Let $H\le G$. Is it the case that for any $x,y\in G$, if $xy\in H$, then $yx\in H$?

Question

This is Exercise 88 of Rose's, "A Course on Group Theory". This Approach0 search was inconclusive due to too many mathematical terms. This MSE search returned nothing.

I have answered my question in typing this up. Now that I've put it together, though, I reckon it's worth sharing so that others may benefit; I could also have made a mistake. It'd help to see a different proof/solution, too, I suppose.

The Question:

Suppose $H\unlhd G$. Show that if $x$ and $y$ are elements of $G$ such that $xy\in H$, then $yx\in H$. Would this be true merely on the hypothesis that $H\le G$? [Emphasis added.]

Thoughts:

I think normality of $H$ in $G$ is necessary.

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Suppose $H\unlhd G$. Let $x,y\in G$ such that $xy\in H$. By normality, $H=yHy^{-1}$, so, in particular, $y(xy)y^{-1}=yx\in H$.

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Now suppose $H\le G$ but $H\not\lhd G$. Let $a,b\in G$ such that $ab\in H$.

My first thought is that, if there is a counterexample, such a group $G$ must not be abelian, since otherwise $ba=ab\in H$ trivially. Also, $a\neq b^{-1}$ and vice versa.

This led me to consider the smallest nonabelian group $S_3\cong D_3$, the dihedral group of order $6$. If we let $H=\{{\rm id}, (12)\}$, maybe we could get somewhere. Nothing leaps out at me as an obvious choice of $a,b$ as above; we would need $ab=(12)$ and $ba\in\{(13), (23), (123), (321)\}$; so let's try, based on experience, $a=(123), b=(23)$; then $ab=(123)(23)=(12)\in H$ and $ba=(23)(123)=(13)\notin H$.

Thus normality is necessary.

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Do you have an alternative approach?

Please help :)

Answer 1

When you say "Let $a,b\in G$ such that $ab\in H$", that doesn't exploit the explicit non-normality of $H$. You want to get right on that from the start. For instance, if you instead say:

Let $a\in G$ be such that $aH\neq Ha$.

Then the proof is over in two short sentences:

Let $b\in aH\setminus Ha$. Then $a^{-1}b\in H$, but $ba^{-1}\notin H$.

Answer 2

Observe that the following holds.

Let$H$ be a subgroup of $G$. Then the following are equivalent.
(a) For all $x,y \in G$ with $xy \in H$, also $yx \in H$.
(b) $H \unlhd G$.

Proof (a) $\implies$ (b): let $h \in H$, $x \in G$, then $(hx)x^{-1}=h \in H$. Hence $x^{-1}hx \in H$.
(b) $\implies$ (a): if $x, y \in G$ with $xy \in H$, then $x^{-1}(xy)x=yx \in H$, since $H$ is normal.