if $$ is a finite abelian group $()$ divides the exponent of $$, for each $\in $

Question

Define the exponent of a group as $\sup o(a)$ ,where $a \in G$. Show that if $G$ is a finite abelian group $o(a)$ divides the exponent of $G$, for each $\in $

Here $o(a)$ denotes the order of an element $a$? How can I show this? Can someone please show the right approach to attempt this question.

Thanks.

Please provide context for your question. What results about groups/finite abelian groups do you know? What do you know about orders of elements in abelian groups? — Arturo Magidin, Mar 17 '21 at 03:57
That is not the standard definition of the exponent of a finite group $G$. The exponent is the least multiple of the orders of the elements. — Derek Holt, Mar 17 '21 at 07:23

score -1 · Answer 1 · answered Mar 17 '21 at 02:59

By the fundamental theorem of finitely generated groups, we can write $G = \bigoplus\limits_{i = 1}^n \bigoplus\limits_{j = 1}^{m_i} \mathbb{Z} / p_i^{e_{ij}} \mathbb{Z}$ with inclusion maps $k_{ij} : \mathbb{Z} / p_i^{e_{ij}} \mathbb{Z} \to G$, such that the $p_i$ are distinct primes and for each $i$, the $e_{ij}$ are sorted in increasing order.

Let $M = \prod\limits_{i = 1}^n p_i^{e_{i m_j}}$. Consider the map $f: G \to G$ defined by multiplication by $M$. It's easy to see that $f = 0$. Thus, we see that for all $a \in A$, $o(a) | M$. And in particular, for all $a \in A$, $o(a) \leq M$.

Now note that for $x = \sum\limits_{i = 1}^n k_{i m_j}(1)$, we have $o(x) = M$.

Then we see that $M$ is the exponent of $G$ and that for all $a \in G$, $o(a)$ is a factor of $M$.

if $$ is a finite abelian group $()$ divides the exponent of $$, for each $\in $

1 Answers1