Question

Let $\mathit{G}$ be a group all of whose subgroups are normal. If $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{G}$, prove that there is an integer$k$ such that $\mathit{a}\mathit{b}\mathbf{=}\mathit{b}{\mathit{a}}^{\mathbf{k}}$.

Short answer

It has been proved that if $a,b\in G$,then there is an integer$k$ such that $ab=b{a}^k$.

Step-by-step solution

Given that

G is a group.

All the subgroups of G are normal.

$a,b\in G$.

Prove that  ab=bak

Consider the subgroup $A=⟨a⟩$

$b^{-1}Ab=A$

Since A is a normal subgroup, so for $b\in G$

$b^{-1}Ab=A$.

This implies that there exist $k\in \mathbb{N}$such that

$b^{-1}ab=a^k$

That is $ab=b{a}^k$

Conclusion

Thus, it can be concluded that if $a,b\in G$,then there is an integer$k$ such that $ab=b{a}^k$.