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 $$a=a^1 \Rightarrow a^2=e$$.