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If $|G|=p^rm$ with $(p,m)=1$, suppose that $x\in G$ is an element such that $o(x)=p^{r_1}m_1$ with $r_1>0$ and $(m_1,p)=1$. I dont understand why exist $a,b\in G$ such that:

1) $a$ has order a power of $p$

2) $b$ has order coprime with $p$

3) $x=ab$ and $[a,b]=1$

This fact is often used in proofs and it is presented as an obviuosly fact.

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1  
I mean, we can just take $a=x$ and $b=1$ right? Or $a=x^{e_1}$ and $b=x^{e_2}$ for any integers $e_1,e_2$? Are you sure there isn't some other part of the question that's missing? –  Alexander Gruber Oct 28 '12 at 17:06
    
I'm sorry there is a mistake in the question, I will edit it. –  Dubious Oct 28 '12 at 17:07
    
What does $[a,b]$ mean in this context? $aba^{-1}b^{-1}$? –  Hagen von Eitzen Oct 28 '12 at 17:18
    
yes $[a,b]$ is the commutator –  Dubious Oct 28 '12 at 17:26

3 Answers 3

up vote 1 down vote accepted

Let $u=x^{m_1}$, $v=x^{p^{r_1}}$. Then $o(u)=p^{r_1}$ and $o(v)=m_1$. The generated group $\langle u,v\rangle$ is a subgroup of $\langle x\rangle$ and its order must be a multiple of both $o(u)$ and $o(v)$, hence we conclude $\langle u,v\rangle = \langle x\rangle$, especially $\langle u,v\rangle$ is abelian. That means that $x=u^k v^n$ for some integers $k,n$. Let $a=u^k$, $b=v^n$. Then

  • The order of $a$ divides the order of $u$, hence is a power of $p$
  • The order of $b$ divides the order of $v$, hence is coprime to $p$
  • Clearly, $x=ab$
  • Since $a$ and $b$ were taken from the cyclic grpup $\langle x\rangle$, they commute.
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Write the cyclic group $\langle x \rangle$ as direct product of a $p$-subgroup and a $p'$-subgroup, and take the two projections of $x$ into the two factors.

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For each $s\in \mathbb N \text{ with }(s,m_1)=1$ the element $a=x^{s\cdot p^{r_1}}$ has order $m_1$. Similarly for each $t\in\mathbb N \text{ with } (t,p)=1$ the element $b=x^{t\cdot m_1}$ has order $p^{r_1}$. From $(p,m_1)=1$ there are $s,t\in\mathbb N$ such that $s\cdot p^{r_1}+t\cdot m_1=1$.

For these $s,t$ you have $x=a\cdot b$ and $[a,b]=1$.

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Your $ab\ne x$. –  Hagen von Eitzen Oct 28 '12 at 17:20

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