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Modern Algebra: Groups, Rings, Modules, Fields

Sam Macdonald

Section 1.2 Order and Power

“Having power is not nearly as important as what you choose to do with it.”
―Roald Dahl
“Good order is the foundation of all things.”
―Edmund Burke

Subsection Power

Though it might seem natural to write the product \(x\cdot x\) using the notation \(x^2\text{,}\) we have no notion of what that means in the world of groups. Do exponents behave the way we want them to in groups, or does something disasterous occur? Do exponents even exist? Do I? In this section, we attempt to answer as many of these questions as is philosophically feasible.

Definition 1.24. \(n\th\) power.

Let \(x\) be an element of a multiplicative group \(G\text{.}\) For \(n\in\Z\text{,}\) the \(n\th\) power \(x^n\) of \(x\) is defined recursively as follows:
  • \(x^0=e\text{,}\) \(x^1=x\text{,}\) and \(x\inv=x\inv\text{;}\)
     1 
    Convenient notation, isn’t it? Almost like we planned it.
  • \(x^{n+1}=x^nx\) when \(n>0\text{;}\) and
  • \(x^n=(x^{-n})\inv\) when \(n\neq0\text{.}\)
If \(G\) was an additive group, then we would write \(nx\) intead of \(x^n\text{.}\) This would be called a multiple of \(x\) instead of a power.

Proof.

  1. We proceed via cases: First, let \(n > 0\) and \(m > 0\text{.}\) We induct on \(m\text{.}\) For the base case, let \(m=1\text{.}\) Thus we have \(x^nx^1=x^{n+1}\) by definition. Additionally, using the Exercise 1.23 we see
    \begin{equation*} x^1x^n = x(\underbrace{x\cdots x}{n}) = (\underbrace{x\cdots x}{n})x = x^nx^1, \end{equation*}
    and thus \(x^nx^1=x^1x^n\text{.}\) Hence \(x^nx^1=x^{n+1}=x^1x^n\text{,}\) satisfying the base case.
    Now we inductively assume \(x^nx^m=x^{n+m}=x^mx^n\) and consider \(x^nx^{m+1}\) Using Definition 1.24 we see \(x^{m+1}=x^mx\text{.}\) Substituting, we have \(x^nx^mx^1\text{.}\) By the inductive hypothesis, we have \(x^nx^mx^1 = x^{n+m}x = x^{n+m+1}\) and \(x^nx^mx^1 = x^mx^nx\text{.}\) Using the Exercise 1.23 again, we have \(x^{m+1}x^n\text{.}\) Thus \(x^nx^{m+1}=x^{n+m+1}=x^{m+1}x^n\text{,}\) completing the induction. Hence, when \(n > 0\) and \(m > 0\) we have \(x^nx^m=x^{n+m}=x^mx^n\text{.}\)
    Now suppose \(n > 0\) and \(m < 0\text{.}\) Thus \(x^m = (x^{-m})\inv\text{,}\) where \(-m > 0\text{.}\) So \(x^nx^{-m}=x^{n+(-m)}\)
    \(x^nx^m = (x^{-n})\inv(x^{-m})\inv = (x^{-m}x^{-n})\inv\)
    Multiplying on both sides by \(x^{-m}\) we deduce that \(x^n =x^{-m}x^{m+n}\) and \(x^m = x^{m + n} x^{-n}\text{.}\) Finally inversion of the equation \(x^mx^n =x^{m + n}\) yields \(x^{-n}x^{-m} = x^{-m + (-n)}\text{.}\) Hence the law is established in all cases.
  2. If \(n > 0\text{,}\) it follows from (1) that \((x^m)^n =x^{mn}\text{.}\) Now assume that \(n < 0\text{;}\) then \((x^m)^n =((x^m)^{-n})\inv =(x^{-mn})\inv =x^{mn}\) since \(x^{-mn}x^{mn}=e\text{.}\)

Exercise 1.26. Abelian Exponents.

Let \(x,y\in G\text{.}\) The equation \((xy)^n=x^ny^n\) holds for all \(n\in\Z\) if and only if \(G\) is abelian.

Subsection Order

Definition 1.27. Order.

  1. In a group \(G\text{,}\) the order of an element \(x\) is the least positive integer \(n\) such that \(x^n = e\text{.}\) If no such \(n\) exists, we say \(x\) has infinite order. We write \(|x|\) for the order of \(x\text{.}\)
  2. The order of a group \(G\) is the Cardinality of the set \(G\text{,}\) denoted \(|G|\text{.}\)

Convention 1.28.

