Just out of curiosity,

If $H$ is a subgroup of finite index in $G$ and $a\in G$, is it always true that $aH$ and $Ha$ are of finite index in $G$?

I think I can prove it as follows: If $a_1H,\ldots,a_kH$ be all the distinct left cosets of $H$, then $(a_1a^{-1})aH,\ldots,(a_ka^{-1})aH$ are distinct left cosets of $H$, and thus $aH$ is of finite index. Similarly, $Ha$ is of finite index.

Is there something wrong in the above proof? I find this result, if correct, extremely natural, but I don't find it in my abstract algebra textbook.

  • $\begingroup$ Can you explain what you mean by "finite index" when $aH$ is not a subgroup? $\endgroup$ – user641 Nov 3 '11 at 4:45
  • 3
    $\begingroup$ You probably don't find it because the index of a subset of $G$ is only defined for subgroups of $G$? Because for arbitrary subsets of $G$, the cosets don't necessarily form partitions, i.e. index seems just wrong. If you've found another definition somewhere perhaps we could try working with it a little. $\endgroup$ – Patrick Da Silva Nov 3 '11 at 4:47
  • $\begingroup$ Ah I see.. Thank you very much! I was trying to prove that $aHa^{-1}$ is of finite index (is this true?), so I took the shortcut which turns out to be wrong. $\endgroup$ – kjkwer Nov 3 '11 at 4:48
  • $\begingroup$ Yes what you said about $aHa^{-1}$ is true. $\endgroup$ – user641 Nov 3 '11 at 5:05
  • $\begingroup$ For an arbitrary subset $H$ of $G$, the cosets of $H$ form a partition of $G$ if and only if $H$ is a coset of some subgroup of $G$. $\endgroup$ – Mikko Korhonen Nov 3 '11 at 17:17

If you are trying to show that $aHa^{-1}$ has the same index in $G$ as $H$, you can define a bijection from $G/H$ to $G/(aHa^{-1})$ by sending $gH$ to $(ga^{-1})aHa^{-1}$. This is a bijection because it has an inverse: right multiplication by $a$. Namely,

$(ga^-1)aHa^{-1}(a) = H$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.