[summary: Test your understanding of the definition of a group with these exercises. ]
Preliminaries
- Show that the identity element in a group is unique. That is, if G is a group and two elements e1,e2∈G both satisfy the axioms describing the identity element, then e1=e2.
%%hidden(Show solution): By definition, an identity element e satisfies eg=ge=g for all g∈G. Hence if e1 is an identity, then e1e2=e2e1=e1. And if e2 is an identity, then e2e1=e1e2=e2. Hence e1=e2. Note that this argument makes no use of inverses, and so is valid for monoids. %%
- Show that inverses are also unique. That is, if g∈G is an element of a group and h1,h2∈G both satisfy the axioms describing the inverse of g, then h1=h2.
%%hidden(Show solution): By definition, an inverse h of g satisfies hg=gh=e. So h1g=gh1=e and h2g=gh2=e. Hence, on the one hand,
h1gh2=(h1g)h2=(e)h2=h2
and, on the other hand,
h1gh2=h1(gh2)=h1(e)=h1.
Hence h1=h2. %%
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.
- The real numbers R together with the addition operation (x,y)↦x+y.
%%hidden(Show answer): Yes, this is a group. The identity element is 0, and inverse is given by x↦−x. %%
- The real numbers R together with the multiplication operation (x,y)↦xy.
%%hidden(Show answer): No, this is not a group. 0∈R has the property that 0×x=0 for all real numbers x, so it can't be invertible no matter what the identity is. %%
- The positive real numbers R>0 together with the multiplication operation (x,y)↦xy.
%%hidden(Show answer): Yes, this is a group. The identity is 1, and inverse is given by x↦1x. In fact this group is isomorphic to (R,+); can you name the isomorphism? %%
- The real numbers R together with the operation (x,y)↦x+y−1.
%%hidden(Show answer): Yes, this is a group (in fact isomorphic to (R,+); can you name the isomorphism?). The identity element is 1, and inverse is given by x↦2−x (can you explain why, conceptually?). %%
- The real numbers R together with the operation (x,y)↦x+y1+xy.
%%hidden(Show answer): 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 [division_by_zero dividing by zero].
This operation is interesting and useful, though, when it is defined. It shows up in [special_relativity special relativity], where it describes how velocities add relativistically (in units where the speed of light is 1). %%