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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.

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Can you explain what you mean by "finite index" when $aH$ is not a subgroup? –  user641 Nov 3 '11 at 4:45
3  
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. –  Patrick Da Silva Nov 3 '11 at 4:47
    
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. –  kjkwer Nov 3 '11 at 4:48
    
Yes what you said about $aHa^{-1}$ is true. –  user641 Nov 3 '11 at 5:05
    
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 $G$. –  Mikko Korhonen Nov 3 '11 at 17:17

1 Answer 1

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,

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

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