Let $G$ be a group, not necessarily finite. If $H$ is a normal subgroup of $G$ of a finite index, say $(G:H)=n$, then for every $g\in G$ we have $g^n\in H$. Does this statement remain valid if do not assume $H$ to be normal?

In particular let $SL_2(\mathbb Z)$ be the modular group, and let $\Gamma\subset SL_2(\mathbb Z)$ be a subgroup of a finite index. Does there exists a positive integer $\ell$ such that $\begin{pmatrix}1&1\\0 & 1\end{pmatrix}^\ell$ lies in $\Gamma$?

  • $\begingroup$ You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$
    – Shaun
    Dec 18, 2018 at 17:26

2 Answers 2



The set $\{ H, gH, g^2H, \dots , g^nH \}$ has $n+1$ elements, so that two of them are equal ( $H$ has only $n$ right cosets).

From $$g^aH=g^bH$$ it follows that $g^{a-b} \in H$.

  • $\begingroup$ But this does not show that $g^n \in H$. It just shows that $g^k \in H$ for some $0 <k \leq n$. $\endgroup$ Oct 5, 2022 at 16:45
  • $\begingroup$ @MetinErsinArıcan Good point. My answer is related to the title of the question which does not specify which exponent $k$ satisfies $g^k \in H$ $\endgroup$
    – Crostul
    Oct 6, 2022 at 10:47

For another proof (though it only answers the title as the $n$ we get is not $[G:H]$)

Let $H$ be a finite index subgroup. Then $G$ acts on $G/H$ by left translation, and so we have a morphism $\rho : G\to \mathfrak{S}G/H$.

Its kernel $K$ is contained in $H$ : indeed if $x\in K$, then $H= \rho(x)(H)= xH$, so $x\in H$. Moreover, $K$ is normal (it's a kernel !), and it has finite index in $G$ (because $G/K\simeq \mathrm{Im}\rho \subset \mathfrak{S}G/H\simeq \mathfrak{S}_{|G/H|}$).

Thus if $x\in G, x^n\in K$ for some $n$, thus $x^n\in H$ for some $n$. Note, however, that $n$ is not necessarily $[G:H]$; and the proof I gave only gives the bound $[G:H]!$ for $n=[G:K]$.

  • $\begingroup$ It's odd how the symbol/rendering of $\mathfrak{S}$ (i.e., $\mathfrak{S}$) looks like a $G$. $\endgroup$
    – Shaun
    Dec 18, 2018 at 17:29
  • 2
    $\begingroup$ @Shaun : that's just the gothic S. It's also easy to see how to deform an S into an $\mathfrak{S}$: stretch the bottom part and take it up above the rest of the S, then retract the top part a bit, and finally straighten the curve $\endgroup$ Dec 18, 2018 at 17:41

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