Basic Theorems Regarding Groups

# Basic Theorems Regarding Groups

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

• 1) $\cdot$ is associative, i.e., for all $a, b, c \in G$, $a \cdot (b \cdot c) = (a \cdot b) \cdot c)$.
• 2) There exists an identity element $e \in G$ such that $a \cdot e = a = e \cdot a$ for all $a \in G$.
• 3) For each $a \in G$ there exists an $a^{-1} \in G$ such that $a \cdot a^{-1} = a^{-1} \cdot a = e$.

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

 Proposition 1: Let $(G, \cdot)$ be a group and let $e$ be the identity for this group. Then: a) The identity element $e \in G$ is unique. b) For each $a \in G$, the corresponding inverse $a^{-1} \in G$ is unique. c) For each $a \in G$, $(a^{-1})^{-1} = a$. d) For all $a, b \in G$, $(a \cdot b)^{-1} = b^{-1} \cdot a^{-1}$. e) For all $a, b \in G$, if $a \cdot b = e$ then $a = b^{-1}$ and $b = a^{-1}$. f) If $a^2 = a$ then $a = e$.
• Proof of a) Suppose that $e$ and $e'$ are both identities for $\cdot$. Then:
(1)
\begin{align} \quad e = e \cdot e = e \cdot e' = e' \cdot e' = e' \end{align}
• Therefore $e = e'$ so the identity for $\cdot$ is unique. $\blacksquare$
• Proof of b) Suppose that $a^{-1} \in G$ and $a^{-1'} \in G$ are both inverses for $a \in G$ under $\cdot$. Then:
(2)
\begin{align} \quad a^{-1} = a^{-1} \cdot e = a^{-1} \cdot (a \cdot a^{-1'}) = (a^{-1} \cdot a)*a^{-1} = e \cdot a^{-1'} = a^{-1'} \end{align}
• Therefore $a^{-1} = a^{-1'}$ so the inverse for $a$ is unique. $\blacksquare$
• Proof of c) Let $a \in G$. Then $(a^{-1})^{-1}$ is the inverse to $a^{-1}$. However, the inverse to $a^{-1}$ is $a$ and by (b) we have shown that the inverse of each element in $G$ is unique. Therefore $a = (a^{-1})^{-1}$. $\blacksquare$
• Proof of d) If we apply the operation $\cdot$ between $b^{-1} \cdot a^{-1}$ and $(a \cdot b)$ we get:
(3)
\begin{align} \quad (a \cdot b) \cdot [b^{-1} \cdot a^{-1}] & = a \cdot [(b \cdot b^{-1}) \cdot a^{-1}] \\ \quad &= a \cdot [e \cdot a^{-1}] \\ \quad &= a \cdot a^{-1} \\ \quad &= e \end{align}
• Therefore the inverse of $(a \cdot b)$ is $b^{-1} \cdot a^{-1}$. We also have that the invere of $(a \cdot b)$ is $(a \cdot b)^{-1}$. By (b), the inverse of $(a \cdot b)$ is unique and so:
(4)
\begin{align} \quad (a \cdot b)^{-1} = b^{-1} \cdot a^{-1} \quad \blacksquare \end{align}
• Proof of e) Suppose that $a \cdot b = e$. Then:
(5)
\begin{align} \quad a \cdot b &= e \\ \quad (a \cdot b) \cdot b^{-1} &= e \cdot b^{-1} \\ \quad a \cdot (b \cdot b^{-1}) &= b^{-1} \\ \quad a \cdot e &= b^{-1} \\ \quad a &= b^{-1} \end{align}
• Similarly:
(6)
\begin{align} \quad a \cdot b &= e \\ \quad a^{-1} \cdot (a \cdot b) &= a^{-1} \cdot e \\ \quad (a^{-1} \cdot a) \cdot b &= a^{-1} \\ \quad e \cdot b &= a^{-1} \\ \quad b &= a^{-1} \quad \blacksquare \end{align}
• Proof of f) Suppose that $a^2 = a \cdot a = a$. Then:
(7)
\begin{align} \quad a^2 &= a \\ \quad a \cdot a &= a \\ \quad a^{-1} \cdot (a \cdot a) &= a^{-1} \cdot a \\ \quad (a^{-1} \cdot a) \cdot a &= e \\ \quad e \cdot a &= e \\ \quad a &= e \end{align}
• Hence $a = e$. Alternatively we see that if $a \cdot a = a$ then the inverse of $a$ with respect to $\cdot$ is $e$, that is $a^{-1} = e$. Multiplying both sides of this equation by $a$ gives us that $a = e$. $\blacksquare$