Lemma 3.4. Let $(G ,*)$ be a group. A nonempty subset $H$ of $G$ is a subgroup of $(G,*)$, iff, for every $a, b\in H$, $a*b^{-1}\in H$.
Proof. First, suppose $H$ is a subgroup. If $b\in H$, then $b^{-1}\in H$ since $(H,*)$ is a group. So, if $a\in H$ as well, then $a*b^{-1}\in H$.
Conversely, suppose that, for every $a,b\in H$, $a*b^{-1}\in H$.
- First, notice that $*$ is associative since $(G,*)$ is a group.
- Let $a\in H$. Then $e=a*a^{-1}$, so $e\in H$.
- Let $b\in H$. Then $b^{-1}=e*b^{-1}$ so inverses exist in $H$.
- Let $a,b\in H$. By the previous step, $b^{-1}\in H$, so $a*(b^{-1})^{-1}=a*b\in H$. Thus, $H$ is closed under $*$
Therefore, $(H,*)$ is a group, which means that $H$ is a subgroup of $G$. $\Box$
The initial part is clear and makes sense, once you assume $H$ to be a subgroup. But the second part, attempting to prove the group properties does not make sense to me.
How did we assume that $a^{-1}$ belongs to $H$ in while showing that identity element ($e$) belongs to it, since the fact that inverses exist is the third point, for which we use the identity element? (second point)
Overall, how does this proof establish the original lemma, since both the parts does not seem to justify "if and only if" part?
I am an absolute beginner here.
Best Answer
If you assume
then if $a\in H$ (since $H$ is nonempty) then $aa^{-1}\in H$. But this is $e_G$. The assumption is that $ab^{-1}\in H$ if $a,b\in H$. So it applies for $b=a$.
It is now established $e_G\in H$ under the above assumption.
But then if $b\in H$ then $e_Gb^{-1}=b^{-1}\in H$, letting $e$ take the roll of $a$ in the assumption. Therefore inverses of elements in $H$ also belong to $H$.