Question

Prove that a cyclic subgroup <a> of a group G is normal if and only if for each $g\in G,ga=a^{k}g$ for some $k\in Z$.

Short answer

It has been proved thata cyclic subgroup <a> of group G is normal if and only if for each $g\in G,ga=a^{k}g$ for some $k\in Z$ .

Step-by-step solution

Suppose that the cyclic subgroup <a>  is normal

It is given that <a> is a cyclic subgroup of G .

Suppose that is <a> normal.

Then for any $g\in G$, $ga{g}^{-1}\in g⟨a⟩g^{-1}\subseteq ⟨a⟩$.

This implies $ga{g}^{-1}=a^k$ for some $k\in Z$.

Multiplying by g on the right, we get $ga=a^{k}g$.

Hence, $ga=a^{k}g$.

Suppose that the cyclic subgroup <a> has the property.

Suppose thathas the property: for each $g\in G,ga=a^{k}g$ for some $k\in Z$.

Now, $ga=a^{k}g$ implies $ga{g}^{-1}=a^k$.

For any $n\in N$,

$\begin{array}{c}g{a}^n{g}^{-1}=(ga{g}^{-1}{)}^n\\ ={\left(a^k\right)}^n\\ =a^{kn}\\ \in ⟨a⟩\end{array}$

Thus, $g⟨a⟩g^{-1}\subseteq ⟨a⟩$for every g .

Therefore, <a>is normal.

Conclusion

Hence, a cyclic subgroup <a> of groupG is normal if and only if for each $g\in G,ga=a^{k}g$ for some $k\in Z$ .