Let $G$ be a group and let $H \subseteq G$ be a finite subset that satisfies properties (H0) and (H1). Prove that $H$ is a subgroup of $G$.

Question

(H0) = set $H$ is non empty.
(H1) = if $a,b \in H$, then $ab \in H$

The only property I need to proof that $H$ is a subgroup is (H2) = if $a \in H$, then $a^{-1} \in H$.

This however is where I do not see the connection.

Given that $H$ is a subset of $G$ I do not see the reasoning that when an element out of $G$ is copied in the set $H$ that also the inverse (of that element) has to be added to $H$.

Also the excerise mentioned Note: the example $\mathbb{Z_{\geq 0}} \in \mathbb{Z^+}$ shows that the finiteness of H is essential!

Due to this and the emphasison the word finite in the excerise I assume that this plays a big part in the proof, but again, I do not see the connection. The only thing I am pretty sure of, is that since $H$ is non empty the identity element of $G$ is in $H$.

Answer 1

Hint:

Let $a \in H$, and consider the set $A = \{a, a^2, a^3, a^4, \ldots\}$.

Since $H$ is finite, what can you say about $A$ (use the pigeonhole principle!)? Can you use this to conclude that $a^{-1} \in H$?

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I hope this helps ^_^

Answer 2

If $a\in H$ by $(H_1)$, $a^n\in H$ for all $n\in \mathbb{N}$. Since $H$ is finite then $\exists n_a\in \mathbb{N}$ such that $a^{n_a}=1$ and of course $a^{n_a-1} \in H$ .That suggests that $1\in H$ and $a^{-1} =a^{n_a-1}\in H$

Answer 3

Let $a\in H$. This exists due to (H0).

If $a=e$, then $a^{-1}=a\in H$.

If $a\neq e$, then consider all the powers of $a$, which must be in $H$ by closure (i.e., (H1)). Since $H$ is finite, by the pigeonhole principle, some of these powers coincide.

Can you continue from here?

Hint: See Gallian's, "Contemporary Abstract Algebra (Eighth Edition)," page 64, Theorem 3.3.

Answer 4

Take $a \in H$. Condition H1 implies that the restriction to $H$ of the map $x \mapsto ax$ is a map $H \to H$. Since $G$ is a group, this map is injective. Since $H$ is finite, the map is surjective. Thus, there is $f \in H$ such that $af=a$, that is, $f=e$. Then there is $b \in H$ such that $ab=e$, that is, $b=a^{-1}$.