# Abstract Algebra: Lagrange's Theorem 2

• Mar 16th 2008, 09:59 AM
Jhevon
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

Thanks for tuning in
• Mar 16th 2008, 10:38 AM
ThePerfectHacker
If $\displaystyle a$ has infinite order than $\displaystyle a,a^2,a^3,...$ must all be distinct. Because otherwise if not then $\displaystyle a^i = a^j \implies a^{i-j}=e$ which is not possible. But then $\displaystyle \{a,a^2,a^3,..\}$ implies $\displaystyle G$ has infinitely many elements, a contradiction.

If $\displaystyle a\in G$ for finite group construct $\displaystyle \left< a\right>$ then this is a subgroup and its order is the order of $\displaystyle a$. Since $\displaystyle |\left< a \right>| | |G|$ it means the order of $\displaystyle a$ divides $\displaystyle |G|$.
• Mar 16th 2008, 10:46 AM
Jhevon
Quote:

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

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

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?