## Section 2.3 Basic Properties of Groups

###### Proposition 2.17.

The identity element in a group \(G\) is unique; that is, there exists only one element \(e \in G\) such that \(eg = ge = g\) for all \(g \in G\text{.}\)

###### Proof.

Suppose that \(e\) and \(e'\) are both identities in \(G\text{.}\) Then \(eg = ge = g\) and \(e'g = ge' = g\) for all \(g \in G\text{.}\) We need to show that \(e = e'\text{.}\) If we think of \(e\) as the identity, then \(ee' = e'\text{;}\) but if \(e'\) is the identity, then \(ee' = e\text{.}\) Combining these two equations, we have \(e = ee' = e'\text{.}\)

Inverses in a group are also unique. If \(g'\) and \(g''\) are both inverses of an element \(g\) in a group \(G\text{,}\) then \(gg' = g'g = e\) and \(gg'' = g''g = e\text{.}\) We want to show that \(g' = g''\text{,}\) but \(g' = g'e = g'(gg'') = (g'g)g'' = eg'' = g''\text{.}\) We summarize this fact in the following proposition.

###### Proposition 2.18.

If \(g\) is any element in a group \(G\text{,}\) then the inverse of \(g\text{,}\) denoted by \(g^{-1}\text{,}\) is unique.

###### Proposition 2.19.

Let \(G\) be a group. If \(a, b \in G\text{,}\) then \((ab)^{-1} = b^{-1}a^{-1}\text{.}\)

###### Proof.

Let \(a, b \in G\text{.}\) Then \(abb^{-1}a^{-1} = aea^{-1} = aa^{-1} = e\text{.}\) Similarly, \(b^{-1}a^{-1}ab = e\text{.}\) But by the previous proposition, inverses are unique; hence, \((ab)^{-1} = b^{-1}a^{-1}\text{.}\)

###### Proposition 2.20.

Let \(G\) be a group. For any \(a \in G\text{,}\) \((a^{-1})^{-1} = a\text{.}\)

###### Proof.

Observe that \(a^{-1} (a^{-1})^{-1} = e\text{.}\) Consequently, multiplying both sides of this equation by \(a\text{,}\) we have

It makes sense to write equations with group elements and group operations. If \(a\) and \(b\) are two elements in a group \(G\text{,}\) does there exist an element \(x \in G\) such that \(ax = b\text{?}\) If such an \(x\) does exist, is it unique? The following proposition answers both of these questions positively.

###### Proposition 2.21.

Let \(G\) be a group and \(a\) and \(b\) be any two elements in \(G\text{.}\) Then the equations \(ax = b\) and \(xa = b\) have unique solutions in \(G\text{.}\)

###### Proof.

Suppose that \(ax = b\text{.}\) We must show that such an \(x\) exists. We can multiply both sides of \(ax = b\) by \(a^{-1}\) to find \(x = ex = a^{-1}ax = a^{-1}b\text{.}\)

To show uniqueness, suppose that \(x_1\) and \(x_2\) are both solutions of \(ax = b\text{;}\) then \(ax_1 = b = ax_2\text{.}\) So \(x_1 = a^{-1}ax_1 = a^{-1}ax_2 = x_2\text{.}\) The proof for the existence and uniqueness of the solution of \(xa = b\) is similar.

###### Proposition 2.22.

If \(G\) is a group and \(a, b, c \in G\text{,}\) then \(ba = ca\) implies \(b = c\) and \(ab = ac\) implies \(b = c\text{.}\)

This proposition tells us that the right and left cancellation laws are true in groups. We leave the proof as an exercise.

We can use exponential notation for groups just as we do in ordinary algebra. If \(G\) is a group and \(g \in G\text{,}\) then we define \(g^0 = e\text{.}\) For \(n \in {\mathbb N}\text{,}\) we define

and

###### Theorem 2.23.

In a group, the usual laws of exponents hold; that is, for all \(g, h \in G\text{,}\)

\(g^mg^n = g^{m+n}\) for all \(m, n \in {\mathbb Z}\text{;}\)

\((g^m)^n = g^{mn}\) for all \(m, n \in {\mathbb Z}\text{;}\)

\((gh)^n = (h^{-1}g^{-1})^{-n}\) for all \(n \in {\mathbb Z}\text{.}\) Furthermore, if \(G\) is abelian, then \((gh)^n = g^nh^n\text{.}\)

