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Let $G$ be a group and $H$ a subgroup with $[G:H]=n$. Is it true that $x^n\in H$ for all $x\in G$?

Remarks. The answer is positive whenever $H$ is normal, e.g., for $n=2$. In general, by using the normal core of $H$, one can find an $m\ge 1$ such that $x^m\in H$ for all $x\in G$.

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I am afraid this need not be the case always... –  Praphulla Koushik Oct 30 '13 at 11:03
    
See also math.stackexchange.com/questions/573050/…. –  lhf Apr 3 at 14:32

4 Answers 4

up vote 8 down vote accepted

For a counterexample, take $S_3$ and the subgroup $H = \{\rm{id},(12)\}$. This has index $3$, but $(13)^3 = (13)\not\in H$.

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Funny enough: I've checked for a counterexample taking $G=D_4$ and $Q_8$ (unfortunately these are not useful), but I skipped the simplest non-abelian case! –  user89712 Oct 30 '13 at 13:29
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@user Indeed, $Q_8$ cannot give a counterexample since all subgroups are normal. And $D_4$ cannot since the index of any non-normal subgroup is equal to the exponent of the group. –  Tobias Kildetoft Oct 30 '13 at 14:24

Trying to get a whole series of counterexamples, I came up with the following, which shows you how to construct these.

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) $H \unlhd G$.

Proof (b)$\Rightarrow$(a) is trivial by Lagrange's Theorem. So let us prove (a)$\Rightarrow$(b) (Sketch) We are going to use induction on $|G|$. To start the induction, we argue that $\operatorname{core}_G(H) \neq \{1\}$. For suppose $\operatorname{core}_G(H) =\{1\}$ and pick $g\in G$ and $h\in H$. By the assumption (a) $(g^{-1}hg)^n=g^{-1}h^ng \in H$, so $h^n \in H^{g^{-1}}$. We conclude that $h^n \in \operatorname{core}_G(H)$, hence $h^n=1$ and the order of $h$ must divide $n$. But the order also divides $|H|$ and since $\gcd(|H|,n)=1$, we conclude $h=1$. But $h$ was arbitrary, so $H$ must be trivial, which contradicts the assumption.

If $H$ is normal there is nothing to prove, so we can safely assume that $\operatorname{core}_G(H)$ is a proper subgroup of $H$. Now write $\bar{G}$ for $G/\operatorname{core}_G(H)$ and $\bar {H}$ for $H/\operatorname{core}_G(H)$, then $\bar {G}$ and $\bar {H}$ satisfy all the conditions of the proposition. By induction we get $\bar {H} \unlhd \bar {G}$, and this implies $H \unlhd G$.

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You are welcome! I was inspired by your question! –  Nicky Hekster Oct 30 '13 at 21:43
    
I hope you understood that from the Proposition it follows that whenever you have a non-normal Sylow $p$-subgroup $P$ of $G$, the pair $(G,P)$ forms a counterexample. See also Tobias' example! –  Nicky Hekster Nov 1 '13 at 13:23

As the answer given by Tobias shows, this is not true in general. However, it is possible to say some things even when $H$ is not a normal subgroup of $G$.

Let $H \leq G$ be a subgroup. For all $x \in G$, there exists $1 \leq r \leq [G:H]$ such that $x^r \in H$.

Proof: Let $r \geq 1$ be the smallest positive integer such that $x^r \in H$. Then $x, x^2, \ldots, x^r$ are in distinct cosets of $H$ and thus $r \leq [G:H]$.

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yes, but this $r$ depends on $x$ unfortunately. –  Nicky Hekster Oct 30 '13 at 20:37
    
@NickyHekster: If $H$ is normal, then $r$ is the order of $xH$ in $G/H$. So it is expected that $r$ should depend on $x$. If you want something that is independent of $x$, note that if $[G:H] = n$, then $x^{n!} \in H$ for all $x \in G$. This can be seen as a corollary of the above result –  Mikko Korhonen Oct 30 '13 at 22:02

Here is series of groups where the property holds, even if the subgroup is not normal.

A subgroup $S$ of a group $G$ is called subnormal (one writes: $S \lhd \lhd G$) if there exists a chain of subgroups $H_i$ of $G$, with $S=H_0$, $H_{i-1} \lhd H_i$, for $i=1, \dots, r$ and $G=H_r$.

Proposition. Let $S \lhd \lhd G$ with index$[G:S]=n$. Then for all $g \in G$ it holds that $g^n \in G$.

Proof Choose subgroups $H_i$ with $S=H_0 \lhd H_1 \dots \lhd H_r=G$, and put index$[H_i:H_{i-1}]=n_i$, $i=1, \dots , r$. Then $n=n_1 \dot n_2 \dots n_r$. If $g \in G$, since $H_{r-1} \lhd H_r$, $g^{n_r} \in H_{r-1}$. Since $H_{r-2} \lhd H_{r-1}$, $(g^{n_r})^{n_{r-1}}=g^{n_r.n_{r-1}} \in H_{r-2}$. Now continue this argument till $S$ is reached.$\square$

Since nilpotent groups are characterized by the fact that all subgroups are subnormal (this can be proved by the “normalizers grow” principle and induction), the proposition allows for a series of examples of a group $G$ and non-normal subgroup $S$, with index$[G:S]=n$, such that $g^n \in G$ for all $g \in G$. For example, take $G$ a $p$-group and $S$ any non-normal subgroup. Smallest example: $G=D_4= \langle a,b:a^4=b^2=1,bab=a^{-1}\rangle$, with $S=\{1,b\}$, which has index 4.

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Thank you again! However, isn't here a better place to post this answer? –  user89712 Nov 20 '13 at 21:37
    
Ah, yes thanks, could find it anymore, will put it there too! –  Nicky Hekster Nov 20 '13 at 21:47

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