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

__Solution:__

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

__Proof:__

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

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

QED

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