Groups in 2016

Algebra Level 3

Let $(G,\circ)$ be a finite abelian group of order $n$ , say $G=\{a_i\}_{i=1}^{i=n}$ , where $n$ is a positive integer. Also, let $x=a_1\circ a_2\circ\cdots\circ a_{n-1}\circ a_n$ .

What is the value of $x^{2016}?$

Details and Assumptions:

$e_G$ denotes the identity element of the group $(G,\circ)$ .
$x^n=\underbrace{x\circ x\circ\cdots\circ x\circ x}_{n\textrm{ times}}$ .

Select one or more

x

n

x^n

n!

x^{n!}

e_G

2016

1 solution

Guillermo Templado
Feb 25, 2016

$G = \{a_i\}_{i=1}^{i= n}$ and G is an abelian group and $x = a_1 \circ a_2 \circ \cdots a_{n-1} \circ a_{n}$ . Since $\circ$ is a conmutative operation and in x are all the elements of G, we have two options:

1.- for $a_i$ its inverse $a_j$ is distinct to $a_i$ and they both are cancelled in x, or

2.- $a_{i}$ is its own inverse,

therefore x powered to 2 is $e_{G}$ since the operation $\circ$ is conmutative, so $x^2 = e_{G} \Rightarrow x^{2016} = (x^2)^{1008} = e_{G}^{1008} = e_{G}$

$x\neq e_G$ in general. A simple counterexample is the group $(\Bbb Z_2,+)$ , the group of integers modulo $2$ under addition modulo $2$ .

$x=0+1=1\neq 0=e_G$

Prasun Biswas - 5 years, 3 months ago

You make reason, I have changed the proof, can you see, if am I right now?

Guillermo Templado - 5 years, 3 months ago

Yes, but there's a much shorter (simpler) argument.

Note that the map $f\colon G\to G$ defined by $f(g)=g^{-1}$ is a bijection (easy to verify) and since $(G,\circ)$ is abelian, we can rearrange and associate the group elements in any order in an expression involving the group elements with the operation $\circ$ , so we can rearrange the elements in $x^2$ such that $x^2=\prod\limits_{i=1}^n (a_ia_i^{-1}) = (e_G)^n=e_G$ and hence conclude that $x^{2m}=e_G~\forall~m\in\Bbb{Z^+}$

Prasun Biswas - 5 years, 3 months ago

G being a finite abelian group is isomorphic to $(\mathbb{Z}_m, +)$ with $m \in \mathbb{N}$ or isomorphic to $(\mathbb{Z}_r \times \ldots \times \mathbb{Z}_s, +)$ with $r, s \in \mathbb{N}$ and the cartesian product is finite.

Guillermo Templado - 5 years, 3 months ago

Groups in 2016

1 solution

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