I have a theorem to establish:

Let $G$ a group and $a\in G$ such that $a$ is an element of finite order, then $\lvert \langle a \rangle\rvert=o(a)$.

We denote $o(a)$ as the order of $a$.

So my question is...

Is it true that $G= \langle a \rangle$ if and only if $\lvert G \rvert=o(a)$?

I am getting confused if it follows by definition or it requires a formal proof.

THANKS, for the help :D

  • $\begingroup$ I'm not sure what you mean. It is always true that $| \langle a \rangle | = o(a)$. Do you mean to ask if $G = \langle a \rangle$ if and only if $| \langle a \rangle | = o(G)$? $\endgroup$ Oct 13, 2020 at 19:38
  • $\begingroup$ No, $|\langle a \rangle|$ is how we define $o(a)$ for $a \in G$, regardless of whether the group $G$ is cyclic. $\endgroup$
    – xxxxxxxxx
    Oct 13, 2020 at 19:38
  • $\begingroup$ In your definition $o(a)$ is defined only if it is finite. If it is finite and $a\in G$ and |$<a>|=|G|$ then $G=<a>$. $\endgroup$
    – Mark
    Oct 13, 2020 at 19:44
  • $\begingroup$ I already do the correction to the text. $\endgroup$
    – Blacks
    Oct 13, 2020 at 19:45
  • $\begingroup$ Yes @HallaSurvivor I tried to say $|G|=o(a)$ $\endgroup$
    – Blacks
    Oct 13, 2020 at 19:48

1 Answer 1


You are correct. If for some $a \in G$ we have $|G| = o(a) < \infty$, then we must have $G = \langle a \rangle$.

To see why, we can use containment and finiteness.

Can you show $\langle a \rangle \subseteq G$? This will use the fact that $G$ is closed under its multiplication, and $a \in G$. If you want to be extra formal, you might show each $a^n \in G$ by induction on $n$.

Next, we use a crucial fact about finite sets. If $|X| = |Y| < \infty$ and $X \subseteq Y$, then $X = Y$. That is, when we're in the finite world, you cannot pull any hilbert's hotel type tricks. So if $X \subseteq Y$ and they are the same size, they must actually be the same.

But we showed earlier that $\langle a \rangle \subseteq G$, and we are assuming that $|G| = | \langle a \rangle | < \infty$. So $G = \langle a \rangle$.

I hope this helps! ^_^

  • 1
    $\begingroup$ Yes! I alredy understood! Thanks @HallaSurvivor $\endgroup$
    – Blacks
    Oct 13, 2020 at 20:22

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