Roots of Elements in Groups

Powers and Roots of Elements in Groups

Definition: Let $(G, *)$ be a group. For any positive integer $n$ we define the $n^{\mathrm{th}}$ Power of $x \in G$ to be $x^n = \underbrace{x * x * ... * x}_{n \: \mathrm{many} \: \mathrm{factors}}$. 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.

Theorem 1: If $(G, *)$ is a group and $m, n \in \mathbb{N}$ then $a^m * a^n = a^{m + n}$.
Theorem 2: If $(G, *)$ is a group and $m, n \in \mathbb{N}$ then $(a^m)^n = a^{mn}$.
Theorem 3: If $(G, *)$ is a group such that $a * b = b * a$ for all $a, b \in G$ then for each positive integer $n$, $(a * b)^n = a^n * b^n$.

Groups $(G, *)$ for which $a * b = b * a$ for all $a, b \in G$ are called Abelian Groups (we'll look at these later).

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

In Theorem 4 above, note that $(-a)^2 = (-a) * (-a) = (-e * -e) * (a * a) = e * a^2 = a^2$. Hence $x = a$ is also a square root of $a$.

Theorem 5: If $(G, *)$ is a group and $e$ is the identity with respect to $*$ then 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 * a^2 = a * e \\ \quad a^3 = a \end{align}
  • Therefore $x = a$, i.e., the element $a$ itself with respect to $*$ is a cube root of $a$. $\blacksquare$
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