# group

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• October 25th 2008, 03:50 AM
petter
group
Let G be a group and $p \in N^*$ s.t equation $x^p=e$ have unique solution . Prove that for every $a\in G$ the equation $x^p=a$ have unique solution
• October 25th 2008, 05:18 AM
HallsofIvy
Quote:

Originally Posted by petter
Let G be a group and $p \in N^*$ s.t equation $x^p=e$ have unique solution . Prove that for every $a\in G$ the equation $x^p=a$ have unique solution

Suppose x^p= a and y^p= a. What can you say about (xy^(-1))^p?
• October 25th 2008, 03:15 PM
ThePerfectHacker
Quote:

Originally Posted by HallsofIvy
Suppose x^p= a and y^p= a. What can you say about (xy^(-1))^p?

But does this only work if $G$ is assumed to be abelian?

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Is $p$ a prime? I will assume that $p$ is a prime.

We will prove this for $S_n$ i.e. if $x^p = y^p \implies x=y$ where $x,y\in S_n$. Now say we can write $x = zx_1$ and $y=zy_1$ where $z$ is a disjoint permutation with $x_1$ and $y_1$. This means $z$ commutes with $x_1$ and $y_1$. Thus, $x^p = z^px_1^p$ and $y^p = z^py_1^p$ henceforth $x_1^p = y_1^p$. We can write $x = \sigma_1 ... \sigma_r$ as a product of disjoint cycles and $y=\tau_1 ... \tau_s$ as a product of disjoint cycles. We want to show $x=y$. Based on the reasoning above we can assume that each $\sigma_i$ and $\tau_j$ are different. It follows that $x^p = \sigma_1^p ... \sigma_r^p$ and $y^p = \tau_1^p ... \tau_s^p$. Since $\sigma_i^p,\tau_j^p$ are non-identity elements it means we have expressed $x^p$ as a product of disjoint cycles (for $\sigma_i^p,\tau_j^p$ are cycles - this is because $p$ is a prime!). Therefore, not only $r=s$ but it must be that we can rearrange them so that $\sigma_i^p = \tau_i^p$. Note, it is impossible that $\sigma_i \not = \tau_i$ and $\sigma_i^p = \tau_i^p$. This means that $x=y$. Now by the Cayley's theorem and finite group and be embedded in the symmetric group. Since this result applies to symmetric groups it will apply to its subgroups as well. Thus, this is true for any finite group $G$ and a prime $p$.

What if $p$ is not a prime? (Thinking)
Perhaps, there is a conterexample - but I am too lazy to try to find one now. (Rofl)
• October 25th 2008, 04:41 PM
ThePerfectHacker
Quote:

Originally Posted by ThePerfectHacker
What if $p$ is not a prime? (Thinking)
Perhaps, there is a conterexample - but I am too lazy to try to find one now. (Rofl)

If $p$ is not a prime then this statement is still true.

Say $x^n = e$ has exactly one solution (trivial one) and $n>1$. Then it must be that $x^p = e$ has exactly one solution for each $p|n$ a prime. Now if $x^n = y^n$ it means $(x^{(n/p)})^p = (y^{(n/p)})^p \implies x^{(n/p)} = y^{(n/p)}$ (by the above post). If $n/p$ is prime then it would be that $x=y$. Otherwise let $q$ be prime divisor of $(n/p)$ and arrive at $x^{n/(pq)} = y^{n/(pq)}$. Keep on doing this until you exhaust all the primes and arrive that $x=y$.