8.3 Elementary consequences of the definition
Throughout this section, \(G\) will be a multiplicatively written group. Remember that being “multiplicatively written” is purely a matter of notation, so everything will apply to any group.
A familiar idea from elementary algebra is that if you have some equation (say involving \(a,b,x\in \mathbb{R}\)) such as \[ax=bx\] then, as long as \(x\neq 0\), you can “cancel” the \(x\) to deduce that \(a=b\). Formally, we multiply both sides by \(x^{-1}\) (which exists by axiom (A11)).
A similar principle applies in a group, because of the fact that elements all have inverses, with one complication caused by the fact that the group operation may not be commutative.
Since multiplying on the left and on the right are in general different, so we need to decide which is appropriate.
Let \(a,b,x\) be elements of a multiplicatively written group \(G\). If \(ax=bx\), then \(a=b\).
Multiply on the right by \(x^{-1}\): \[\begin{align*} ax=bx&\Rightarrow (ax)x^{-1}=(bx)x^{-1}\\ &\Rightarrow a(xx^{-1})=b(xx^{-1})\\ &\Rightarrow ae=be\\ &\Rightarrow a=b. \end{align*}\]
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Let \(a,b,x\) be elements of a multiplicatively written group \(G\). If \(xa=xb\), then \(a=b\).
Multiply on the left by \(x^{-1}\): \[\begin{align*} xa=xb&\Rightarrow x^{-1}(xa)=x^{-1}(xb)\\ &\Rightarrow (xx^{-1})a =(xx^{-1})b\\ &\Rightarrow ea=eb\\ &\Rightarrow a=b. \end{align*}\]
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This simple principle has some nice consequences that make studying groups easier. One is that the defining property of identity element \(e\) is enough to identify it: no other element has the same property.
Let \(a,x\) be elements of a multiplicatively written group. If \(ax=a\) then \(x=e\). Similarly, if \(xa=a\) then \(x=e\).
If \(ax=a\) then \(ax=ae\). By “left cancellation”, we can cancel \(a\) to deduce \(x=e\). Similarly, if \(xa=a\) then \(xa=ea\), so by “right cancellation” we can cancel \(a\) to deduce \(x=e\).
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A similar proof shows that an element of a group can only have one inverse.
Let \(x,y\) be elements of a multiplicatively written group. If \(xy=e\) then \(x=y^{-1}\) and \(y=x^{-1}\).
If \(xy=e\) then \(xy=xx^{-1}\), and so, by left cancellation \(y=x^{-1}\). Similarly, if \(xy=e\) then \(xy=y^{-1}y\), and so by right cancellation \(x=y^{-1}\).
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This means that, to prove that one element \(x\) of a group is the inverse of another element \(y\), we just need to check that their product (either way round: \(xy\) or \(yx\)) is equal to the identity. Here are some examples of useful facts that we can prove like this:
Let \(x\) be an element of a multiplicatively written group. Then the inverse of \(x^{-1}\) is \(x\): \((x^{-1})^{-1}=x\).
By uniqueness of inverses, we just need to check that \(xx^{-1}=e\), which is true.
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Let \(x,y\) be elements of a multiplicatively written group. Then the inverse of \(xy\) is \((xy)^{-1}=y^{-1}x^{-1}\).
By uniqueness of inverses, we just need to check that \((xy)(y^{-1}x^{-1})=e\). But \[(xy)(y^{-1}x^{-1})=((xy)y^{-1})x^{-1}=(x(yy^{-1}))x^{-1}=(xe)x^{-1}=xx^{-1}=e.\]
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Make sure you understand how each step of the previous proof follows from the definition of a group, and in particular how we have used the associative property. The notes will be less explicit about this in future proofs, leaving out brackets.
Next, some notation. If \(x\in G\), we write write \(x^2\) for the product \(xx\) of \(x\) with itself, \(x^3\) for the product \(x(x^2)\) (which is the same as \((x^2)x\) by associativity), and so on.
Note that for \(n=-1\) we also have a meaning for \(x^n\), since \(x^{-1}\) is notation we use for the inverse of \(x\).
We extend this even further by defining \(x^{-n}\) to be \((x^n)^{-1}\) if \(n>0\) (which gives us a meaning for \(x^n\) for any non-zero integer \(n\), positive or negative) and defining \(x^0\) to be the identity element \(e\) (so that we now have a meaning for \(x^n\) for every \(n\in \mathbb{Z}\).
We call \(x^n\) the \(n\)th power of \(x\).To justify why this is a sensible notation, let us see what happens when we multiply powers. First:
If \(n>0\) then \(x^{-n}=(x^{-1})^n\).
By definition, \(x^{-n}=(x^n)^{-1}\). To prove this is the same as \((x^{-1})^n\) we just have to show \((x^{-1})^nx^n=e\), by uniqueness of inverses. But \[(x^{-1})^nx^n=x^{-1}\cdots x^{-1}x^{-1}xx\cdots x= x^{-1}\cdots x^{-1}ex\cdots x=x^{-1}\cdots x^{-1}x\cdots x=\cdots =e,\] cancelling each \(x^{-1}\) with an \(x\).
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If \(x\) is an element of a multiplicatively written group \(G\), and \(m\) and \(n\) are integers, then \[(x^m)(x^n)=x^{m+n}.\]
Let us fix \(n\in \mathbb{Z}\) and prove it is true for all \(m\in \mathbb{Z}\).
We first prove this by induction for when \(m\geq 0\). It is true when \(m=0\) since \[(x^0)(x^n)=e(x^n)=x^n.\] Suppose it is true for \(m=k-1\) [that is \((x^{k-1})(x^n)=x^{k-1+n}\)]. We show it is true for \(m=k\). We have \[(x^k)(x^n)=\left(x(x^{k-1})\right)(x^n)=x\left((x^{k-1}(x^n)\right)=x(x^{k+n-1})=x^{k+n}.\] So by induction it is true for all \(m\geq 0\).
If \(m<0\), and \(y=x^{-1}\), then by the lemma \((x^m)(x^n)=(y^{-m})(y^{-n})\) which is equal to \(y^{-(m+n)}=x^{m+n}\) by applying what we’ve already proved with \(y\) in place of \(x\).□
We’ve already proved the formula \((xy)^{-1}=y^{-1}x^{-1}\). What about \((xy)^n\) for other values of \(n\)? In a non-abelian group, there is no simple formula. In particular: