Suppose that $G$ is a group and $H$ a nonempty finite sub- set of $G$ closed under the product in $G$. Then $H$ is a subgroup of $G$.

Question

I am reading the book: I.N Herstein's Abstract Algebra, 3rd edition.

I couldn't understand the proof to the following lemma.

Where Lemma 2.3.1 is the following...


My doubt
Why is it being assumed that all the elements in the subgroup $H$ are integral powers of some element $a \in H$ i.e. $H$ is cyclic?

Answer 1

In the proof, an element $e\ne a\in H$ is chosen and then the powers of $a$ are formed: $a,a^2,a^3,\ldots$. Since $H$ is finite and closed under multiplication, these powers must lie in $H$. Thus two elements in this list must be equal if the series has more than $|H|$ elements. In the proof, $a^i=a^j$ for some $1\leq i < j\leq |H|+1$. The rest should be clear.

Answer 2

It is not assumed that every element of $H$ is a power of $n$. To find the inverse of $a \in H$, $a \neq e$ you consider the $n+1$ elements $a,a^{2},...,a^{n+1}$ and then argue that at least two of these power must be equal. And this helps us to get an inverse for $a$.

Answer 3

The text tells that $\forall a\in H$ his powers must be in $H$ since $H$ si closed. So for every element of $H$ you can do this statement, obtaining $e\in H$