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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 $<a>$, 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

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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.

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  • $\begingroup$ but $(a^{-1})^{-1}=a \in H$. According to supposition, I don't see any contradiction $\endgroup$ Apr 20 '20 at 8:56
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    $\begingroup$ 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 $b^{-1} = a$. $\endgroup$
    – poopist
    Apr 20 '20 at 9:02
  • $\begingroup$ oh yes, $b=a^{-1}$, sparked it. Thanks poopist! Haha $\endgroup$ Apr 20 '20 at 9:08
  • $\begingroup$ how do I close this question? $\endgroup$ Apr 20 '20 at 9:09
  • $\begingroup$ Thanks of the \lanlge and \rangle symbols. That was extremely relevant for clean-latex-writing. $\endgroup$ Apr 20 '20 at 9:12

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