• May 25th 2010, 08:54 PM
simplependulum
Hi I am new to Group Theory and I am confused by a problem :

Let $\displaystyle G$ be any group of order $\displaystyle n$ and $\displaystyle k$ be an integer prime to $\displaystyle n$ .

Is it true that $\displaystyle x^k = y^k ~ \implies x = y ~~~ x,y \in G$ ?

I know there exists $\displaystyle A,B ~\in \mathbb{N}$ such that $\displaystyle Ak - Bn = 1$

Then $\displaystyle (x^k)^A = (y^k)^A$

$\displaystyle x^{Ak} = y^{Ak}$

$\displaystyle x^{1 + Bn} = y^{1 + Bn}$

$\displaystyle x (x^n)^B = y (y^n)^B$

I think if $\displaystyle a^n = e$ , then i can prove it but is it really true ? Or my proof has some mistakes ? Actually , it is what i guess and i don't know if it is correct .

Thanks !
• May 25th 2010, 09:12 PM
chiph588@
Quote:

Originally Posted by simplependulum
Hi I am new to Group Theory and I am confused by a problem :

Let $\displaystyle G$ be any group of order $\displaystyle n$ and $\displaystyle k$ be an integer prime to $\displaystyle n$ .

Is it true that $\displaystyle x^k = y^k ~ \implies x = y ~~~ x,y \in G$ ?

I know there exists $\displaystyle A,B ~\in \mathbb{N}$ such that $\displaystyle Ak - Bn = 1$

Then $\displaystyle (x^k)^A = (y^k)^A$

$\displaystyle x^{Ak} = y^{Ak}$

$\displaystyle x^{1 + Bn} = y^{1 + Bn}$

$\displaystyle x (x^n)^B = y (y^n)^B$

I think if $\displaystyle a^n = e$ , then i can prove it but is it really true ? Or my proof has some mistakes ? Actually , it is what i guess and i don't know if it is correct .

Thanks !

Looks fine to me!
• May 25th 2010, 09:16 PM
simplependulum
Quote:

Originally Posted by chiph588@
Looks fine to me!

But how to prove it true ? $\displaystyle a^n =e$

$\displaystyle e$ is the neutral element of $\displaystyle G$
• May 25th 2010, 09:20 PM
chiph588@
Quote:

Originally Posted by simplependulum
But how to prove it true ? $\displaystyle a^n =e$

$\displaystyle e$ is the neutral element of $\displaystyle G$

See here.
• May 25th 2010, 09:33 PM
simplependulum
Quote:

Originally Posted by chiph588@
See here.

Oh i see . If the group is finite , then we would find that

$\displaystyle \{\ e,a,a^2,... \}\$ contains some repeated elements and write $\displaystyle a^s = a^t ~ \implies a^{s-t} = e ~~ s>t$

$\displaystyle \{\ e , a , a^2 ,..., a^{s-t-1} \}\$ is a subgroup of $\displaystyle G$ then use Lagrange's Theorem to show that $\displaystyle a^n = e$ .

Thank you very much (Happy)
• May 26th 2010, 01:36 AM
Swlabr
Quote:

Originally Posted by simplependulum
Hi I am new to Group Theory and I am confused by a problem :

Let $\displaystyle G$ be any group of order $\displaystyle n$ and $\displaystyle k$ be an integer prime to $\displaystyle n$ .

Is it true that $\displaystyle x^k = y^k ~ \implies x = y ~~~ x,y \in G$ ?

I know there exists $\displaystyle A,B ~\in \mathbb{N}$ such that $\displaystyle Ak - Bn = 1$

Then $\displaystyle (x^k)^A = (y^k)^A$

$\displaystyle x^{Ak} = y^{Ak}$

$\displaystyle x^{1 + Bn} = y^{1 + Bn}$

$\displaystyle x (x^n)^B = y (y^n)^B$

I think if $\displaystyle a^n = e$ , then i can prove it but is it really true ? Or my proof has some mistakes ? Actually , it is what i guess and i don't know if it is correct .

Thanks !

The property $\displaystyle 1 \neq x^n = y^n \Rightarrow x=y$ is called the `unique roots property', and it does not always hold. It does hold in, for example, free groups (infinite groups that can be viewed as the sort of top-down building blocks of group theory).

Such groups have the property that $\displaystyle g^mh^n = h^ng^m \Rightarrow gh=hg$.

However, this property does not always hold. A finite example would be the elements $\displaystyle \alpha = (123)(45)$ and $\displaystyle \beta = (123)$. Then $\displaystyle \alpha^2 = \beta^2$ but $\displaystyle \alpha \neq \beta$.

What I believe you proved in your post is that if $\displaystyle |G| = n$ and $\displaystyle gcd(n, k) = 1$ then if $\displaystyle a^k=b^k$ we have that $\displaystyle a=b$.

(An infinite example of a group without the unique roots property is the group given by the presentation (rules) $\displaystyle \langle a, b; a^2=b^2\rangle$. This is non-trivial and infinite as it has a homomorphism onto the infinite dihedral group, $\displaystyle C_2 \ast C_2$)