Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have somewhere the concepts of left groups (A non empty set with an associative binary operation, a left identity and each element has left inverse) and right groups. I have also studied some basic properties of identities and inverses here.

Now I have a question.

Does there exist a left group which is not right group?

share|improve this question
    
I made a mistake in my answer, so I thought I should point this out so that others do not make the same mistake. Injective functions have right-inverses, but the inverses are left-invertible (they are surjections). Similarly, surjections have left-inverses, which are injections. In general, an operation which yields either a left group or a right group is actually a group. –  user1729 Apr 30 at 11:04
    
(That should be, "I made a mistake in my answers and I have now fixed it".) –  user1729 Apr 30 at 11:21

2 Answers 2

up vote 5 down vote accepted

A set under an operation $(S, \ast)$ which is either a left group or a right group is actually a group. I'll prove this for left groups. The right-group proof is analogous.

Let $s\in S$ be arbitrary, and write $1$ for the left-identity. Then $s$ has a left-inverse $t$. We prove that $st=1$. We have the following. $$stst=s(ts)t=st$$ Then $st$ has a left-inverse, and multiplying both sides of $stst=st$ by this we see that $st=1$ as required. Therefore, we have two-sided inverses.

It now suffices to prove that $s\ast1=s$ (where $s$ is arbitrary). This follows from the following. $$s=(st)\ast s=sts=s\ast(ts)=s\ast1$$ The proof is complete.

share|improve this answer

In fact a left group is a group and so is a right group. The lemmas below, which I have cut and pasted from some lecture notes, prove this.

Definition. A left group is a set $G$ together with a binary operation $\circ: G \times G \rightarrow G$ that satisfies the following properties:

(i) For all $g,h \in G$, $g \circ h \in G$;

(ii) For all $g,h,k \in G$, $(g \circ h) \circ k = g \circ (h \circ k)$;

(iii) There exists an element $e \in G$, called a left identity element, such that:

(a) for all $g \in G$, $e \circ g = g$; and

(b) for all $g \in G$ there exists $h \in G$ (a left inverse of $g$) such that $h \circ g = e$.

Lemma 1. Let $G$ be a left group, let $e \in G$ be a left identity element, and for $g \in G$, let $h \in G$ be a left inverse element of $g$. Then $g \circ e = g$ and $g \circ h = e$ (i.e. $e$ is also a right inverse element and $h$ is also a right inverse of $g$).

Proof. We have $h \circ (g \circ e) = (h \circ g) \circ e = e \circ e = e = h \circ g.$ Now let $h'$ be an inverse of $h$. Then multiplying the left and right sides of this equation on the left by $h'$ and using associativity gives $(h' \circ h) \circ (g \circ e) = (h' \circ h) \circ g$. But $(h' \circ h) \circ (g \circ e) = e \circ (g \circ e) = g \circ e$, and $(h' \circ h) \circ g = e \circ g = g$, so we get $g \circ e = g$.

We have $h \circ (g \circ h) = (h \circ g) \circ h = e \circ h = h$, and multiplying on the left by $h'$ gives $(h' \circ h) \circ (g \circ h) = h' \circ h$. But $(h' \circ h) \circ (g \circ h) = e \circ (g \circ h) = g \circ h$ and $(h' \circ h) = e$, so $g \circ h = e$.

Lemma 2. Let $G$ be a left group. Then $G$ has a unique identity element, and any $g \in G$ has a unique inverse.

Proof. Let $e$ and $f$ be two identity elements of $G$. Then, $e \circ f = f$, but by Lemma 1, we also have $e \circ f = e$, so $e=f$ and the identity element is unique.

Let $h$ and $h'$ be two inverses for $g$. Then $h \circ g = h' \circ g = e$, but by Lemma 1 we also have $g \circ h = e$, so $$h = e \circ h = (h' \circ g) \circ h = h' \circ (g \circ h) = h' \circ e = h'$$ and the inverse of $g$ is unique.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.