I have the following question:

Let $G$ be a group and let $H$ be a subgroup of finite index of $G$. Let $|G:H|=n$ Then it holds: $g^{n!}\in H$ for all $g\in G$.

Why is this true?

I think, that's not very difficult, but I have no idea at the moment.


  • 1
    $\begingroup$ Have you tried the Pigeonhole principle? $\endgroup$ – agtortorella Feb 13 '12 at 17:49
  • $\begingroup$ Thank you for these nice solutions! $\endgroup$ – Peter Feb 15 '12 at 13:44
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    $\begingroup$ It seems you have asked many questions but have accepted answers to none of them. If you are satisfied with the answer of a question you have asked, you can accept it by clicking on the check mark below the up/down arrows. It lets other users know that this question has a good answer. $\endgroup$ – Najib Idrissi May 31 '12 at 15:58
  • $\begingroup$ If $H$ is normal in $G$, we actually have $g^n \in H$ (see also math.stackexchange.com/questions/545417). $\endgroup$ – Watson Nov 26 '18 at 20:32
  • $\begingroup$ (See also math.stackexchange.com/questions/472672, if $H$ is normal in $G$). $\endgroup$ – Watson Nov 28 '18 at 9:41

Let $G$ act on the left cosets of $H$ by left multiplication, $$g: aH\longmapsto gaH.$$ This gives an action of $G$ on a set of $n$ elements, hence induces a homomorphism $G\to S_n$. In particular, for every $g\in G$, $g^{n!}$ must lie in the kernel, since $|S_n|=n!$. But the kernel is contained in $H$: if $g$ lies in the kernel, then $gaH = aH$ for all $a\in G$, hence in particular $gH=H$, so $g\in H$. Thus, for every $g\in G$, $g^{n!}\in H$.

  • $\begingroup$ One may observe that the $n!$ can be replaced by the exponent of $S_n$ (which is considerably lower than $n!$) and the proof still works. $\endgroup$ – Andrea Mori Feb 13 '12 at 18:11
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    $\begingroup$ @Andrea: It may be lower than $n!$ and usually is (except for small $n$). Of course, the above does even more, since we are showing that $g^{n!}\in \cap_{x\in G}xHx^{-1}$. $\endgroup$ – Arturo Magidin Feb 13 '12 at 19:40
  • $\begingroup$ @AndreaMori, what is the exponent of $S_n$? $\endgroup$ – lhf Feb 13 '12 at 20:24
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    $\begingroup$ @lhf: See the OEIS, sequence A003418. $\endgroup$ – Arturo Magidin Feb 13 '12 at 20:29
  • $\begingroup$ @ArturoMagidin, thanks. I knew that already but I had forgotten: $\exp S_n = \operatorname{lcm}(1,2,\dots,n)$. $\endgroup$ – lhf Feb 13 '12 at 20:35

In case anyone wanted a nice name for the minimal number, it is basically the obvious guess, though I think it might be a little surprising that the obvious guess is actually right.

Proposition: If G is a group, and H is a subgroup of G, then the exponent of $G/\operatorname{Core}_G(H)$ is the smallest positive integer n such that for all g in G, $g^n \in H$.

Proof: Suppose n is a positive integer such that $g^n \in H$ for all g in G. Let k in G, then $(g^{k^{-1}})^n \in H$ as well. Conjugating by k, $g^n \in H^k$. In particular, $g^n$ is contained in every conjugate of H, not just H itself. Hence $g^n$ is contained in $\operatorname{Core}_G(H) = \displaystyle \bigcap_{g\in G}H^g$, the largest G-normal subgroup of H. In particular, n is a multiple of the exponent of the group $G/\operatorname{Core}_G(H)$. Conversely, the exponent n of the group $G/\operatorname{Core}_G(H)$ definitely satisfies $g^n \in \operatorname{Core}_G(H) \leq H$. $\qquad \square$

Equivalently, if G is a transitive permutation group, then the smallest positive integer n such that the nth power of every element of G fixes the first point is also the smallest positive integer n such that the nth power of every element of G fixes every point.


Take $g\in G.$ Apply the pigeonhole principle to the $n+1$-tuple $g_0,\ldots,g_n,$ inductively defined by $g_0=e,$ and $g_{i+1}=g.g_i.$

You find $g_i$ and $g_j$ in the same left coset of $H$ in $G,$ for some $i,j=0,1,\ldots,n$ with $i<j.$

Consequently $g^{j-i}=g_jg_i^{-1}\in H,$ therefore a fortiori $g^{n!}\in H.$


See also my post here

Proposition. Let $H$ be a non-trivial subgroup of the finite group $G$, with $n = [G:H]$. Assume that $\gcd(|H|,n)=1$. Then the following are equivalent.

(a) For all $g \in G$: $g^n \in H$.
(b) $H \unlhd G$.


This is an easy question if you start by showing that $g^n \in H$. Consequently, $g$ raised to the power of any multiple of $n$ is also in H (since $[G:H] = n$). Just an idea.

  • 4
    $\begingroup$ This only works when $H$ is normal in $G$. Read the other proofs. $\endgroup$ – Martin Brandenburg May 31 '12 at 15:44

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