Can $a, b\in H\Rightarrow (ba)^{-1}\in H$ be a subgroup test?

Consider the following conditions:

Let $H$ be a nonempty subset of group $G$. $a, b\in H\Rightarrow (ba)^{-1}\in H$. ($*$)

Since $H\neq\emptyset$, there exits $a\in H$. Then one can use ($*$) $a$ to "generate" the elements in $H$. It seems that one can not finally generate $a^{-1}$ in $H$, though ($*$) is so close to the one-step subgroup test. Can one come up with examples in finite and infinite groups that $H$ is not a subgroup?

• If you want an infinite group as a counter example, take a finite group example and put infinitely many copies of it in a direct product, but consider the same elements. It can give you either finite or infinite $H$ as counter examples if you consider only the element in the first copy of your group $G$ or in the diagonal subgroup (i.e. the elements of the form $(g,g,g,g,g,g,g,\dots)$). Sep 16, 2012 at 5:34
• the trouble is you can't show the identity of $G$ lies in $H$. to use this to show $e \in H$, we'd need inverses in $H$ first, and to show inverses are in $H$, you need the identity to be in $H$. the reason why the "one step" test works, is because it combines closure and inversion in "one step". Sep 16, 2012 at 6:38

How about the (additive) group $\mathbb{Z}/(6)$, and the subset $H=\{1,4\}$? Then $-(1+1)=-2\equiv4$; $-(1+4)=-5\equiv1$, and $-(4+4)=-8\equiv4$. And of course $H$ isn’t a subgroup.
• Indeed, thinking things over, I realize that $\mathbb{Z}/(3)$ with the non-subgroup subset $\{1\}$ is a minimal (counter) example. And in turn, this shows clearly that $\mathbb{Z}$ and the set of all integers congruent to $1$ modulo $3$ is an infinite counterexample. Sep 16, 2012 at 5:20
The key point to find such counterexamples can be looking for a non-identity element in $G$ such that $a=a^{-2}$. Then $H=\{a\}$ is a counterexample.