# Does every subgroup of finite index contain a power of each element of the group?

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 $$\Gamma$$?

• 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. Dec 18, 2018 at 17:26

Yes.

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 $$g^{a-b} \in H$$.

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 $$n=[G:K]$$.

• It's odd how the symbol/rendering of $\mathfrak{S}$ (i.e., $\mathfrak{S}$) looks like a $G$. Dec 18, 2018 at 17:29
• @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 Dec 18, 2018 at 17:41