We will leave the proof of this theorem as an exercise. Notice that \((gh)^n \neq g^nh^n\) in general, since the group may not be abelian. If the group is \({\mathbb Z}\) or \({\mathbb Z}_n\text{,}\) we write the group operation additively and the exponential operation multiplicatively; that is, we write \(ng\) instead of \(g^n\text{.}\) The laws of exponents now become

\(mg + ng = (m+n)g\) for all \(m, n \in {\mathbb Z}\text{;}\)

\(m(ng) = (mn)g\) for all \(m, n \in {\mathbb Z}\text{;}\)

\(m(g + h) = mg + mh\) for all \(n \in {\mathbb Z}\text{.}\)

It is important to realize that the last statement can be made only because \({\mathbb Z}\) and \({\mathbb Z}_n\) are commutative groups.

### Reading Questions Reading Questions

###### 1.

For any group \(G\) with elements \(a\) and \(b\text{,}\) is \((ab)\inv = a\inv b\inv\text{?}\) Explain why or give a counterexample.

###### 2.

In any group \(G\) with elements \(a\) and \(b\text{,}\) is there a unique solution in the group to the equation \(ax = b\text{?}\) Explain why or give a counterexample.

###### 3.

What are the *right and left cancellation laws*? Do these hold for all groups, some groups, or no groups? Explain.

###### 4.

After reading the section, what questions do you still have? Write at least one well formulated question (even if you think you understand everything).

### Exercises Practice Problems

###### 1.

Consider a group \(G\) which contains elements \(a\text{,}\) \(b\text{,}\) and \(c\text{,}\) with identity \(e\) (among others, possibly). Solve each of the equations below for \(x\text{.}\)

- \(\displaystyle axb=c\)
- \(x^2a = bxc\inv\) given \(acx=xac\text{.}\)
- \(x^2 = a^2\) given \(x^5 = e\text{.}\)

###### 2.

For each of the following rules, either prove that it is true in every group \(G\) or give a counterexample.

If \(x^2 = e\text{,}\) then \(x=e\text{.}\)

\((ab)^2 = a^2b^2\text{.}\)

If \(x^2 = x\) then \(x = e\text{.}\)

For every \(x \in G\text{,}\) there is some \(y\in G\) such taht \(x = y^2\) (in other words, every \(x\) has a square root).

###### 3.

If \(a,b \in G\) and \(ab = ba\text{,}\) we say that \(a\) and \(b\) commute. Assuming that \(a\) and \(b\) commute, prove the following.

\(a\inv\) and \(b\inv\) commute.

\(a\) and \(b\inv\) commute.

\(a\) commutes with \(ab\text{.}\)

\(a^2\) commutes with \(b^2\text{.}\)

\(xax\inv\) commutes with \(xbx\inv\) for any \(x \in G\text{.}\)

###### 4.

Let \(G\) be a group. Let \(a, b, c\) denote elements of \(G\text{,}\) and let \(e\) be the identity in \(G\text{.}\)

Prove that if \(ab = e\text{,}\) then \(ba = e\text{.}\)

Prove that if \(abc = e\text{,}\) then \(cab = e\) and \(bca = e\text{.}\)

###### 5.

Prove that if \(xay = a\inv\text{,}\) then \(yax = a\inv\text{.}\)

###### 6.

Let \(a\text{,}\) \(b\text{,}\) and \(c\) each be equal to its own inverse. Prove that if \(ab = c\text{,}\) then \(bc = a\) and \(ca = b\text{.}\)

###### 7.

Prove that if \(abc\) is its own inverse, then \(bca\) is its own inverse, and \(cab\) is its own inverse.

###### 8.

Let \(a\) and \(b\) each be equal to its own inverse. Prove that \(ba\) is the inverse of \(ab\text{.}\)

### Exercises Collected Homework

###### 1.

Let \(a\) be an element of a group \(G\) with identity \(e\text{.}\) Prove that \(a = a\inv\) if and only if \(a^2 = e\text{.}\)

You need to prove two directions here: if \(a = a\inv\) then \(a^2 = e\) and if \(a^2 = e\) then \(a = a\inv\text{.}\)

###### 2.

Let \(G\) be a group and suppose that \((ab)^2 = a^2 b^2\) for all \(a, b \in G\text{.}\) Prove that \(G\) is abelian.

You are trying to prove \(xy = yx\) for all \(x, y \in G\text{.}\) Don't assume this! It should be the last line of your proof.

###### 3.

True or false: if \(x^2 = a^2\) then \(x = a\text{.}\) Prove your answer.