# Math Help - Abstract Algebra: Lagrange's Theorem 2

1. ## Abstract Algebra: Lagrange's Theorem 2

Hail Mathematicians,

My last question for the day... I think. Check my solution please.

Problem:

The exponent of a group $G$ is the smallest positive integer $n$ such that $a^n = e$ for each $a \in G$ ( $e$ is the identity element), if such an integer $n$ exists. Prove that every finite group has an exponent, and that this exponent divides the order of the group. [Suggestion: Consider the LCM of $\{ o(a)~:~a \in G \}$.]

Solution:

We need to show: (1) If $G$ is finite, then $G$ has an exponent; and (2) we need to find such an exponent.

Proof:

(1) By the contrapositive, assume that $G$ does not have an exponent. Then there exists some $a \in G$ such that $a^n \ne e$ for all $n \in \mathbb{Z}^+$. But that would mean $a$ is of infinite order (by the definition of $o(a)$), and so, $G$ would have to be infinite.

(2) Consider $n = \mbox{LCM}( \{ o(a)~:~a \in G \})$. Now, for any $a \in G$, we have $a^k = e$ for some $k \in \mathbb{Z}^+$ by (1). But since $k \mid n$, we can write $n = mk$ for some $k \in \mathbb{Z}$. Thus, $a^n = a^{mk} = (a^k)^m = e^m = e$. And since $n$ is the LCM of the orders of all $a \in G$, it is the smallest integer for which this is true. Moreover, $n \mid |G|$ as a corollary of Lagrange's Theorem.

QED

Thanks for tuning in

2. If $a$ has infinite order than $a,a^2,a^3,...$ must all be distinct. Because otherwise if not then $a^i = a^j \implies a^{i-j}=e$ which is not possible. But then $\{a,a^2,a^3,..\}$ implies $G$ has infinitely many elements, a contradiction.

If $a\in G$ for finite group construct $\left< a\right>$ then this is a subgroup and its order is the order of $a$. Since $|\left< a \right>| | |G|$ it means the order of $a$ divides $|G|$.

3. Originally Posted by ThePerfectHacker
If $a$ has infinite order than $a,a^2,a^3,...$ must all be distinct. Because otherwise if not then $a^i = a^j \implies a^{i-j}=e$ which is not possible. But then $\{a,a^2,a^3,..\}$ implies $G$ has infinitely many elements, a contradiction.

If $a\in G$ for finite group construct $\left< a\right>$ then this is a subgroup and its order is the order of $a$. Since $|\left< a \right>| | |G|$ it means the order of $a$ divides $|G|$.
i thought of doing the proof you gave, pretty much exactly, but (1) i didn't want to single out <a>, i tend to have a problem choosing specific examples when i am trying to prove something in general. and (2) the question suggested using the LCM of the orders, so i wanted to do it that way. does my proof do the job with those two in mind?

going back to what i said in (1) above: are there subgroups of G that are not generated? that is, they are not of the form <a> for any a in G?