Let $G$ be a finite group. Suppose that $G$ has a normal Sylow $p-$subgroup $P$ such that $|P|=p^2$ where $p\neq 2$, but $P$ does not contain an element of order $p^2$ or equivalently $P$ is not cyclic.

Is it the case that if $g \in G$ has order $k$ such that $\gcd(k,p)=1$ then there is an element of $G$ of order $kp$? Thanks in advance.

  • 1
    $\begingroup$ Certainly not for every $k$ that's not a multiple of $p$, since it's a finite group and there are infinitely many such $k$. Or did you mean something else? $\endgroup$ – Robert Israel Jul 27 '12 at 5:03
  • $\begingroup$ @Robert Israel: In the finite group $G$ there are finite suck $k$. $\endgroup$ – T R Jul 27 '12 at 5:30
  • $\begingroup$ Oh, now that the "$g \in G$ has order $k$" has been inserted, the question makes some sense. $\endgroup$ – Robert Israel Jul 27 '12 at 6:43
  • $\begingroup$ Also the $\,p\neq 2\,$ was erased ,@RobertIsrael...this is odd as the edit wasn't from the OP. $\endgroup$ – DonAntonio Jul 27 '12 at 6:48
  • $\begingroup$ @Donanotonio I rewrote the question a little bit to make it easier to read and accidentally removed the $p\neq 2$ condition. If you look at the revision history, my answer is mainly rephrasing and the only thing that I added that wasn't really there before was "or equivalently is not cyclic". Anyway, if my edit went overboard feel free to rollback to the fourth revision. $\endgroup$ – JSchlather Jul 27 '12 at 7:17

The alternating group of degree 4 provides a model for counterexamples for all primes $p$ (and all powers $p^f$, not just $p^2$). Rather than viewing the alternating group acting on 4 unstructured points, think of it acting on the underlying set of a finite field. Rather than viewing it as all possible “even” permutations of the set, view it as all possible “affine” permutations of the set.

$$\newcommand{\AGL}{\operatorname{AGL}} G=\AGL(1,K) = \{ f : K \to K : x \mapsto \alpha x + \beta ~\mid~ \alpha,\beta \in K, \alpha \neq 0 \}$$

In the case of $A_4$, we take $K$ of order $4$. The subgroup $$K_+ = \{ f : K \to K : x \mapsto x + \beta ~\mid~ \beta \in K \}$$is a normal, elementary abelian, Sylow $p$-subgroup of $\AGL(1,K)$ whenever $K$ is a a finite field of characteristic $p$. The subgroup $$K^\times = \{ f : K \to K : x \mapsto \alpha x ~\mid~ \alpha \in K, \alpha \neq 0\}$$ is a cyclic subgroup of order $k_0=|K|-1$, relatively prime to the order of $K$.

If $K$ is chosen to be a field of size $p^2$, then $K_+$ is a normal, noncyclic Sylow $p$-subgroup of $G$ of order $p^2$, and $k$ divides $|G|=kp^2$, but of course $G$ has no subgroup of order $k_0p$, much less an element of order $k_0p$.

Proposition: The orders of elements of $G$ are exactly $\{p\} \cup \{ k : k \text{ divides } k_0 \}$.

Proof: Clearly $K_+$ has elements of order $p$, and the cyclic group $K^\times$ has elements of the other orders. If $g$ has order $kp$ for $k$ dividing $k_0$, then because $\gcd(k,p)=1$ we can write $1=uk+vp$, and so $g=g^1 = g^{(uk)} g^{(vp)}$ and $b=g^{(uk)}$ has order $p$ while $a=g^{(vp)}$ has order $k$. Hence we have an element $a$ of order $k$ commuting with an element $b$ of order $p$. All such elements $b$ lie in $K_+$. Write $a = (x\mapsto \alpha x + \beta)$ and $b=(x\mapsto x+\gamma)$. Then one product of $a$ and $b$ is $x\mapsto \alpha x + (\beta+\gamma)$ and the other is $x\mapsto \alpha x + (\beta+\alpha \gamma)$. For these two maps to be the same, they must act the same on $0 \in K$, and in particular $\beta+\gamma = \beta+\alpha\gamma$ so that $\alpha=1$ or $\gamma=0$. In the first case, this means $\alpha =1$ and $a \in K_+$ has order dividing both $p$ and $k_0$, so $k=1$. In the second case, this means $b$ is the identity, contradicting $g$ having order a multiple of $p$. $\square$

| cite | improve this answer | |
  • $\begingroup$ Thank you very much for your nice example. Now I realized the direct product of $P$ and any $p^{^{\prime }}$-group (of order coprime with $p$) is a group with above properties. $\endgroup$ – T R Jul 27 '12 at 14:49
  • $\begingroup$ Yes, the direct product always has elements of order $kp$, but this sort of "regular" semi-direct product has no elements of order $kp$. The key feature is whether some element of order $k$ commutes with some element of order $p$. $\endgroup$ – Jack Schmidt Jul 27 '12 at 15:08

The answer is no in general. Consider $G$ to be the semi-direct product of an elementary Abelian group of order $121$ acted on by ${\rm SL}(2,5)$ (there is a faithful such action). Each element of $G$ has order $11$ or some divisor of $120.$ There is no element of order $11k$ for any $k >1.$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.