Question

Prove theorem 7.7

Short answer

If G is a group and $a\in G$ then for all $m,n$ in$\mathrm{\mathbb{Z}}$ , $a^m{a}^n=a^{m+n}$ and ${\left(a^m\right)}^n=a^{mn}$.

Step-by-step solution

Theorem 7.7

Let G be a group and let $a\in G$then for all $m,n$ in $\mathrm{\mathbb{Z}}$, $a^m{a}^n=a^{m+n}$and ${\left(a^m\right)}^n=a^{mn}$ .

Proof 7.7

Let G be a group and $a\in G$ then for all $m,n$in $\mathrm{\mathbb{Z}}$,

$\begin{array}{c}{\mathrm{a}}^{\mathrm{m}}{\mathrm{a}}^{\mathrm{n}}=\left({\prod }_{\mathrm{i}=1}^{\mathrm{m}}\mathrm{a}\right)\left({\prod }_{\mathrm{j}=1}^{\mathrm{n}}\mathrm{a}\right)\\ =\left({\prod }_{\mathrm{i}=1}^{\mathrm{m}+\mathrm{n}}\mathrm{a}\right)\\ ={\mathrm{a}}^{\mathrm{m}+\mathrm{n}}\end{array}$

Thus, role="math" localid="1654256575895" $a^m{a}^n=a^{m+n}$.

Similarly, let G be a group and role="math" localid="1654256775039" $a\in G$ then for all role="math" localid="1654256767089" $m.n$ in role="math" localid="1654256772005" $\mathrm{\mathbb{Z}}$,

$\begin{array}{c}{\left({\mathrm{a}}^{\mathrm{m}}\right)}^{\mathrm{n}}={\prod }_{\mathrm{i}=1}^{\mathrm{n}}\left({\prod }_{\mathrm{j}=1}^{\mathrm{m}}\mathrm{a}\right)\\ =\left({\prod }_{\mathrm{i}=1}^{\mathrm{mn}}\mathrm{a}\right)\\ ={\mathrm{a}}^{\mathrm{mn}}\end{array}$

Thus,${\left(a^m\right)}^n=a^{mn}$.

Therefore, if G is a group and $a\in G$ then for all $m,n$ in $\mathrm{\mathbb{Z}}$, $a^m{a}^n=a^{m+n}$ and ${\left({\mathrm{a}}^{\mathrm{m}}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{mn}}$.