Set with Associativity, Left Identity, Left Inverse is a Group – Abstract Algebra

abstract-algebragroup-theoryintuition

I couldn't unravel the 3rd para. in a similar post. Proof that left identity element = right identity element:

$\begin{align} \color{#1E90FF}e*e &= \color{darkorange}
{e} \\
\color{#1E90FF}{a^{-1}*a}*e& =\color{darkorange}
{a^{-1}*a} \\
\color{purple}{(a^{-1})^{-1}*}\color{#1E90FF}{a^{-1}*a}*e& =\color{purple}{(a^{-1})^{-1}*}\color{darkorange}
{a^{-1}*a} \\
\color{purple}{[(a^{-1})^{-1}*}\color{#1E90FF}{a^{-1}] *a}*e& =\color{purple}{[(a^{-1})^{-1}*}\color{darkorange}
{a^{-1}]*a} \\
\color{#1E90FF}{a}*e& = \qquad \qquad \qquad \quad\color{darkorange}
{a}
\end{align}$

Proof that left inverse = right inverse,

$\begin{align}a^{-1} * a & = e \\
a^{-1} * a \color{#1E90FF}{* a^{-1}} &= e \color{#1E90FF}{* a^{-1}} \\
a^{-1} * a \color{#1E90FF}{* a^{-1}} &= \color{#1E90FF}{a^{-1}} \\
\color{magenta}{[(a^{-1})^{-1}*}a^{-1}] * a \color{#1E90FF}{* a^{-1}} &= \color{magenta}{(a^{-1})^{-1}*} \color{#1E90FF}{a^{-1}}
\\
a \color{#1E90FF}{* a^{-1}} &= e
\end{align}$

  1. How can you prognosticate the tricky algebra here? You must rewrite $e$ as $a*a^{-1}$ and must know what to multiply by. Can someone please make this less prescient?

  2. Why does a one-sided definition of a group have to be all left sided or right sided?
    I'm NOT asking about the algebra…I'm asking for the intuition?

  3. If one-sided definitions are correct for groups, why not use them instead of the standard two-sided definitions?

Reference: Fraleigh, A First Course in Abstract Algebra, p. 49 Question 4.38.

Best Answer

$$\begin{align}a\cdot e &= e\cdot(a\cdot e)&\text{ a left identity is all I have at our disposal}\\ &=(\color{darkcyan}{e}\cdot a)\cdot e&\text{associativity is the only apparent option}\\ &=(\color{darkcyan}{(x^{-1}\cdot x)}\cdot a)\cdot e&\text{let's introduce something new for $e$, maybe later pick a nice $x$}\\ &=(\color{magenta}{x}^{-1}\cdot(\color{magenta}x\cdot a))\cdot e&\text{again, what else but associativity is possible?}\\ &=???&\text{IDEA! If $x$ happens to be $a^{-1}$, we can continue}\\ &=(\color{magenta}{(a^{-1})}^{-1}\cdot(\color{magenta}{a^{-1}}\cdot a))\cdot e&\text{... like this}\\ &=((a^{-1})^{-1}\cdot e)\cdot e&\text{ so that some expressions cacel}\\ &=(a^{-1})^{-1}\cdot (e\cdot e)&\text{associativity, what else?}\\ &=(a^{-1})^{-1}\cdot e&\text{now use the $e$ is left neutral - gee, the last two steps in effect ...}\\ &=(a^{-1})^{-1}\cdot(a^{-1}\cdot a)&\text{got rid of an $e$ on the right! So lets undo all steps until that point}\\ &=((a^{-1})^{-1}\cdot a^{-1})\cdot a\\ &=e\cdot a\\ &=a\end{align} $$

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