Hw4 Problem 9

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Problem 9

(*) The following problem shows that we can loosen the requirements in the last two group axioms and still get a group. Consider a set $G$ with a binary operation $*$ such that

• $*$ is associative
• there exists a left identity element $e \in G$ such that $e * x = x$ for all $x \in G$
• for each $a \in G$ there exists a left inverse $a' \in G$ such that $a' * a = e$.

(a) Show that left cancellation holds. That is, if $a * b = a * c$, then $b = c$.
(b) Show that the left identity element $e$ is also a right identity element for all $x \in G$.
(c) Show that the left inverse $a'$ for $a$ is also a right inverse for $a$.

Solution

For all $a, a', b, c, x, x', e \in G$.

(a)

(1)

\begin{equation} a * b = a * c \end{equation}

(2)

\begin{equation} a' * (a * b) = a' * (a * c) \end{equation}

(3)

\begin{equation} (a' * a) * b = (a' * a) * c \end{equation}

(4)

\begin{equation} e * b = e * c \end{equation}

(5)

\begin{equation} b = c \end{equation}

So left cancellation holds.

(b)

(6)

\begin{equation} x' * (x * e) = (x' * x) * e \end{equation}

(7)

\begin{equation} = e * e \end{equation}

(8)

\begin{equation} = e \end{equation}

(9)

\begin{equation} = x' * x \end{equation}

Since

(10)

\begin{equation} x' * (x * e) = x' * x \end{equation}

(11)

\begin{equation} x * e = x \end{equation}

Then $e$ is a right identity element for all $x \in G$.

(c)

(12)

\begin{equation} a' * (a * a') = (a' * a) * a' \end{equation}

(13)

\begin{equation} = e * a' \end{equation}

(14)

\begin{equation} = a' \end{equation}

(15)

\begin{equation} = a' * e \end{equation}

Since

(16)

\begin{equation} a' * (a * a') = a' * e \end{equation}

(17)

\begin{equation} a * a' = e \end{equation}

Then $a'$ is a right inverse for $a$.