# A noneempty subset $H$ of a finite goup $G$ is a subgroup if and only if $H$ is closed

I need to prove proposition 1, and this is what I have.

Definition 1: If there exists a nonzero integer such that $a^m=e$, then the order of the element $'a'$ is defined to be the least positive integer $'n'$ such that $a^n=e$

If there does not exists a nonzero integer $m$ such that $a^m=e$, then the order is infinty

Definition 2: A subset $H$ of a group $G$ is a subgroup if

1. $e\in H$
2. if $x,y\in H$, then $xy\in H$.
3. if $x\in H$, then $x^{-1}\in H$

Proposition 1: A noneempty subset $H$ of a finite goup $G$ is a subgroup if and only if $H$ is closed

proof: $\Rightarrow )$ Let H be a noneempty set of a finite grout $G$, then we know by definition 2(2) that $H$ is closed under group multiplication.

$\Leftarrow$): $H$ is finite therefore if $a\in H$, then $\text{ord}(a)=n$. Thus $e=a^n\in H$. We have shown that $e\in H$. If we can show that for every $h\in H$ that $h^{-1}\in H$, then we can conclude that $H$ is a subgroup of $G$. So let $h\in H$, then we know that there exists a element $k\in \Bbb{N}$ such that $h^k=e$. But then $h^{k-1}=h^{-1}$ and $h^{k-1} h=h h^{k-1}=e$. So $h^{-1}\in H$.

Therefore $H$ is a subgroup of $G$

My question is:

• Is this a correct proof, because I dont know if $\text{ord}(a)=n$ when $H$ is finite?
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You don't need the part about $n$, which anyway you didn't specify (unless you meant it is $\operatorname{ord}(a)$ by definition). The existence of $k>0$ with $h^k=e$ evoked a bit further on suffices. And that existence is clear, the infinitely many values $h^i$ for $i\in\Bbb N$ cannot all be different, and once $h^i=h^j$ with $i>j$ you can take $k=i-j$. – Marc van Leeuwen Nov 27 '12 at 13:47
Thank you, I get it now. This is much better. – Onur Nov 27 '12 at 17:12