Relationship between periodic groups and cyclic groups

Question

Suppose $G$ is a finite abelian group with period equal to $|G|$. Prove $G$ is cyclic. We define the period of $G$ as the minimum $n \in \mathbb{Z}^{+}$ with $n\cdot g = 0$ for all $g \in G$.

So far I've tried to prove that $G$ is not isomorphic to any direct sum of finite abelian groups except for $\mathbb{Z}_{n}$ by means of The Fundamental Theorem of Finite Abelian Groups.

Answer 1

By assumption of period of $G$ is equal to $n:=|G|$, there is an element $g\in G$ s.t. $g,2g,\cdots, (n-1)g\neq 0.$

And we can prove $ig\neq jg$ for $1\le i\neq j\le n-1$.

So we get $G=\{ ig\in G| i=0,1,2,\cdots,n-1\}$. This means $g$ is just generator of $G$.

Answer 2

First note that your definition of period (which, as noted in a comment, is more often called exponent) can be rephrased as saying that it is the lcm of the orders of all the elements of the group. This is because for a given $g \in G$ and $n$ we have

$n \cdot g = 0$ iff the period of $g$ divides $n$.

Note that a non-abelian finite group need not have an element whose period equals the period of the whole group, think $S_{3}$.

However we have the following.

Let $G$ be a finite abelian group of period $e$. Then $G$ contains an element of period $e$.

First note that if $G$ contains an element $a$ of period $k$, then it contain an element $b$ of order any divisor $h$ of $k$, just take $b = (k/h ) \cdot a$.

If $e = \displaystyle\prod_{i=1}^{t} p_{i}^{n_{i}}$ is the decomposition of $e$ as a product of powers of distinct primes, with each $n_{i} > 0$, then, by the definition of $e$ as a lcm, $G$ contains an element $a_{i}$ of order a multiple of $p_{i}^{n_{i}}$ for each $i$.

By the previous argument, $G$ contains an element $b_{i}$ of period exactly $p_{i}^{n_{i}}$, for each $i$.

Since $G$ is abelian, $b = b_{1} + \dots + b_{t}$ has order $e$.

Your problem is now the special case $e = \lvert G \rvert$.

Answer 3

Since a finite abelian group is a product of cyclic groups (structure theorem), one form being

$$\Bbb Z_{a_1}×\Bbb Z_{a_2}×\dots ×\Bbb Z_{a_n},$$ where $a_i\mid a_{i+1}\,,\forall i,$

the exponent (period) is just the order of the cyclic factor of greatest order.

In this case that factor must have order $\lvert G\rvert. $ Or, $$G=\Bbb Z_{\lvert G\rvert}.$$