Lets look at this exercise:

$G$ is a finite group and $p$ is a prime that divides $|G|$. If for every element $x\in G$ such that $g.c.d.(o(x),p)=1$, we have that $g.c.d.\left([G:C_G(x)],p\right)=1$, prove that $G$ is the direct product of a $p$-Sylow with a group that has order coprime with $p$.

(notation: clearly $C_G(\cdot)$ is the centralizer in $G$)

If $|G|=p^rm$ ($p$ is coprime with $m$), to solve the exercise, I think that it is enough to find a normal subgroup of $G$ that has order $m$ and such that centralize a $p$-Sylow.

  • $\begingroup$ Where is this problem from? $\endgroup$ Oct 27, 2012 at 15:54
  • $\begingroup$ It is from an exam covering the finite group theory. $\endgroup$
    – Dubious
    Oct 27, 2012 at 16:22
  • 1
    $\begingroup$ Is it from an exam you are currently taking? Because then you should be solving it yourself... $\endgroup$ Oct 28, 2012 at 17:42
  • $\begingroup$ It is an old problem, and I'm trying to solve it. $\endgroup$
    – Dubious
    Oct 28, 2012 at 19:13
  • $\begingroup$ no the p-Sylow is not abelian $\endgroup$
    – Dubious
    Oct 31, 2012 at 15:43

1 Answer 1


I'd translate your condition on $G$ to "every $p'$-element centralizes some $p$-Sylow subgroup". As all $p$-Sylow subgroups are conjugate, this is equivalent to saying that every $p'$-element has a conjugate lying in the centralizer of a fixed $p$-Sylow subgroup $S$.

For finite groups the only subgroup intersecting each conjugacy class is the whole group (see https://mathoverflow.net/questions/26979/generating-a-finite-group-from-elements-in-each-conjugacy-class).

So if you can show that the normalizer $N_G(S)$ contains a conjugate to every element of $G$ that has order a multiple of $p$, you know that $S$ is normal in $G$. Then by Schur-Zassenhaus there exists a $p'$-subgroup $H$ such that $G = S\rtimes H$. But as $H$ centralizes $S$ by the assumptions, the product is direct.

To close the gap, by Sylow's theorem it is enough to show that given an element $x \in G$ of order divisible by $p$ it normalizes some $p$-Sylow subgroup. Write $x = y\cdot z$ with $y$ a $p$-element, $z$ a $p'$-element and $y$ and $z$ commuting (look at $\langle x \rangle$ if you are unsure about the existence of such a decomposition). If $z=1$ then $x$ is contained in some $p$-Sylow subgroup. Otherwise the centralizer of $z$ contains a $p$-Sylow $T$ that wlog contains $y$. As $z$ centralizes $T$ it also normalizes it, and we are done.

  • $\begingroup$ very well, thanks! However I found previously, another but longer solution (it is very similar to yours). I will post my solution for completenes $\endgroup$
    – Dubious
    Oct 31, 2012 at 16:06
  • $\begingroup$ @Galoisfan: You can even streamline my proof a little by unifying the argument that every element normalizes some Sylow subgroup. $\endgroup$
    – j.p.
    Oct 31, 2012 at 19:01

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