I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
The following lemma is in this book:
LEMMA 2.4.1
A nonempty subset $H$ of the group $G$ is a subgroup of $G$ if and only if
- $a,b\in H$ implies that $ab\in H.$
- $a\in H$ implies that $a^{-1}\in H.$
Proof. If $H$ is a subgroup of $G$, then it is obvious that (1) and (2) must hold.
Surely, I think it is easy to prove that (2) must hold if $H$ is a subgroup of $G$.
Let $e_H$ be an identity element of $H$. $e_H$ is an element of $G$. $e_H=e_G\cdot e_H=e_H\cdot e_H$.
So, by right cancellation law in $G$, $e_G=e_H$. Let $a_{H}^{-1}$ be the inverse element of $a$ in $H$. $a_{H}^{-1}$ is an element of $G$. $a\cdot a_{G}^{-1}=e_G=e_H=a\cdot a_{H}^{-1}$. So, by left cancellation law in $G$, $a_{G}^{-1}=a_{H}^{-1}$.
Proof of (2): Let $a\in H$. $a^{-1}=a_{G}^{-1}=a_{H}^{-1}\in H$.
Surely, (2) was easy to prove.
But is (2) really obvious?
Best Answer
$2$ is obvious only when $G$ is a group of finite order. Otherwise you need to verify $2$ and there are $H\subset G$ which is closed but not closed under inverse.
Consider, $G=(\Bbb{Z}, +) $ and $H=\{0, 1,2,...\}$