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Problem 53 (Finite Subgroup Test)

Let $G$ be a group. Suppose that $H$ is a nonempty finite subset of $G$ and that $H$ is closed under the operation of $G$ (so if $a,b\in H$, then we must have $ab\in H$). Prove that $H$ is a subgroup of $G$.

Hint. If you use the subgroup test (show that $H$ is a nonempty subset of $G$ that is closed under the operation and taking inverses), then we get to assume it's a nonempty subset of $G$ that is closed under the operation. All you have to do is explain why it's closed under taking inverses. The work you did in the problem The Inverse In A Finite Group Is A Power Of The Element should help you quite a bit. However, you can't use this theorem directly because you do not know that $G$ is a finite group. You'll want to reuse your work from that problem, not refer to the problem.

Solution

Let $H$ be a nonempty subset of $G$ that is closed under the operation of $G$. We will use the subgroup test to prove that $H$ is a subgroup of $G$. We have assumed that $H\neq \emptyset$ and that $H\subseteq G$, and that $H$ is closed under the operation of $G$. The only thing that remains is to prove that $H$ is closed under taking inverses. Let $x\in H$. If $x=e$, then $x^{-1}=e\in H$, and we're done. Suppose that $x\neq e$. Let $ S=\{x^n\mid n\in\mathbb{N} \} $. Since $H$ is closed under the operation, we know $S\subseteq H$. We will prove that $x^{-1}\in S$, which means $x^{-1}\in H$ as desired.

Since $H$ is finite, we know that there must exist positive integers $i$ and $j$ such that $x^i=x^j$, otherwise $S$ would be an infinite subset $H$ (impossible). Without loss of generality, suppose $i<j$. The fact $x^i=x^j$ implies $e=x^{j-i}$, or more importantly that $$x^{-1} = x^{j-i-1}.$$ Since we know $i<j$, this means that $j-i\geq 1$, or instead $j-i-1\geq 0$. To show that $x^{-1}\in S$, we just need to prove that $j-i-1\geq 1$. Suppose instead that $j-i-1=0$. We have $e=x^{j-i}=x^{-1}$, which means $x=e$. This is a contradiction, as we assumed $x\neq e$. In summary, we have shown that $x^{-1} = x^{j-i-1}$ where $j-i-1\geq 1$ which means $x^{-1}\in S\subseteq H$. This proves that $H$ is closed under taking inverses.

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