Can subgroups of subgroups be normal?

Question

Prove that if $N$ is a normal subgroup of $G$ and $N$ is cyclic, then if $K<N$, $K$ is a normal subgroup of $G$.

I understand that I have to show that $gKg^{-1}=K$ $\forall g \in G$. I also know that the fact that $N$ is cyclic plays a part in this because if $N$ is not cyclic, then it doesn't work. Any tips?

Answer 1

Let $N=\langle x\rangle$. Let $x^k\in K$ be arbitrary for some $k\in\mathbb{Z}$. Let $g\in G$ be arbitrary, so you want to show $gx^kg^{-1}\in K$ to show $K$ is normal in $G$. Since $x\in N$, you know $gxg^{-1}\in N$, so $gxg^{-1}=x^j$ for some $j$. Use this to show $gx^kg^{-1}$ is a power of $x^k$, to see $gx^kg^{-1}\in K$.

Mouse over below if you get stuck.

Observe that $gx^kg^{-1}=(gxg^{-1})^k=(x^j)^k=(x^k)^j\in K$

Answer 2

For $N$ finite:

A characteristic subgroup $K$ of $N$, where $N$ is a normal subgroup of $G$ is always normal in $G$. Cyclic subgroups have exactly one subgroup of each order that divides the group's order.

Since isomorphisms send subgroups to subgroups with the same order we conclude all subgroups of a finite cyclic subgroup are characteristic.

Since $K$ is characteristic in $N$ and $N$ is normal in $G$ we conclude $K$ is normal in $G$

Answer 3

A characteristic subgroup $K$ of a normal subgroup $N$ of a group $G$ is normal in $G$.

A characteristic subgroup $K$ is one fixed under any automorphism of the group $N$ in which it is characteristic.

Conjugation by elements of $G$ fixes $N$ so acts as an automorphism of $N$, and therefore fixes $C$. Since $C$ is fixed under conjugation in $G$ it is normal in $G$.

So if $N$ has a single subgroup of a given order, that must be fixed under an automorphism - often Sylow subgroups can be shown to be characteristic by this route. Subgroups like the Centre of $N$ and the Commutator Subgroup are characteristic too.

Note: If $K$ is not a characteristic subgroup of $N$, one can build a group $G$ in which $N$ is normal and in which conjugation induces an automorphism of $N$ which does not fix $K$.

Answer 4

suppose that N=(a) . for every x in K we have that x=na for some intiger n. now for every g in G we have (-g)xg=n(-gag). we know that N is normal in G then we have -gag is in N so we can write -gag=ma for some intiger m. Finaly we have : (-g)xg=n(ma)=m(na)=mx belongs to K. Know K is normal in G. NOTE: In this note I suppose that the action in G is additive.