = (a^i)^n = (a^n)^i = e^i = e,$

showing that ord$(b) | n$.

That is, ord$(b) |$ ord$(a)$.