1. ## Finding a group

anyone knows a non abelian group $\displaystyle G$ such that for every $\displaystyle n\geq 2$ and $\displaystyle g\in G$ the equation $\displaystyle x^n=g$ has a unique solution (i.e. $\displaystyle x^n=g$ always has a solution and that solution is unique).

2. Originally Posted by Andres Perez
anyone knows a non abelian group $\displaystyle G$ such that for every $\displaystyle n\geq 2$ and $\displaystyle g\in G$ the equation $\displaystyle x^n=g$ has a unique solution (i.e. $\displaystyle x^n=g$ always has a solution and that solution is unique).
Hint: Take $\displaystyle G$ to be any finite group.

Hint2: What must the element $\displaystyle g$ be? Remember, the equation must hold for every element in your group...

3. Originally Posted by Swlabr
Hint: Take $\displaystyle G$ to be any finite group.

Hint2: What must the element [tex]g[\math] be? Remember, the equation must hold for every element in your group...

I don't follow your hints: if $\displaystyle G=S_3$ , say and $\displaystyle n=3\,,\,\,g=(1)$ then there is NO unique solution to the equation $\displaystyle x^3=g$ in S_3....so why do you hint to take $\displaystyle G$ to be any finite group? Unless you meant that ANY finite group will fail the test...but then this hardly answers the OP's question, doesn't it?

I, for one, cannot think of any group at all, abelian or not, finite or not, that the required condition is true...

Tonio

4. Originally Posted by tonio
I don't follow your hints: if $\displaystyle G=S_3$ , say and $\displaystyle n=3\,,\,\,g=(1)$ then there is NO unique solution to the equation $\displaystyle x^3=g$ in S_3....so why do you hint to take $\displaystyle G$ to be any finite group? Unless you meant that ANY finite group will fail the test...but then this hardly answers the OP's question, doesn't it?

I, for one, cannot think of any group at all, abelian or not, finite or not, that the required condition is true...

Tonio
oooooo-I was thinking the g was unique, not the x...

Taking it your way, I will give a hint as to why no such groups exists (abelian or non-abelian): Take g=1. Then what?

5. Originally Posted by Swlabr
oooooo-I was thinking the g was unique, not the x...

Taking it your way, I will give a hint as to why no such groups exists (abelian or non-abelian): Take g=1. Then what?

Exactly: there can't be any group that fulfills that condition. Perhaps the OP did mean something else...

Tonio

6. Originally Posted by Swlabr

Taking it your way, I will give a hint as to why no such groups exists (abelian or non-abelian): Take g=1. Then what?
actually there are infinitely many such abelian groups. for example $\displaystyle \mathbb{Q}$ or, more generally, any field with characteristic zero considered as additive groups.

finding a non-abelian group with that property doesn't seems to be that easy! what is obvious is that such groups would have to be infinite.

7. Originally Posted by Andres Perez
anyone knows a non abelian group $\displaystyle G$ such that for every $\displaystyle n\geq 2$ and $\displaystyle g\in G$ the equation $\displaystyle x^n=g$ has a unique solution (i.e. $\displaystyle x^n=g$ always has a solution and that solution is unique).
consider $\displaystyle \mathbb{Q}$ and $\displaystyle \mathbb{R}$ as additive groups and define the map $\displaystyle \theta : \mathbb{Q} \longrightarrow \text{Aut}(\mathbb{R})$ by $\displaystyle \theta(q)(r)=2^qr, \ \forall \ q \in \mathbb{Q}, r \in \mathbb{R}.$ see that $\displaystyle \theta$ is a group homomorphism. let $\displaystyle G=\mathbb{R} \rtimes_{\mathbb{\theta}} \mathbb{Q}.$ then $\displaystyle G$ is non-abelian.

to prove that it has the above property see that for any $\displaystyle x=(r,q) \in G$ and integer $\displaystyle n \geq 1$ we have $\displaystyle x^n=((1+2^q+ \cdots + 2^{(n-1)q})r, nq).$ so if $\displaystyle g=(r',q') \in G$ and $\displaystyle n \geq 1$ are given, then the

only $\displaystyle x \in G$ which satisfies the equation $\displaystyle x^n=g$ is $\displaystyle x=(r,q),$ where $\displaystyle q=\frac{q'}{n}$ and $\displaystyle r=\frac{r'}{1+2^q + \cdots + 2^{(n-1)q}}.$