Decomposition into $p$ part and $p'$ part

Question

I am doing a small presentation in group theory and I am struggling with something.

We have the definitions:

Let $G$ be a finite group and $p$ a prime such that $p||G|$. An element $g \in G$ is called a $p$-element if $o(g) = p^t$ for some $t \in\mathbb{N}$. An element $h \in G$ is called a $p'$-element if $\gcd(ord(h), p) = 1$.

I cannot prove the following:

If $g\in G$, it follows that $\exists x,y\in G$ such that $g = xy=yx$, where $x$ is a $p$-element and $y$ is a $p'$-element.

I suppose that $x,y$ are powers of $g$. If $ord(g)=p^rs$, then $g^{p^rs+1}=g$. I tried different combinations for exponents but I could not get to the result.

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

2

$p^r$ and $s$ are relatively prime, so there are suitable $l$ and $k$ so that $lp^r+ks=1$. Let $x=g^{ks}$ and let $y$ be $g^{lp^r}$.

To prove $g^{ks}$ is a $p$ element and to prove that $g^{lp^r}$ is a $p'$ element you are going to need this result: $|g^k|=\dfrac{|g|}{\gcd(|g|,k)}$

To see why a $k$ and $l$ exist you have to use the euclidean algorithm and work backwards. Here is an example

edited Apr 13 '17 at 12:21

Community

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answered Jan 06 '15 at 07:00

Asinomás

105,651

I use $|g|$ to denote the order of $g$. – Asinomás Jan 06 '15 at 07:01
Apparently, I have not thought enough about it. Thank you. – Iulia Jan 06 '15 at 07:34

Decomposition into $p$ part and $p'$ part

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