The Inverse Element in a Group

Within any group $ (G,*) $, for a given element $a \in G$, its inverse element $a^{-1}$ is defined as the one that, combined with $a$ through the group's operation $ * $, yields the group's identity element $e$. This means: $$ a * a^{-1} = a^{-1} * a = e $$ where $ * $ symbolizes the group's binary operation.

This principle underscores that every element within the group is paired with a unique inverse.

Essential to the group's structure, the inverse element ensures there's a mechanism to "neutralize" any element, effectively resetting the operation to its identity element.

The hallmark of inverses, their existence, and uniqueness, are fundamental distinctions that separate groups from other algebraic structures.

Group refresher: Defined as a collection $G$ coupled with a binary operation, a group associates each pair of its elements with another element of $G$, adhering to four core properties: closure, associativity, identity, and the presence of inverse elements.

Let's delve into some illustrative examples.

Example

Consider the group $ (\mathbb{Z} , + ) $, which represents the integers $\mathbb{Z}$ under addition $ + $.

In this scenario, the identity element is $0$, as adding $0$ to any integer nets the original integer.

Thus, the inverse element $a^{-1}$ for any integer $a$ is its negative $-a$, ensuring $a + (-a) = 0$.

Specific examples: For the integer $a = 5$, its inverse is $-5$, because $5 + (-5) = 0$. For $a = -3$, the inverse is $3$, as $(-3) + 3 = 0$. Both cases demonstrate that combining an element with its inverse results in the identity element of $0$.

Here, the opposite element $ a^{-1} $ serves to "cancel out" an element $ a $, seamlessly reverting the operation back to the identity element $ 0 $ through the group's operation.

Example 2

Next, let's explore the group of $2 \times 2$ invertible matrices under multiplication.

A matrix is considered invertible (or nonsingular) if it pairs with another matrix to produce the identity matrix upon multiplication. Such a matrix is invertible if its determinant is not zero, indicating not all $2 \times 2$ matrices are invertible.

In this instance, the identity matrix $I_2$ stands as the group's identity element.

$$ I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$

Take the matrix $A$ as a practical $2 \times 2$ matrix example:

$$ A = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} $$

To find its inverse $A^{-1}$, we seek:

$$ A \cdot A^{-1} = A^{-1} \cdot A = I_2 $$

The formula for inverting a $2 \times 2$ matrix, $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, goes as follows:

$$ A^{-1} = \frac{1}{ad-bc} \cdot \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} $$

For matrix $A$, we calculate the determinant $ \Delta = (ad-bc)$:

$$ \Delta \ A = 2 \cdot 3 - 5 \cdot 1 = 6 - 5 = 1 $$

With the determinant clearly nonzero, matrix $A$ is confirmably invertible, with its inverse being:

$$ A^{-1} = \frac{1}{1} \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} $$

Ensuring $A \cdot A^{-1} = I_2$:

$$ A \cdot A^{-1} = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \cdot \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} $$

$$ A \cdot A^{-1} = \begin{pmatrix} 2\cdot3 + 5\cdot(-1) & 2\cdot(-5) + 5\cdot2 \\ 1\cdot3 + 3\cdot(-1) & 1\cdot(-5) + 3\cdot2 \end{pmatrix} $$

$$ A \cdot A^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I_2 $$

This clearly shows that $A^{-1}$ effectively reverses $A$, as their multiplication leads back to the identity matrix $I_2$, affirming $ A \cdot A^{-1} = A^{-1} \cdot A = I_2 $.