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

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- Oct 25th 2008, 03:50 AMpettergroup
Let G be a group and $\displaystyle p \in N^* $ s.t equation $\displaystyle x^p=e$ have unique solution . Prove that for every $\displaystyle a\in G$ the equation $\displaystyle x^p=a$ have unique solution

- Oct 25th 2008, 05:18 AMHallsofIvy
- Oct 25th 2008, 03:15 PMThePerfectHacker
But does this only work if $\displaystyle G$ is assumed to be

*abelian*?

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

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

What if $\displaystyle p$ is not a prime? (Thinking)

Perhaps, there is a conterexample - but I am too lazy to try to find one now. (Rofl) - Oct 25th 2008, 04:41 PMThePerfectHacker
If $\displaystyle p$ is not a prime then this statement is still true.

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