# If $H$ has finite index there are finitely many distinct subgroups of form $aHa^{-1}$

If $$H$$ has finite index there are finitely many distinct subgroups of form $$aHa^{-1}$$.

I tried the following:

Let the distinct left cosets of $$H$$ in $$G$$ be $$a_1H ,a_2H, \dots, a_nH$$. Then the distinct right cosets are $$Ha_1^{-1}, Ha_2^{-1}, \dots, Ha_n^{-1}$$. Consider any subgroup $$aHa^{-1}$$. Now $$aH=a_iH$$ for some $$i=1,2,\dots, n$$. Then $$Ha^{-1}=Ha_i^{-1}$$. So $$aHa^{-1}=a_iHa_i^{-1}$$, proving these subgroups are finite in number.

In my proof I have used the following argument: \begin{align}aH=bH &\iff a^{-1}b\in H \\ &\iff a^{-1}(b^{-1})^{-1}\in H\\ &\iff Ha^{-1}=Hb^{-1} \end{align}

Is my proof okay?

P.S. I understand that all of the subgroups $$a_iHa_i^{-1}$$ need not be unique. For instance, the subgroup $$3\Bbb Z$$ has 3 distinct left/right cosets in $$\Bbb Z$$. However the only subgroup of the form $$a+3\Bbb Z -a$$ is $$3\Bbb Z$$.

• It looks fine to me. You did not justify some of the implications but they're correct. Commented May 27, 2018 at 11:16
• @Yanko Thank you. I have added another line. Commented May 27, 2018 at 11:21