Is there an equivalence between $a$ and $b$ when $ab \in H$, $H \subset G$

Question

$$ab \in H \iff a \equiv b \pmod H$$

It abides by reflexivity, symmetry and transitivity.

Proof of reflexivity: (taken from this site)

$H \subset G \\ a,e \in G \\ e = aa^{-1} \in H$

Proof of symmetry:

$ab \in H \\ (ab)^2 \in H \\ a^{-1}(ab)^2b^{-1} = a^{-1}(abab)b^{-1} \in H = (a^{-1}a)ba(bb^{-1}) \in H = ebae \in H = ba \in H$

Proof of transitivity:

$ab, bc \in H \\ abbc \in H \\ abb^{-1}b^{-1}bc \in H = aeec \in H = ac \in H$

However, I was told in this post that if $ab$ is in a group, it does not mean that $a$ is equivalent to $b$ in that group. However, it seems to me that the three criteria for equivalence has been met. What is correct?

EDIT:

Apparently, for the equivalence relation between $a$ and $b$ to be reflexive, both $a$ and $b$ must be equivalent to themselves (respectively) within the sub-group (if I understand correctly). Now, for something to be equivalent to itself is easy, because it is equal to itself, and equality is an equivalence. So, isn't the relation between $a$ and $b$ reflexive?

Answer 1

Consider $\{[0]_3\}=H\le G=\Bbb Z_3$. In order for your relation, call it $\sim$, to be an equivalence relation, we require $a\sim a$ for all $a\in G$ (reflexivity); but this means $a+_3a=2a\in H$ for all $a\in G$, and yet

$$\begin{align} [1]_3+_3[1]_3&=[1+1]_3\\ &=[2]_3\\ &\neq [0]_3, \end{align}$$

where

$$[x]_3=\{ y\in\Bbb Z\, :\, 3\mid x-y\}.$$

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What is an equivalence relation, though, for a general group $G$, is

$$a\sim_Hb\iff ab^{-1}\in H,$$

for all $a,b\in G$ and $H\le G$.

Answer 2

There is very little in this post that makes any sense. It seems to be utterly confused as to what it means for a binary relation to be an equivalence relation, let alone facts about groups.

Remember: given any set $S$, a binary relation $\sim$ is an equivalence relation on $S$ if and only if three things happen:

1. For every $s\in S$, $s\sim s$; every element is related to itself.
2. For every $s,t\in S$, if $s\sim t$ (if $s$ is related to $t$), then $t\sim s$ (then $t$ is related to $s$).
3. For every $s,t,v\in S$, if $s\sim t$ and $t\sim v$ (if $s$ is related to $t$ and $t$ is related to $v$) then $s\sim v$ (then $s$ is related to $v$).

We say the relation is reflexive if it satisfies (1); symmetric if it satisfies (2); and transitive if it satisfies (3).

Now, you are proposing a relation on the elements of $G$ (relative to a given fixed subgroup $H$) defined as follows: $$ a\sim b\iff ab\in H.$$ Now, this is a perfectly fine binary relation on elements of $G$. The question is whether it is, generally, an equivalence relation. (Spoiler: no, it almost never is).

What would it mean for this relation to be reflexive? We would need to show that for every $g\in G$, $g\sim g$ holds. When does $g\sim g$ hold? When $gg=g^2\in H$. That is: $$g\sim g\iff g^2\in H.$$ So in order to prove that this relation is reflexive, you would need to show that the square of every element is in $H$. You did nothing anywhere near this. Your argument is nonsensical, in the sense that it literally makes no sense. Picking $a,e\in G$... why? Reflexivity is about a single element. And what does it matter that $aa^{-1}\in H$? That product is not one that we care about in this relation. Your argument could be used to show that for every $g\in G$ we have $g\sim g^{-1}$ (every element is related to its inverse), but that is not what we need to show, so there is little point to showing it in this case. There are, of course, plenty of subgroups of a group that do not contain every square, so in general this relation will not be reflexive. It is reflexive if and only if $G^2=\langle g^2\mid g\in G\rangle\leq H$.

What would we need for this relation to be symmetric? We would need to show that if $a\sim b$ holds, then $b\sim a$ holds. So we would need to show that if we know that $ab\in H$, then it will follow that $ba\in H$. Now you correctly note that if $ab\in H$ then $(ab)^2\in H$. But then you make a computation, $a^{-1}(ab)^2b^{-1}$ that makes absolutely no sense, and for which you have absolutely no warrant to think it will lie in $H$. Multiplying an element of $H$, namely $ab$, by something that you do not know whether it lies in $H$ or not (namely, $a^{-1}$) tells you absolutely nothing about whether it is in $H$ or not. In fact, the only way you can guarantee that $a^{-1}(ab)b^{-1}$ lies in $H$ is if you know that both $a$ and $b$ lie in $H$... and you most certainly do not.

I already gave you, elsewhere, an explicit example of a case in which you have $H\leq G$, elements $a,b\in G$, such that $ab\in H$ but $ba\notin H$. Namely, take $G=S_3$, $H=\{ e,(12)\}$, and $a=(123)$, $b=(23)$. Then $ab=(123)(23)=(12)\in H$, but $ba=(23)(123) = (13)\notin H$. (I compose my permutations right-to-left). So this condition will not hold in general either.

As for transitivity, what we need to show is that if $a\sim b$ and $b\sim c$ (that is, if we already know that $ab$ and $bc$ both are in $H$), then $a\sim c$ (then $ac\in H$). Again, you did not do anything related to this. Instead, you considered $(ab)b^{-2}(bc)$. While the two outside factors lie in $H$, the middle factor does not, in general, lie in $H$, so you do not know whether this product lies in $H$. Thus, your argument fails again for the same reason as your other arguments: you seem to think that you can just multiply by whatever you want and the result will be in $H$. That is simply not the case.

I strongly urge you to write out exactly what you need to prove, then translate it into their group theory meaning, and then try to prove them carefully. Your telegraphic method here makes it impossible to tell exactly where your errors come from, other than from a general misunderstanding of how groups work.

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We might note that if $G^2\leq H$, then this relation is identical as a relation to the usual one, $$a\equiv b\iff ab^{-1}\in H;$$ identical in the sense that $a\sim b\iff a\equiv b$. Indeed, under the general assumption that every square is in $H$, we have that $ab\in H$ if and only if $ab^{-1}=(ab)(b^{-1})^2\in H$.

Since the condition that $G^2\leq H$ is necessary in order for this relation to be reflexive, it follows that this relation is an equivalence relation on $G$ if and only if $G^2\leq H$ if and only if this is just the usual relation "equivalent modulo $H$ on the right" in disguise.