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

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?

• @AaronMaroja: Since $G=\langle a \rangle$ and the OP speaks about the order of $a$, it is more than reasonable to assume so. Jun 8, 2015 at 15:30

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.

• Is it possible to prove without Lagrange's theorem? That's a few chapters ahead of this exercise's chapter. In other words, I'm assuming there's another way to prove given that Lagrange's theorem has yet to be mentioned. Jun 8, 2015 at 15:47
• You can utilize first principles that more or less amounts to the same thing. What can you say about $b^{|\langle a \rangle|}$ using the fact that the group is finite? Jun 8, 2015 at 15:49
• @KevinMeredith I gave an alternative proof. Jun 8, 2015 at 16:03

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 $|G| = n$.

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.

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.

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.

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$(a)$.