2
$\begingroup$

Let $G$ be a finite abelian group. If $m=$exponent of $G$, then by definition $m=lcm(|g_1|,...,|g_n|)$ for all $g_i \in G$ . It suffices to show if $g \in G$ has the maximum order, then $\forall g_i \in G$, $|g_i|$ divides $|g|$. But then I'm stuck.

$\endgroup$
2
$\begingroup$

(1). $g^n=1\implies |g|$ divides $n.$ Proof: Otherwise $n=m|g|+r$ for some integers $m,r$ with $1\leq r<|g|.$ But then $g^r=g^n (g^{|g|})^m=1$ with $1\leq r<|g|$, contrary to the def'n of $|g|.$

(2).. If $n>0$ and $n$ divides $|g|$ then $|g^n|=|g|/n.$ Proof: We have $(g^n)^{|g|/n}=g^{|g|}=1$, so by (1), $|g^n|$ divides $|g|/n.$ Therefore $$|g^n|\leq |g|/n.$$ But $g^{n|g^n|}=(g^n)^{|g^n|}=1,$ so by (1), $|g|$ divides $n|g^n|,$ so $|g|\leq n|g^n|$. Therefore $$|g|/n\leq |g^n|.$$

(3).(i). If $\gcd (|g|,|g'|=1)$ then $|gg'|=|g|\cdot |g'|.$

Proof: $(gg')^{|g|\cdot |g'|}=(g^{|g|})^{|g'|}\cdot (g'^{|g'|})^{|g|}=1$. So by (1) $|gg'|$ divides $|g|\cdot |g'|.$ $$ \text { Therefore }\bullet \quad |gg'|\leq |g|\cdot |g'|.$$ And we have $g^{|gg'|}=g'^{(-|gg'|)}$ so $$ \quad g^{|gg'|\cdot |g'|}= (g^{|gg'|})^{|g'|}= (g'^{(-|gg'|)})^{|g'|}= g'^{(-|gg'|\cdot |g'|)}= (g'^{|g'|})^{(-|gg'|)}=1.$$ So by (1), $|g|$ divides $|gg'|\cdot |g'|.$ Since $\gcd(|g|,|g'|)=1$ this implies $$|g| \text { divides } |gg'|.$$ Interchanging $g$ with $g'$ we also have $$|g'|\text { divides } |gg'|.$$ Now since $|g|$ and $|g'|$ both divide $|gg'|$ with $\gcd (|g|,|g'|)=1,$ we have: $|g|\cdot |g'|$ divides $|gg'|.$ $$\text {Therefore }\bullet \bullet \quad|g|\cdot |g'|\leq |gg'|.$$ By $\bullet$ and $\bullet \bullet$ we have $|gg'|=|g| \cdot |g'|$.

(3).(ii). If $g_1,...,g_k$ are members of $G$ such that $\gcd (|g_i| ,|g_j|)=1$ when $i\ne j,$ then $$|\prod_{i=1}^kg_i|=\prod_{i=1}^k|g_i|.$$

Proof: $k=1$ is trivial. If $k>1$ use induction on $k$: Let $g=\prod_{i=1}^{k-1}g_i$ and $g'=g_k.$ If $|g|=\prod_{i=1}^{k-1}|g_i|$ then $\gcd(g,g')=1$ so by (3).(i) we have $|gg'|=|g|\cdot |g'|=\prod_{i=1}^k|g_i|.$

(4). Let $\{p_1,...,p_k\}$ be a set of $k$ distinct primes such that for every $g\in G$ the order of $g$ is $\prod_{i=1}^kp_i^{e_i(g)}$ where each $e_i(g)$ is a non-negative integer.

For each $i$ let $x_i\in G$ such that $e_i(x_i)=\max \{e_i(x):x\in G\}.$

Let $n_i=|x_i|/p_i^{e_i(x_i)}.$ Let $y_i=x_i^{n_i}.$

By (2) we have $|y_i|=|x_i|/n_i=p_i^{e_i( x_i)}.$

Let $z=\prod_{i=1}^k y_i.$ Then $|y|=$ $|\prod_{i=1}^ky_i|=$ $\prod_{i=1}^k|y_i|=\prod_{i=1}^kp_i^{e_i(x_i)}$ by (3).(ii).

For every $g\in G$ we have $e_i(g)\leq e_i(x_i)$ by def'n of $x_i,$ so $|g|\leq |y|.$ In fact $|g|$ divides $|y|.$

REMARK. In the case where $G$ is the set of non-zero members of finite field $F$, where the group operation of $G$ is multiplication in $F$: Let $|G|$ be the number of members of $G.$ Let $y$ be as in (4). Now $|g|$ divides $|y|$ for every $g\in G$. So in F the number of solutions of the polynomial $x^{|y|}-1=0$ is $|G|$, so $|y|\geq |G|.$ But (Lagrange's theorem) $|y|$ divides $|G|.$ So $|y|=|G|$, implying that $G$ is a cyclic group.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.