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?

  • $\begingroup$ 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. $\endgroup$ – user1729 Apr 30 '14 at 11:04
  • $\begingroup$ (That should be, "I made a mistake in my answers and I have now fixed it".) $\endgroup$ – user1729 Apr 30 '14 at 11:21

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.

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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.

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