The Ab. Group of Inv. Elements that are Equal to Themselves

The Abelian Group of Invertible Elements that are Equal to Themselves

Recall from the Abelian Groups page that an Abelian group is a group with the extra property of commutativity on its operation $\cdot$. That is, an Abelian group is a set $G$ paired with a binary operation $\cdot : G \times G \to G$ where:

• 1) For all $a, b, c \in G$ we have that $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ (Associativity of $\cdot$).
• 2) There exists an element $e \in G$ such that $a \cdot e = a$ and $e \cdot a = a$ (The existence of an identity for $\cdot$).
• 3) For all $a \in G$ there exists a $a^{-1} \in G$ such that $a \cdot a^{-1} = e$ and $a^{-1} \cdot a = e$ (The existence of inverses for each element in $G$).
• 4) For all $a, b \in G$ we have that $a \cdot b = b \cdot a$ (Commutativity of $\cdot$).

We will now look at the Abelian (sub)group of invertible elements that are equal to themselves.

Let $(G, \cdot)$ be an abelian group and consider the following subset $H \subseteq G$ of elements $x \in G$ that are equal to the inverses, that is:

(1)
\begin{align} \quad H = \{ x \in G : x = x^{-1} \} \end{align}

We will now show that $(H, \cdot)$ is actually a group itself - namely an abelian one. Since $H \subseteq G$ we only need to check whether $H$ is closed under $\cdot$ and whether for all $x \in H$ there exists $x^{-1} \in H$ such that $x \cdot x^{-1} = e$ and $x^{-1} \cdot x = e$.

Let $x, y \in H$. Then $x = x^{-1}$ and $y = y^{-1}$. Using the fact that $x \cdot y = y \cdot x$ (due to $\cdot$ being commutative from $(G, \cdot)$ being an abelian group) we have that:

(2)
\begin{align} \quad x \cdot y = x^{-1} \cdot y^{-1} = (y \cdot x)^{-1} = (x \cdot y)^{-1} \end{align}

Therefore $(x \cdot y) \in H$ and so $H$ is closed under the operation $\cdot$.

Now clearly if $x \in H$ then by definition since $x = x^{-1}$ we have that $x^{-1} \in H$ is such that $x \cdot x^{-1} = e$ and $x^{-1} \cdot x = e$.

Therefore $(H, \cdot)$ is an abelian group