If $H$ is a cyclic subgroup of $G$ and $H$ is normal in $G$, then every subgoup of $H$ is normal in $G$.

Exercise 11, page 45 from Hungerford's book Algebra.

If $H$ is a cyclic subgroup of $G$ and $H$ is normal in $G$, then every subgroup of $H$ is normal in $G$.

I am trying to show that $a^{-1}Ka\subset K$, but I got stuck. What I am supposed to do now?

Thanks for your kindly help.

Solution 1:

Suppose $H = \langle h \rangle$ is normal in $G$ and that $K$ is a subgroup of $H$. Any subgroup of a cyclic group is cyclic, so $K = \langle h^d \rangle$ for some integer $d$.

Let $g \in G$. Since $H$ is normal, $g^{-1}hg = h^i$ for some integer $i$. Then for any integer $k$ you get $g^{-1}(h^d)^kg = (g^{-1}hg)^{dk} = (h^i)^{dk} = (h^d)^{ik}$. This shows that for any $k \in K$, the element $g^{-1}kg$ is in $K$. Therefore $K$ is normal.

Solution 2:

Here is a somewhat more general fact which seems useful enough to keep in mind:

If $G$ is a group, $H$ is a normal subgroup of $G$ and $K$ is a characteristic subgroup of $H$, then $K$ is a normal subgroup of $G$.

The proof is almost immediate if you know the definitions: for any $x \in G$, since $H$ is normal in $G$, conjugation by $H$ induces an automorphism $\varphi_x$ of $H$, but not necessarily an "inner" automorphism: i.e., if $x \notin H$, $\varphi_x$ need not be conjugation by any element of $H$. Thus we have assumed that $K$ is just not normal but characteristic as a subgroup of $H$, i.e., stable under all automorphisms of $H$. Done.

For much more detail, see e.g. here.

As others have pointed out, we also need to see that any subgroup of a cyclic group $H$ is characteristic. Well, any subgroup which is the unique subgroup of its order is characteristic -- this takes care of the case in which $H$ is finite. And any subgroup which is the unique subgroup of its index is characteristic -- this takes care of the case in which $H$ is infinite. (Alternately, if $H \cong (\mathbb{Z},+)$, the only nontrivial automorphism is multiplication by $-1$, which evidently stabilizes all the subgroups $n \mathbb{Z}$.)

Solution 3:

since $H$ is normal in $G$ you get $a^{-1}Ka \subset H$, for all $a\in G$. Now use the fact that $H$ is cyclic (there is only one subgroup of $H$ such that $\dots$)

Solution 4:

I'll give a try. If $H=\langle h \rangle$ , then $H$ is an abelian group and $K$ is a normal subgroup of $H$. Let $d$ the lowest positive integer such that $h^{d}\in K$. Then $K=\langle h^{d} \rangle$ and we have $H/K=\{K,hK,\cdots,h^{d-1}K\}$. Let $g\in G$ and $k=h^{dn}\in K$. Then $gkg^{-1}K=(ghg^{-1})^{dn}K=K$. Thus $gKg^{-1}\subset K$, for all $g\in G$.

I hope that it is correct.