Proof that a subset closed under group operation of a finite group is a subgroup

I have some idea, lets take any element $a$ from that subgroup and repeatedly apply group operator over it, since group is finite that element $a$ will appear again after some time, $\{a, a^2, a^3,...,a\}.$

Then element previous to 2nd $a$ should be identity element. And prior to it is $a$'s inverse. Hence it has all property of groups. Is this proof correct.

I am not sure about that can 2nd $a$ appear only if we have identity element in it. Please help.

Reference: Fraleigh p.58 Question 5.50 in A First Course in Abstract Algebra

Solution 1:

While you are working in the right direction, it is a priori not clear that $a$ itself will repeat. In principle we might have a sequence of powers like $a,a^2,a^3, a^4, a^2, a^3, a^4, a^2, \ldots$.

However, if $G$ is a group and $A$ is a finite nonempty subset of $G$ that is closed under multiplication, then for any $a\in A$ the map $\mathbb N\to G$, $n\mapsto a^n$ is in fact a map $f\colon \mathbb N\to A$ (by induction on $n$) and cannot be injectve because $A$ is finite, so you will have some natural numbers $n,m$ with $n<m$ and $a^n=a^m$. Then $m=n+k$ with $k\in \mathbb N$ and we conclude $1=a^{k}\in A$. If $a=1$ then trivially $a^{-1}\in A$ as well. If $a\ne 1$, on the other hand, then clearly $k>1$ and we find $a^{-1}=a^{k-1}\in A$. So we see that $A$ is indeed a subgroup.