Powers and Roots of Elements in Groups

Powers and Roots of Elements in Groups

Definition: Let $(G, \cdot)$ be a group. For any positive integer $n$ we define the $n^{\mathrm{th}}$ Power of $x \in G$ to be $x^n = \underbrace{x \cdot x \cdot ... \cdot x}_{n \: \mathrm{many} \: \mathrm{factors}}$. Furthermore, if $a \in G$ is such that $a = x^n$ then $x$ is said to be an $n^{\mathrm{th}}$ Root of $a$.

If $a = x^2$ then $x$ is said to be a Square Root of $a$, and if $a = x^3$ then $x$ is said to be a Cube Root of $a$.

For example, consider the set of positive real numbers $\mathbb{R}^+$ and let $\cdot$ be the operation of standard multiplication. Then $(\mathbb{R}^+, \cdot)$ is a group as $\mathbb{R}^+$ is closed under standard multiplication, standard multiplication is associative, the identity element is $e = 1 \in \mathbb{R}^+$ and for each element in $x \in \mathbb{R}^+$ we have that the inverse of $x$ with respect to $\cdot$ is $\frac{1}{x} \in \mathbb{R}^+$.

Consider any number $x \in \mathbb{R}^+$. Then for each positive integer $n$, we have that:

(1)
\begin{align} \quad x^n = \underbrace{x \cdot x \cdot ... \cdot x}_{n \: \mathrm{many} \: \mathrm{factors}} \end{align}

For the element $4 \in \mathbb{R}^+$ we have that $2 \in \mathbb{R}^+$ is a square root of $4$ since $4 = 2^2$. For the element $16 \in \mathbb{R}^+$ we have that $4 \in \mathbb{R}^+$ is a square root of $16$ since $16 = 4^2$ and furthermore, $2$ is a fourth root of $16$ since $16 = 2^4$.

In fact, for this group, there exists exactly one $n^{\mathrm{th}}$ root for each element $x \in \mathbb{R}+$ and for each positive integer $n$.

Of course, if we instead consider the set $\mathbb{R} \setminus \{ 0 \}$ of all real numbers excluding $0$, then $(\mathbb{R} \setminus \{ 0 \}, \cdot )$ is also a group (namely, a group extension of $\mathbb{R}+$). We see that $4 \in \mathbb{R} \setminus \{ 0 \}$ now has two square roots, namely $x = 2$ since $4 = 2^2$ and $x = -2$ since $4 = (-2)^2$.

Therefore it's important to recognize that the number of $n^{\mathrm{th}}$ roots an element might have depends on the set itself and the operation defined on that set.

Proposition 1: Let $(G, \cdot)$ be a group and $m, n \in \mathbb{N}$. Then:
a) $a^m \cdot a^n = a^{m + n}$.
b) $(a^m)^n = a^{mn}$.
c) If $G$ is an abelian group then $(a \cdot b)^n = a^n \cdot b^n$.
  • Proof of c) We will carry this out by induction. Let $(G, \cdot)$ be a group such that $a \cdot b = b \cdot a$ for all $a, b \in G$. For each positive integer $n$ let $S(n)$ be the statement that $(a \cdot b)^n = a^n \cdot b^n$.
  • For the base step, consider $n = 1$. Then we have that $(a \cdot b)^1 = a \cdot b$ so $S(1)$ is true.
  • Assume that for some positive integer $k > 1$ that $S(k)$ is true, that is $(a \cdot b)^k = a^k \cdot b^k$. We want to therefore show that $S(k+1)$ is true, that is $(a \cdot b)^{k+1} = a^{k+1} \cdot b^{k+1}$. Since $a \cdot b = b \cdot a$ we have that $\cdot$ is a commutative operation and so:
(2)
\begin{align} \quad a^{k+1} \cdot b^{k+1} = (a^{k} \cdot a) \cdot (b^{k} \cdot b) = (a \cdot b) \cdot (a^k \cdot b^k) \overset{I.H.} = (a \cdot b) \cdot (a \cdot b)^k = (a \cdot b)^{k+1} \end{align}
  • Therefore $S(k+1)$ is true. By the principle of mathematical induction we therefore have that for each positive integer $n$ that if $a \cdot b = b \cdot a$ then for all $a, b \in G$:
(3)
\begin{align} \quad (a \cdot b)^n = a^n \cdot b^n \quad \blacksquare \end{align}
Proposition 2: If $(G, \cdot)$ is a group and $e$ is the identity with respect to $\cdot$ then if $a \in G$ is such that $a^3 = e$ then there exists an $x \in G$ such that $a = x^2$.
  • Proof: Suppose that $a^3 = e$. Then:
(4)
\begin{align} \quad (-a)^2 \cdot a^3 &= (-a)^2 \cdot e \\ \quad [(-a)^2 \cdot a^2] \cdot a &= (-a)^2 \\ \quad e \cdot a &= (-a)^2 \\ \quad a &= (-a)^2 \end{align}
  • Therefore $x = -a$, i.e., the inverse of $a$ with respect to $\cdot$ is a square root of $a$. $\blacksquare$

In Propositio 2 above, note that $(-a)^2 = (-a) \cdot (-a) = (-e \cdot -e) \cdot (a \cdot a) = e \cdot a^2 = a^2$. Hence $x = a$ is also a square root of $a$.

Proposition 3: If $(G, \cdot)$ is a group and $e$ is the identity with respect to $\cdot$ and if $a \in G$ is such that $a^2 = e$ then there exists an $x \in G$ such that $a = x^3$.
  • Proof: Suppose that $a^2 = e$. Then:
(5)
\begin{align} \quad a^2 = e \\ \quad a \cdot a^2 = a \cdot e \\ \quad a^3 = a \end{align}
  • Therefore $x = a$, i.e., the element $a$ itself with respect to $\cdot$ is a cube root of $a$. $\blacksquare$
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