Abelian group of finite order

Question

This problem was submitted to me as part of Abstract Algebra Homework.

Let $(G,\cdot)$ be an abelian group of finite order. Show that if $o(G)=n$ then $a^n=e$ for all $a\in G $. Hint : Show that if $G=\{e,a_1,a_2,\cdots,a_{n-1} \}$ then $G=\{a,a\cdot a_1, a \cdot a_2 , \cdots , a \cdot a_{n-1} \}$

with $o(\cdot)$ is the cardinality and $e$ is the identity element.

My first idea is that it reminded me of Cauchy's theorem for Abelian groups of finite order. However I did not manage to link it to it. Edit 1: It was actually Lagrange theorem I was thinking of.

A colleague proposed :

• to prove that $\exists k \in [[1,n]], a^k =e$
• to prove the hint when $a \neq e$ (it is trivial otherwise), $\forall k < n, a^k \neq e$, "since otherwise the powers of $a$ form a smaller set of numbers than $G$ so you have to have $a^n =e$.

Edit 2: The proof must not use Lagrange's theorem.

Answer 1

Let $(G,*)$ a group of finite order. We have that $o(G)=n$ with $n \in \mathbb{N}$. We have that if $a^{m+m'}=a^{m}$ then $a^{m'}=e$ with $m,m' \in \mathbb{N}$ (we only need multiplicate for a^{-m} in both terms of equality). Then we have that $a^{n}=a^{i}$ with $i \in \{0,...,n-1\}$ because $G$ is a finite group of order $n$. Then the set $H := \{1,a,...,a^{n-i-1}\}$ is a subgroup. Then we aplicate Lagrange's Theorem and we have that $n=o(G)=o(H)*l=(n-i)*l$ with $l \in \mathbb{N}$. So $a^{n}=a^{(n-i)*l}=a^(n-i)*...*a^(n-i)=e*...*e=e$

Sorry I do not read that the proof can not use Lagrange's Theorem. Sorry I am not sure but... Do you say that if $a \neq e$ then $a^{k} \neq e \forall k \in \{1,...,n-1\}$? That is false. For example take $(\mathbb{Z}/4\mathbb{Z},+)$ and take the subgroup $\{\bar{0},\bar{2}\}$ then you have that $\bar{2}^{2}=\bar{4}=\bar{0}$ and $2 < 4$.

Answer 2

It is $G=aG$; in fact, the map $G \to aG$, defined by $g \mapsto ag$, is a bijection and $aG\subseteq G$ by closure under multiplication. If $a=e$, then trivially $a^n=e$. Assume now $a \ne e$ and call $a_n:=e$; then, $G=aG \Rightarrow \forall i \in \{1,\dots,n\}, \exists j \in \{1,\dots,n\}, j \ne i$, such that $a_i=aa_j$; therefore:

$$\prod_{i=1}^{n}a_i=\prod_{i=1}^{n}aa_{\sigma(i)} \tag 1$$

where $\sigma \in S_{n}$ (and such that $\sigma(i) \ne i, \forall i \in \{1,\dots,n\}$, though this seems not to be relevant in proving the claim). By commutativity, $(1)$ reads:

$$\prod_{i=1}^{n}a_i=\prod_{i=1}^{n}aa_{\sigma(i)}=a^n\prod_{i=1}^{n}a_{\sigma(i)}=a^n\prod_{i=1}^{n}a_i \tag 2$$

whence $a^n=e$.

Remark. In order $(1)$ to hold, it is essential that the product extend till $n$.

Answer 3

Use the hint: let $b=ea_1a_2\dotsm a_{n-1}$. Then $$ b=ae\cdot aa_1\cdot \dotsm aa_{n-1}=a^nb $$ (because the group is abelian). Hence $a^n=e$ by cancelling $b$.

Why can we write $G=\{a,aa_1,\dots,aa_{n-1}\}$? Consider the map $\mu_a\colon G\to G$, $\mu_a(x)=ax$. This map is injective, hence also surjective.