Some authors use the notation \(\#G\) or \(o(G)\) to refer to the order of a group. We won’t name names, but they know who they are.

Example 1.29. Order.

  1. \(|e|=1\) in every group \(G\text{.}\)
  2. \(|1|=n\) in \(\Z/n\)
  3. In the additive groups \(\Z\text{,}\) \(\Q\text{,}\) \(\R\text{,}\) and \(\C\) every nonzero (i.e., nonidentity) element has infinite order.
     2 
    Groups with this property are known as torsion-free, a concept explored in more advanced contexts.
  4. In the additve group \(\Z/6\text{,}\) the element \(2\) has order \(3\text{,}\) as \(2+2=4\) and \(2+2+2=0\text{.}\)

Exercise 1.30. Groups of Order \(4\).

Every group of order \(4\) is abelian.
Solution.
Let \(G\) be a group with \(4\) elements, and suppose there are elements \(a,b\in G\) such that \(ab\neq ba\) for some elements \(a, b\in G\text{.}\) Since \(ab \neq ba\text{,}\) we must have \(a \neq b\text{,}\) \(a \neq e\text{,}\) and \(b\neq e\text{.}\) Since \(G\) has only \(4\) elements and \(ab \neq ba\text{,}\) either \(ab \in \{e, a, b\}\) or \(ba \in\{e, a, b\}\text{.}\) Without loss, say the former occurs. But \(ab = e\) implies \(b=a\inv\) and we know \(a\inv\) commutes with \(a\text{,}\) and hence this is not possible. If \(ab = a\text{,}\) then \(b = e\) and if \(ab = b\) then \(a = 1\text{,}\) both of which are impossible. Since \(a, b\) were arbitrary, \(G\) must be abelian.

Proof.

  1. Let \(G\) be a group and consider an element \(a\) in \(G\) with order \(1\text{.}\) By definition, the order of an element is the smallest positive integer \(n\) such that \(a^n = e\text{,}\) where \(e\) is the identity element of \(G\text{.}\) Since the order of \(a\) is \(1\text{,}\) we have \(a^1 = a = e\text{.}\) This implies that \(a\) is equal to the identity element \(e\text{.}\)

Exercise 1.32. \(b\) there or \(b^2=e\).

If \(G\) is a group such every non-identity element has order \(2\text{,}\) then \(G\) is abelian.

Proof.

Let \(G\) be a group and \(x \in G\) any element.
  1. Let \(|x| = n\text{,}\) and suppose by way of contradiction \(x^j=x^k\) for some \(j,k\) such that \(0\leq j < k\leq n \text{.}\) Mutliplying \(x^j=x^k\) on both sides by \(x^{-j}\) yields \(e=x^{k-j}\text{.}\) As \(k-j < n\text{,}\) this contradicts the assumption \(|x| = n\text{.}\)
  2. If \(x^i\neq x^j\) for integers \(i\) and \(j\) with \(i \neq j\text{,}\) then \(x^{j-i} = 0 = x^{i-j}\text{.}\) Since \(i \neq j\text{,}\) one of \(j - i\) and \(i - j\) must be strictly positive, contrary to the assumption ath \(|x| = \infty\text{.}\)
  3. If \(|x| < \infty\text{,}\) then part (1) shows that \(G\) has at least \(|x|\) elements; if \(|x| = \infty\text{,}\) then part (2) shows that \(G\) has infinitely many elements.
  4. Let \(n = |x|\text{.}\) We have \(m = nq + r\) for some \(0 \leq r < n\) by the Division Algorithm We have \(x^r = (x^n)^qx^r = x^m = e\) and so, by the definition of “order”, it must be that \(r = 0\)

Exercise 1.34. Order Hors d’oeuvre.

Let \(G\) be a group.
  1. Let \(g \in G\) be an element of finite order. Show that \(g^m\) has finite order for any integer \(m \geq 0\text{,}\) and in fact
    \begin{equation*} |g^m| = \frac{\lcm(m,|g|)}{m} = \frac{|g|}{\gcd(m, |g|)}. \end{equation*}
  2. Prove that for all \(g, h\) in \(G\text{,}\) \(|gh| = |hg|\) holds.
The central theorem of group order, (perhaps the central theorem of finite group theory), Lagrange’s Theorem, requires more machinery than we currently posess. We will get there evantually, but first, it would be useful to examine some more concrete groups and their interactions with what we have constructed thus far.
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