# Preliminaries

1. Show that the identity element in a group is unique. That is, if $$G$$ is a group and two elements $$e_1, e_2 \in G$$ both satisfy the axioms describing the identity element, then $$e_1 = e_2$$.

By definition, an identity element $$e$$ satisfies $$eg = ge = g$$ for all $$g \in G$$. Hence if $$e_1$$ is an identity, then $$e_1 e_2 = e_2 e_1 = e_1$$. And if $$e_2$$ is an identity, then $$e_2 e_1 = e_1 e_2 = e_2$$. Hence $$e_1 = e_2$$. Note that this argument makes no use of inverses, and so is valid for monoids.

1. Show that inverses are also unique. That is, if $$g \in G$$ is an element of a group and $$h_1, h_2 \in G$$ both satisfy the axioms describing the inverse of $$g$$, then $$h_1 = h_2$$.

By definition, an inverse $$h$$ of $$g$$ satisfies $$hg = gh = e$$. So $$h_1 g = g h_1 = e$$ and $$h_2 g = g h_2 = e$$. Hence, on the one hand,

$$h_1 g h_2 = (h_1 g) h_2 = (e) h_2 = h_2$$

and, on the other hand,

$$h_1 g h_2 = h_1 (g h_2) = h_1 (e) = h_1.$$

Hence $$h_1 = h_2$$. <div><div>

# Examples involving numbers

Determine whether the following sets equipped with the specified binary operations are groups. If so, describe their identity elements (which by the previous exercise must be unique) and how to take inverses.

1. The real numbers $$\mathbb{R}$$ together with the addition operation $$(x, y) \mapsto x + y$$.

Yes, this is a group. The identity element is $$0$$, and inverse is given by $$x \mapsto -x$$.

1. The real numbers $$\mathbb{R}$$ together with the multiplication operation $$(x, y) \mapsto xy$$.

No, this is not a group.$$0 \in \mathbb{R}$$ has the property that $$0 \times x = 0$$ for all real numbers $$x$$, so it can’t be invertible no matter what the identity is.

1. The positive real numbers $$\mathbb{R}_{>0}$$ together with the multiplication operation $$(x, y) \mapsto xy$$.

Yes, this is a group. The identity is $$1$$, and inverse is given by $$x \mapsto \frac{1}{x}$$. In fact this group is isomorphic to $$(\mathbb{R}, +)$$; can you name the isomorphism?

1. The real numbers $$\mathbb{R}$$ together with the operation $$(x, y) \mapsto x + y - 1$$.

Yes, this is a group (in fact isomorphic to $$(\mathbb{R}, +)$$; can you name the isomorphism?). The identity element is $$1$$, and inverse is given by $$x \mapsto 2 - x$$ (can you explain why, conceptually?).

1. The real numbers $$\mathbb{R}$$ together with the operation $$(x, y) \mapsto \frac{x + y}{1 + xy}$$.

No, this is not a group. It’s easy to be tricked into thinking it is, because if you just work through the algebra, it seems that all of the group axioms hold. However, this operation is not an operation! It’s not defined if the denominator is $$0$$, because then we’d be dividing by zero.

This operation is interesting and useful, though, when it is defined. It shows up in special relativity, where it describes how velocities add relativistically (in units where the speed of light is $$1$$). <div><div>

Parents:

• Group

The algebraic structure that captures symmetry, relationships between transformations, and part of what multiplication and addition have in common.