# Let H be a non-empty subset with property P closed under the group G, can H be a subgroup?

Let $$H$$ be non-empty subset of $$G$$ and closed under its groups operations. Let the subset $$H$$ be defined by the property that if $$a \notin H$$ then $$a^{-1} \notin H$$. Is H a subgroup?

Questions:

Can this property show the existence of the identity in $$H$$?

Can this property be used to find if $$ab^{-1} \in H$$ for $$a \in H$$ and $$b \in H$$ or simply that $$b^{-1} \in H$$ whenever $$b \in H$$?

Extra:

I reached a conclusion, that, from the information given in the question the property of $$H$$ does not show the existence of $$e$$ and $$ab^{-1} \in H$$ and is not a subgroup.

Then, I recalled that there exists, cylic subgroups $$$$, generated by an element $$a \in G$$, which applies this property in a way.

Example: Let's take a simple group $$Z_{10}$$ and its cyclic subgroup generated by 4, is <4> = {4,8,2,6,0}, applies property if $$5 \notin <4>$$, then $$5^{-1} \notin <4>$$.

Similarly, $$3 \notin <4>$$, then $$3^{-1} \notin <4>$$

$$1 \notin <4>$$, then $$1^{-1} \notin <4>$$

Where I am right now:

I am not able to answer the above two questions and this example has shown a case where it is possible for such a set $$H$$ to exist.

Question reference: Gallian - Contemporary Abstract Algebra, Chapter 3, Question 13

Yes the identity element belongs in $$H$$. The core idea is that $$a = (a^{-1})^{-1}$$ in $$G$$. Suppose $$a \in H$$ and assume that $$a^{-1} \not \in H$$, then by the assumption we have $$a = (a^{-1})^{-1} \not \in H$$, which is a contradiction. So it must be that $$a \in H$$ implies $$a^{-1} \in H$$. Hence the identity element is in $$H$$ (since it is non-empty, pick any element and multiply by its inverse).
Something irrelevant to the question. For the subgroup generation symbol, you can write $$\langle ... \rangle$$ (\langle and \rangle, which is also the symbols used for inner products), it looks better than $$<>$$ and avoids ambiguity.
• but $(a^{-1})^{-1}=a \in H$. According to supposition, I don't see any contradiction Apr 20 '20 at 8:56
• Suppose $a \in H$. If $a^{-1} \not \in H$ then $a = (a^{-1})^{-1} \not \in H$. So $a \in H$ but also $a \not \in H$, this is the contradiction. To make it more clear, let $b = a^{-1}$, then $b \not \in H$ implies $b^{-1} \not \in H$, but $b^{-1} = a$. Apr 20 '20 at 9:02
• oh yes, $b=a^{-1}$, sparked it. Thanks poopist! Haha Apr 20 '20 at 9:08