# proof that a finite abelian group has element with $\delimiter 69640972g\delimiter 86418188=\mathop{exp}\nolimits(G)$

###### Theorem 1

If $G$ is a finite abelian group, then $G$ has an element of order $\exp(G)$.

Proof. Write $\exp(G)=\prod p_{i}^{k_{i}}$. Since $\exp(G)$ is the least common multiple of the orders of each group element, it follows that for each $i$, there is an element whose order is a multiple of $p_{i}^{k_{i}}$, say $\lvert c_{i}\rvert=a_{i}p_{i}^{k_{i}}$. Let $d_{i}=c_{i}^{a_{i}}$. Then $\lvert d_{i}\rvert=p_{i}^{k_{i}}$. The $d_{i}$ thus have pairwise relatively prime orders, and thus

 $\left\lvert\prod d_{i}\right\rvert=\prod\left\lvert d_{i}\right\rvert=\exp(G)$

so that $\prod d_{i}$ is the desired element.

Title proof that a finite abelian group has element with $\delimiter 69640972g\delimiter 86418188=\mathop{exp}\nolimits(G)$ ProofThatAFiniteAbelianGroupHasElementWithlvertGrvertexpG 2013-03-22 16:34:05 2013-03-22 16:34:05 rm50 (10146) rm50 (10146) 4 rm50 (10146) Proof msc 20A99