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Problem 46 (The Subgroup Test - Subgroups Are Subsets That Are Closed Under Products And Taking Inverses)

Suppose that $G$ is a
group
Definition (Group)

Let $G$ be a nonempty set, and let $*$ be a binary operation on $G$, which means for every $x,y\in G$ we have $x*y\in G$ $\textbf{[Closure]}$. The structure $\mathbb{G} = (G,*)$ is called a $\textdef{group}$ if the following hold.

  1. $\textbf{[Associativity]}$ For all $x,y,z\in G$ we have $(x* y)* z = x* (y* z)$.
  2. $\textbf{[Identity]}$ There is an $e\in G$ such that for all $x\in G$ we have $x * e = e* x = x$.
  3. $\textbf{[Inverses]}$ For all $x\in G$ there is a $y\in G$ such that $x* y = y* x = e$.

We usually simply write $G$ when referring to the entire structure $\mathbb{G}=(G,*)$. The element $e$ from the second point is called the $\textdef{identity}$. The element $y$ from the third point is called the $\textdef{inverse}$ of $x$ and is usually denoted $x^{-1}$. One often simply writes $xy$ in place of $x*y$, and for every positive integer $n$, we'll write $x^n$ as shorthand for $x* x* \cdots * x$ ($n$ times).

and $H$ is a nonempty subset of $G$ that satisfies the following properites:
  1. If $a,b\in H$, then $ab\in H$. (We say $H$ is closed under the group operation.)
  2. If $a\in H$, then $a^{-1}\in H$. (We say $H$ is closed under taking inverses.)
Prove that $H$ is a
subgroup
Definition (Subgroup)

Let $(G,\cdot)$ be a group, and let $H$ be a nonempty subset of $G$. Then $H$ is called a $\textdef{subgroup}$ of $G$ if the following hold:

  1. $\textbf{[Closure]}$ for all $h,k\in H$ we have $h\cdot k\in H$, and
  2. $(H,\cdot)$ is a group.

When $H$ is a subgroup of $G$, we write $H\le G$. Any subgroup of $G$ that is not equal to $G$ itself we call a $\textdef{proper subgroup}$. The subset of $G$ consisting of just the identity we call the $\textdef{trivial subgroup}$.

of $G$.

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