# Any set with Associativity, Left Identity, Left Inverse, is a Group. - Fraleigh p.49 4.38

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.

• Why don't you use "educated guess" rather than "prognosticate"? Dec 23 '13 at 15:58
• @AlexBecker Isn't it wrong to say "person educated guessed"? And I want to stay far away from the word "guess". Dec 23 '13 at 16:03
• The grammatically correct way would be to say "made an educated guess". And the word "prognosticate" isn't quite a synonym for making an educated guess - it's more like predicting the future. Dec 23 '13 at 16:08
• I think the main reason for not using the one-sided characterization as the definition of "group" is that it would require doing this imprognosticable deduction right at the beginning of the development of group theory. Dec 23 '13 at 17:12
• The preference for two-sided over one-sided axioms is the same as the preference for multiplication axioms over the single division axiom: $x / ((((x / x) / y) / z) / (((x / x) / x) / z)) = y$ -- one uses whatever one finds easiest. The division thing is probably never easiest, though showing the one-sided axioms are satisfied is sometimes way less effort (though proving abstract facts is easier with the two-sided characterization). Dec 23 '13 at 22:42

\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}

• Thanks a lot. (1). What do you mean by 'a nice $x$'? (2). What do you mean by 'gee, the last two steps in effect'? (3). Why 'undo all steps until that point'? What's 'that point'? Dec 28 '13 at 9:01
• (1) The $x$ was introduced as something new; a few lines later it turns out that the IDEA of picking $x=a^{-1}$ is "nice" as it allows cncellation (2) [the sentence is continued on the next line] We were able to transform $\mathrm{foo}\cdot e$ to $\mathrm{foo}$, which is what the whole problem is about (just with $\mathrm{foo}=a$ instead of $\mathrm{foo}=(a^{-1})^{-1}\cdot e$); therefore (3) I suggest to do what we did to get from $a\cdot e$ to $\mathrm{foo}\cdot e$ in reverse order, which should take us from $\mathrm{foo}$ to the desired $a$. Dec 28 '13 at 13:39

There are two key points behind the tricky algebra that you want to prognosticate. Both come from the fact that groups were abstracted from the study of actions, and that most of the constructions in group theory are really special cases of constructions from group actions (in fact, normal subgroups were defined in terms of group actions before abstract groups themselves were even defined).

The first key point is that the inverse of an inverse of a thing should behave like the original thing, since undoing an action and the undoing that, should result in just doing the action. Formally, the relevant definitions are the following.

A magma is a set $M$ together with a binary operation, i.e. a function $M\times M\to M$ (which I will denote by juxtaposition: $(m_1,m_2)\mapsto m_1m_2$). An action of a magma $M$ on a set $X$, (also called an $M$-set) is a two-variable function $M\times X\to X$ (denoted by $(m,x)\mapsto m\cdot x$) and satisfying the axiom $(m_1m_2)\cdot x=m_1\cdot(m_2\cdot x)$. The intuition here is supposed to be that the elements of the magma "move around" the elements of the set in a certain specified way.

One of the most important constructions in the theory of actions is that of the the stabilizer of an element $x$ of an $M$-set $X$. Concretely, the stabilizer $\DeclareMathOperator{\Stab}{Stab}\Stab(x)$ of $x\in X$ (for a magma action $M\times X\to X$) is defined as $\Stab(x)=\{m\in M\colon m\cdot x=x\}$ (side-note: $\Stab(x)$ is a "sub-magma": if $m_1,m_2\in\Stab(x)$, then $m_1m_2\in\Stab(x)$).

With this out of the way, we can now say what it means for $m'$ to be an inverse to $m$, relative to a magma-action $M\times X\to X$. It means that $m'm\in\Stab(x)$ for every $x$ (i.e. $(m'm)\cdot x=m'\cdot(m\cdot x)=x$). Similarly, an element $e$ is an identity for the action if $e\in\Stab(x)$ for every $x$, i.e. $e\cdot x=x$.

The key point behind the algebra that you wish to prognosticate is then the following. If $m''$ is an inverse for $m'$, and $m'$ is an inverse for $m$, then since $m''$ is supposed to be undoing what $m'$ is doing, and $m'$ is undoing what $m$ is doing, $m''$ should be doing what $m$ is doing. In equations, $m''\cdot x=m''\cdot (m'\cdot(m\cdot x))=(m''m')\cdot(m\cdot x)=m\cdot x$ for every $x\in X$. In particular, since $m''$ is an inverse of $m'$, $m$ should also be an inverse of $m'$, and it is: replacing $x$ with $m'\cdot x$ above reveals that $x=(m''m')\cdot x=m''\cdot(m' x)=(mm')\cdot x$ so that $m$ is indeed an inverse (i.e. $mm'$ is an identity).

