Proving $g( (x^{k})^{n})g^{-1}\in K$

Question

If we are given groups $K$, $H$ and $G$, such that $K \leq H\trianglelefteq G $ and $H$ is cyclic. Prove that $K\trianglelefteq G $.

My work:
Let $H=\langle x \rangle, x \in G$ and $K=\langle x^{k} \rangle$ for some int. k.

if we pick $(x^{k})^{n} \in K$ for some n how do we prove that:

$g \in G$,
$g( (x^{k})^{n})g^{-1} \in K$, since $x^{k} \in H$, can we say $g(gx^{k}g^{-1})^{n}g^{-1}$ and move forward from here?

Answer 1

Hint:

$$gx^{nk}g^{-1}=\left(gx^kg^{-1}\right)^n$$

Also, observe that a general element in $\;\langle x^k\rangle\;$ is of the form $\;(x^k)^n=x^{kn}\neq x^{k^n}\;$

Answer 2

Lemma: In cyclic group for every possible order, there is a uniqe subgroup.

$$K^g\leq H^g=H$$ and since $K^g$ is also a group, we must have $K^g=K$ ....

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In general, if $K$ is characteristic in $H$ and $H$ is normal in $G$ then $K$ is normal in $G$ and every subgroup of a cyclic group is characteristic.

Answer 3

I assume your groups are finite, but the argument can be adapted to work in general.

Have you seen the result that a cyclic group of order $n$ has exactly one subgroup of order $m$ for each divisor $m$ of $n$?

Then if $K$ has order $m$, and $g \in G$, $g^{-1} K g$ is another subgroup of $H$ of order $m$, and thus...