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.