Solutions To Mathematics Textbooks/Topics In Algebra (2nd) 9788126510184/Group Theory/Page 35

2

As $a$ and $b$ are abelian, $ab=ba$.

$(a.b)^{2}=(a.b).(a.b)=a.(b.a).b=a.(a.b).b=a^{2}b^{2}$

$(a.b)^{3}=(a.b).(a.b).(a.b)=a.(b.a).(b.a).b=a.(a.b).(a.b).b=a.a.(b.a).b.b=a.a.(a.b).b.b=a^{3}b^{3}$

Similarly progressing we obtain $(a.b)^{n}=a^{n}b^{n}$

3

${\begin{array}{lcll}(a.b)^{2}&=&a^{2}.b^{2}&\\(a.b).(a.b)&=&(a.a).(b.b)&\\a.(b.(a.b))&=&a.(a.(b.b))&{\text{since . is associative}}\\b.(a.b)&=&a.(b.b)&{\text{by left cancellation law}}\\(b.a).b&=&(a.b).b&{\text{since . is associative}}\\b.a&=&a.b&{\text{by right cancellation law}}\\\end{array}}$

4

${\begin{array}{lcl}(a.b)^{i}=a^{i}b^{i}\ (1)\\(a.b)^{i+1}=a^{i+1}b^{i+1}\ (2)\\(a.b)^{i+2}=a^{i+2}b^{i+2}\ (3)\\\\Using(2),\\\\(a.b)^{i+1}=a^{i+1}b^{i+1}\\(a.b)^{i}.(a.b)=a^{i+1}.(b^{i}.b)\\(a^{i}b^{i}).(a.b)=(a^{i+1}.b^{i}).b,from(1)\\(a^{i}.b^{i}).a=(a^{i}.a).b^{i}\\b^{i}.a=a.b^{i}\ (4)\\\\Similarly\ using(3),we\ obtain,\\\\b^{i+1}.a=a.b^{i+1}\\b.(b^{i}.a)=a.b^{i+1}\\b.(a.b^{i})=a.b^{i+1},using\ (4)\\(b.a).b^{i}=(a.b).b^{i}\\b.a=a.b\end{array}}$

8

${\begin{array}{lcl}Let\ g\in G,\ where\ g\ is\ a\ group\\So,\ G=\{g,g^{2},g^{3},...\}\\As\ G\ is\ a\ finite\ group,\ for\ some\ i,j,\\g^{i}=g^{j}\\g^{i-j}=e\\where\ i-j\ is\ any\ integer\ N\\Hence\ a^{N}=e\ for\ some\ N\end{array}}$

10

${\begin{array}{lcl}Let\ a,\ b\in \ G,\ a=a^{-1},\ b=b^{-1}\\ab=(ab)^{-1}=b^{-1}a^{-1}=ba\end{array}}$

11

Let $G$ be a group of $2n$ elements, where $n$ is a natural number. Except $e$ there is an odd number of elements in $G$.

Those elements who's order is greater than 2 can be paired with their inverse.

Since we have started with an of number of elements, we must end with at least one unpaired element $a$ that satisfies $a=a^{1}\Rightarrow a^{2}=e$.