# 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 $*$. That is, an Abelian group is a set $G$ paired with a binary operation $* : G \times G \to G$ where:

- For all $a, b \in G$ we have that $(a * b) \in G$ (Closure under $*$).

- For all $a, b, c \in G$ we have that $(a * b) * c = a * (b * c)$ (Associativity of $*$).

- There exists an element $e \in G$ such that $a * e = a$ and $e * a = a$ (The existence of an identity for $*$).

- For all $a \in G$ there exists a $a^{-1} \in G$ such that $a * a^{-1} = e$ and $a^{-1} * a = e$ (The existence of inverses for each element in $G$).

- For all $a, b \in G$ we have that $a * b = b * a$ (Commutativity of $*$).

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

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

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

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

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

Therefore $(H, *)$ is an abelian group