The second key point is that a group acts on itself in two ways: on the left, and on the right, and that inverses allow you to transfer information from one side of the action to the other.

A magma is called a semi-group if the binary operation is associative, i.e. $(m_1m_2)m_3=m_1(m_2m_3)$. This is the same as requiring that the binary operation of the magma is itself an action of the magma on itself, i.e. if we set $m_1\cdot m_2:=m_1m_2$, associativity says that $(m_1m_2)\cdot m_3=m_1\cdot(m_2m_3)$. Let me repeat: associativity of the binary operation corresponds to the algebraic structure acting on itself via the binary operation.

A semi-group is called a monoid if there is an element $e\in M$ such that $em=m=me$ for every $m\in M$. We call such an $e$ a two-sided identity, while a left-sided identity is any element $e$ satisfying $em=m$ for every $m\in M$, i.e. if $e\in\Stab(m)$ for every $m\in M$, so an identity for the action of $M$ on itself. We can't, however, easily express the idea that $e$ is a right-identity, i.e. that $me=m$ for every $m\in M$ using the obvious action of $M$ on itself. What is missing here is the that a semi-group also has a contravariant action on itself. A contravariant action $M\times X\to X$ satisfies $(m_1 m_2)\cdot x=m_2\cdot(m_1\cdot x)$ (whereas a usual (covariant) action satisfies $(m_1m_2)\cdot x=m_1\cdot(m_2\cdot x))$. For a semi-group associativity says not only that $m_1\cdot_l m_2:=m_1m_2$ defines an action of $M$ on itself, but that $m_1\cdot_{r}m_2:=m_2m_1$ defines a contravariant action (side-note: a semi-group is commutative, i.e. $m_1m_2=m_2m_1$ if and only if the left- and right-actions $\cdot_l$ and $\cdot_r$ are the same action).

To summarize, a semi-group is a magma $M$ with an associative operation, which automatically implies that $M$ acts on itself in two ways: on the left and on the right. A left-identity is an identity for the left-action, a right-identity is an identity for the right-action, a left-inverse is an inverse for the left-action, and a right-inverse is an inverse for the right-action.

Now, it is easy to see how inversion relates the left- and right-actions. If $m_1'$ and $m_2'$ are left-inverses for $m_1$ and $m_2$, then $m_1'm_2'$ is a left-inverse for $m_2m_1$. Furthermore, the left-inverse of any element of $\Stab_l(m)$ also has to be in $\Stab_l(m)$. This implies that the inverses of elements in $\Stab_l(m)$ are all contained in $\Stab_r(m')$ for $m'$ a left-inverse of $m$. Finally, since a left-identity is tautologically its own left-inverse, it follows that any left-identity is already contained in the right-stabilizer of any left-inverse.

Thus, it is enough to show that every element of $M$ is a left-inverse, which from the first key point would follow from knowing that any left-inverse has a left-inverse, which is implied by requiring every element to have a left-inverse. Thus, any left-identity is a right-identity, and every element is both a left and a right-inverse.

1) It might be a bit more believable if you imagine it being worked out backwards from how you've written it, and keep in mind that it is a "standard arithmetic trick" to use inverses to introduce terms without affecting the identity. So the use of $e = a^{-1} a$ and variants of it is not surprising, if you've seen many arithmetic or basic calculus results proven.

The double-inverses showing up is probably the more "oh interesting" part. But, it seems easy to imagine working on these proofs for a long time, getting stuck, and noticing "Ah, if I had used $(a^{-1})^{-1}$ and $a^{-1}$ instead, I'd be able to make this cancellation here..."

I expect a combination of time spent attempting the proof and familiarity with other proofs is how you get these, much like any other proof that seems to pull something out of nowhere.

2) Perhaps because viewing it all as left-sided things fits with an interpretation in terms of functions (one for each element) & composition. This view makes it clear that a group is a set acting on itself as a group. If you're familiar with the idea of group actions, then it seems natural to try defining it this way. These proofs you posted point out the bonus that the group acts back on itself from the right side as well, and elements have the same roles from that side.

3) No idea. I think it's standard to learn that you can check the group properties just on one side though. So, in some sense, it doesn't matter which you consider "the definition". Something just ended up being more popular.

I have seen a more comprehensible proof that the one sided axioms are sufficient. It starts with an easy to prove lemma: a*a=a implies that a=e. (Just multiply both sides by the inverse of a.) Once you have this, the rest is easy. For what it's worth, the Fraleigh proof totally frustrated me fifty years ago. You are not alone!