# Let $H\subset G$, for a group $G$, where for every $a\notin H$, $a^{-1}\notin H$, and $H$ is closed under $G$'s operation. Is $H\le G?$

Suppose $$H\subset G$$, where $$G$$ is a group and H is nonempty and closed under the binary operation of $$G$$.

Additionally, assume $$H$$ has the property that if $$a$$ is not in $$H$$ then $$a^{-1}$$ is not in $$H$$.

It seems rather trivial to see the given condition: $$\left(a\notin H\implies a^{-1}\notin H\right)\iff \left(a^{-1}\in H \implies a\in H\right)$$ by contraposition. The latter of the two above is simply one of the hallmark conditions of a subgroup (among closure under the binary operation of $$G$$) i.e. there's a theorem we've proved previously in class:

Theorem: If $$ab\in H$$ when $$a\in H$$ $$\land$$ $$b\in H$$ and if $$a^{-1}\in H$$ when $$a\in H$$, then $$H$$ is a subgroup of $$G$$ assuming $$G$$ is a group with $$\emptyset \neq H\subset G$$.

Thus since in the problem we're given $$H$$ is closed under the operation of $$G$$ and is a nonempty subset of $$G$$, then doesn't seeing the given property's equivalence with its contrapositive as I've shown finish the proof by showing $$a^{-1}\in$$ when $$a\in H$$?

Seems too easy.

• Could you please put brackets around the implications in your „condition“ :) On a first glance I dont see that your argument fails and it feels right to me... Commented Jan 20, 2020 at 20:42
• $(p\implies q)\iff(\neg q\implies\neg p)$. Commented Jan 20, 2020 at 20:44
• Did that make it clearer? @PrudiiArca
– user689775
Commented Jan 20, 2020 at 20:46
• Yes, the argument works. Explicitly: let $a\in H$; we need to show that $a^{-1}\in H$. If $a^{-1}\in H$, you are done; if $a^{-1}\notin H$, then by the condition we have $a=(a^{-1})^{-1}\notin H$, a contradiction. Thus, $a^{-1}\in H$, as required. Commented Jan 20, 2020 at 20:48
• It was clear to me, but missing brackets in logic freak me out :D Commented Jan 20, 2020 at 20:51

You are in the right track, but it is not just the contrapositive of the statement, so perhaps you missed a small detail.

Note that the contrapositive of "$$a \not\in H$$ implies $$a^{-1} \not\in H$$" is the following statement:

$$a^{-1} \in H$$ implies $$a \in H$$.

With this, use $$a = (a^{-1})^{-1}$$ to conclude that $$H$$ is closed under inverses.

No. Let G be integers mod 2, let H = {1}. The the only element not in H is 0, whose inverse is also not in H.

Edit. Did not read that H is closed.

• Yeah, the lack of that condition in the title threw me at first too. Commented Jan 20, 2020 at 21:37