# 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. – Shaun Dec 18 '18 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 $G$. – Shaun Dec 18 '18 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 – Max Dec 18 '18 at 17:41