A subgroup of a cyclic normal subgroup of a Group is Normal Is it true that Subgroup of a Cyclic Normal subgroup of a Group is again Normal ? If so any hints for the proof?
 A: Sure. It follows from a more general fact: a characteristic subgroup of a normal subgroup of $G$ is also a normal subgroup of $G$. It's even easier to think about the question in these general terms.
A: If $H<N\lhd G$ and $N$ is cyclic, then $H$ is a characteristic subgroup of $N$, i.e. it is left invariant by all automorphism of $N$, thus including those outer automorphisms of $N$ that are actually inner automorhisms of $G$, i.e. $H$ is conjugation invariant.
A: Yes. This is one of the several cases when normality is transitive. If
$$N\triangleleft C\triangleleft G\,\,\,,\,\,C=\text{cyclic, then}\;\; C=\langle c\rangle\Longrightarrow N=\langle c^k\rangle\Longrightarrow$$
$$\Longrightarrow \,\,\forall g\in G\,\,,\,\,x^{-1}c^{rk}x=(x^{-1}c^kx)^r\in N\,....$$
A: Suppose $G$ is the group and $N = \langle x \rangle $ is a cyclic normal subgroup of $G$.
Note that for different values of $k$, $H_k= \langle x^{gcd(k,o(x)} \rangle$ are subgroups of $N$.
Step 1:
$N$ is normal $\Rightarrow$ $\forall g \in G:\ gxg^{-1} \in N$ $\Rightarrow$ For all $g \in G$, there exist $\alpha$ (namely $\alpha_g$) such that $gxg^{-1}=x^{\alpha_g}$.
Step 2:
Now, I claim $\forall g \in G, \forall y\in N:\ gy=y^{\alpha_g}g$
Proof:
Since $y \in N$, there exists $l$ such that $y=x^l$.
$\forall g \in G:\ gy=gx^l=gxx^{l-1}=x^{\alpha_g}gx^{l-1}=x^{\alpha_g}gxx^{l-2}=x^{2\alpha_g}gx^{l-2}= \cdots = x^{l\alpha_g}g=y^{\alpha_g}g$.
Step 3:
$\forall g \in G, \forall y\in H_k:\ gy=y^{\alpha_g}g$ $\Rightarrow$ $\forall g \in G, \forall y\in H_k:\ gyg^{-1}=y^{\alpha_g}$
Since $y\in H_k$, $y^{\alpha_g}\in H_k$ $\Rightarrow$ $\forall g \in G: gH_kg^{-1} \subset H_k$
Consequently $H_k$ are normal in $G$.
