I was solving the book Abstract Algebra by Charles Pinter. In one of the questions I was stuck.

The question is as follows :-

$H$ is a subgroup of group $G$.Two equivalence relations on $G$ are defined as follows :-

1) $a \sim b$ iff $ab^{-1}\in H$

2) $a \sim b$ iff $a^{-1}b\in H$

Are the two equivalence relations the same. If so prove , if not find a counterexample.

I tried to first prove that it is not the case and tried a few counter examples but they didn't work. I tried $H = \{e\}$ or $H = \{G\}$ which are the trivial subgroups. I also tried to use some common examples of sub groups such as the set of even integers under addition which is a subgroup of set of integers. The set of positive rationals under multiplication which is a subgroup of positive reals. It didn't work. And I also tried to prove they are the same. But couldn't do so. My approach was to somehow derive $a^{-1}b\in H$ from $ab^{-1}\in H$ . I also tried to use a third element $c$ and use the transitivity law but that also didn't work. Can you point out a direction in which I can proceed to try to solve it?


  • $\begingroup$ Would the downvoter care to share their opinion of how the OP could improve this question? (Personally, I can see nothing wrong with it...) $\endgroup$
    – user1729
    Jul 28 '13 at 14:48
  • 1
    $\begingroup$ @user1729 oh yeah, i'm sorry, deleting the comment $\endgroup$
    – W_D
    Jul 28 '13 at 14:59

They are not equal. To see that they are non-equal, note that $ab^{-1}\in H\Leftrightarrow ba^{-1}\in H$. This does not imply $a^{-1}b\in H$, as otherwise every subgroup of every group is normal! (Why?)

So, find a subgroup $H$ of a group $G$ which is not normal. Then this does not hold. (Note: the subgroups you tried were all normal). For example, Take $G=S_3$ and $H=\{id, (1, 2)\}$. Then let $a=(1, 3)$ and $b=(1, 2, 3)$. What happens?

Note: This is very much related to normality. It turns out that these equivalence relations are equal if and only if the subgroup $H$ is normal. This is because $ab^{-1}\in H\Leftrightarrow Ha=Hb$ while $a^{-1}b\in H\Leftrightarrow aH=bH$. (Why?) So, for these equivalence relations to be equal you need the left and right cosets of $H$ to be equal, which happens if and only $H$ is normal in $G$.

  • $\begingroup$ Ya I am yet to study Normality . I browsed over that section sometime ago but haven't reached it as such. Will try to read that section to be able to get a clearer picture behind the motivation of this question. But its good , such questions give a glance of concepts to come later in the book. Thanks again for the reply. $\endgroup$
    – ameyask
    Jul 28 '13 at 16:36
  • $\begingroup$ You are welcome. Normality isn't necessary to understand this question (which is why I gave the counter-example). I mentioned normality because this is what the question is getting at! $\endgroup$
    – user1729
    Jul 28 '13 at 16:42
  • $\begingroup$ Just a confirmation . (1,2) means the bijection which interchanges element 1 and 2 ? and (1,2,3) means one which replaces element 1 by element 2 , element 2 by element 3 , element 3 by element 1 ? Thanks again. $\endgroup$
    – ameyask
    Jul 28 '13 at 16:54
  • $\begingroup$ @ameyask86 Yes, that is correct. $\endgroup$
    – user1729
    Jul 28 '13 at 17:13

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