Basic Theorems Regarding Groups

# Basic Theorems Regarding Groups

Recall from the Groups page that a group $(G, *)$ is a set $G$ with a binary operation $*$ such that:

- $G$ is closed under $*$, i.e., $a, b \in G$ implies $(a * b) \in G$.

- $*$ is associative, i.e., for all $a, b, c \in G$, $a * (b * c) = (a * b) * c)$.

- There exists an identity element $e \in G$ such that $a * e = a = e * a$.

- For each $a \in G$ there exists a $a^{-1} \in G$ such that $a * a^{-1} = a^{-1} * a = e$.

We will now look at some rather basic theorems regarding groups which we can derive from the group axioms above.

Theorem 1: Let $(G, *)$ be a group and let $e$ be the identity for $*$. Then $e$ is unique. |

**Proof:**Suppose that $e$ and $e'$ are both identities for $*$. Then:

\begin{align} \quad e = e * e = e * e' = e' * e' = e' \end{align}

- Therefore $e = e'$ so the identity for $*$ is unique. $\blacksquare$

Theorem 2: Let $(G, *)$ be a group. Then for every $a \in G$ there exists a unique $a^{-1} \in G$ that is the inverse of $a$ under $*$. |

**Proof:**Suppose that $a^{-1} \in G$ and $a^{-1'} \in G$ are both inverses for $a \in G$ under $*$. Then:

\begin{align} \quad a^{-1} = a^{-1} * e = a^{-1} * (a * a^{-1'}) = (a^{-1} * a)*a^{-1} = e * a^{-1'} = a^{-1'} \end{align}

- Therefore $a^{-1} = a^{-1'}$ so the inverse for $a$ is unique. $\blacksquare$

Corollary 1: Let $(G, *)$ be a group. Then for all $a \in G$, $(a^{-1})^{-1} = a$. |

**Proof:**Let $a \in G$. Then $(a^{-1})^{-1}$ is the inverse to $a^{-1}$. However, the inverse to $a^{-1}$ is $a$ and by Theorem 2 we have shown that the inverse of each element in $G$ is unique. Therefore $a = (a^{-1})^{-1}$. $\blacksquare$

Corollary 2: Let $(G, *)$ be a group, $a, b \in G$, and let $e$ be the identity for $*$. Then $(a * b)^{-1} = b^{-1} * a^{-1}$. |

**Proof:**If we apply the operation $*$ between $b^{-1} * a^{-1}$ and $(a * b)$ we get:

\begin{align} \quad (a * b) * [b^{-1} * a^{-1}] \\ \quad = a * [(b * b^{-1}) * a^{-1}] \\ \quad = a * [e * a^{-1}] \\ \quad = a * a^{-1} \\ \quad = e \end{align}

- Therefore the inverse of $(a * b)$ is $b^{-1} * a^{-1}$. We also have that the invere of $(a * b)$ is $(a * b)^{-1}$. By Theorem 2, the inverse of $(a * b)$ is unique and so:

\begin{align} \quad (a * b)^{-1} = b^{-1} * a^{-1} \quad \blacksquare \end{align}

Theorem 3: Let $(G, *)$ be a group, $a, b \in G$, and let $e$ be the identity for $*$. If $a * b = e$ then $a = b^{-1}$ and $b = a^{-1}$. |

**Proof:**Suppose that $a * b = e$. Then:

\begin{align} \quad a * b = e \\ \quad (a * b) * b^{-1} = e * b^{-1} \\ \quad a * (b * b^{-1}) = b^{-1} \\ \quad a * e = b^{-1} \\ \quad a = b^{-1} \end{align}

- Similarly:

\begin{align} \quad a * b = e \\ \quad a^{-1} * (a * b) = a^{-1} * e \\ \quad (a^{-1} * a) * b = a^{-1} \\ \quad e * b = a^{-1} \\ \quad b = a^{-1} \quad \blacksquare \end{align}

Theorem 4: Let $(G, *)$ be a group and let $a \in G$. If $a^2 = a * a = a$ then $a = e$. |

**Proof:**Suppose that $a^2 = a*a = a$. Then:

\begin{align} \quad a^2 = a \\ \quad a * a = a \\ \quad a^{-1} * (a * a) = a^{-1} * a \\ \quad (a^{-1} * a) * a = e \\ \quad e * a = e \\ \quad a = e \end{align}

- Hence $a = e$. Alternatively we see that if $a * a = a$ then the inverse of $a$ with respect to $*$ is $e$, that is $a^{-1} = e$. Multiplying both sides of this equation by $a$ gives us that $a = e$. $\blacksquare$