Let G be a set equipped with an associative binary operatio

. Let G be a set equipped with an associative binary operation satisfying the following two properties:

Left surjectivity: left multiplication by any element gives a surjective map; in other words, for each g G, the left translation g : G G dened by g(h) = g h is surjective.

Right surjectivity: right multiplication by any element gives a surjective map; in other words, for each g G, the right translation µg : G G dened by µg(h) = h g is surjective.

(a) Show that each element in G has a left identity (which a priori may depend on the element). I.e., for each g G there is an element eg so that eg g = g.

(b) Let g be an element of G, and eg the left identity for g guaranteed by part (a). Show that eg is actually a left identity for all elements in G. (Hint: Use left surjectivity and associativity.)

(c) Show that G has a right identity.

(d) Show that the left identity and right identity are equal, so G simply has an identity

. (e) Show that each element in G has a left inverse and a right inverse. 1

(f) Show that the left inverse equals the right inverse, so each element in G simply has an inverse.

(g) Prove that G is a group.

Solution

(a) Let G be a set equipped with an associative binary operation satisfying the following two properties:

Left surjectivity: left multiplication by any element gives a surjective map; in other words, for each g G, the left translation g : G G dened by g(h) = g h is surjective.

Right surjectivity: right multiplication by any element gives a surjective map; in other words, for each g G, the right translation µg : G G dened by µg(h) = h g is surjective.

Let g G

Applying right surjectivity i.e. right multiplication by any element gives a surjective map, we say that the map µg : G G dened by µg(h) = h g is surjective.

Therefore, h g G for all h G

So, there exist some h = eg G such that h g = g

or, eg g = g

This element eg is said to be left identity of g. Since, g G is arbitrary; we say that each element in G has a left identity. (Proved)

(b) To prove that eg is the left identity for all elements of G

Proof: eg is left identity of g eg g = g

g eg

= (eg g) eg

= eg (g eg)                 Applying associativity

eg (g eg) = g eg eg is the left identity for g eg

Therefore, eg is the left identity for all elements of G

(c) Let g G

Applying left surjectivity i.e. left multiplication by any element gives a surjective map, we say that the map g : G G dened by g(h) = g h is surjective.

Therefore, g h G for all h G

So, there exist some h = ge G such that g h = g

or, g ge = g

This element ge is said to be right identity of g. Since, g G is arbitrary; we say that each element in G has a right identity.

Similarly, we can show that ge is the right identity for all elements of G.

(d) eg ge = ge (because eg is the left identity)

eg ge = eg (because ge is the right identity)

ge = eg

The left identity and right identity are equal, so G simply has an identity denoted as e

ge = eg = e (Proved)