# 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)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)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