24
$\begingroup$

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.

$\endgroup$
6
  • 2
    $\begingroup$ Use the fact that if $H$ is cyclic and $k|\sharp H$, then $H$ has a unique subgroup of order $k$. $\endgroup$
    – user10676
    Feb 5, 2012 at 20:18
  • 5
    $\begingroup$ You might want to do it by proving the following two useful statements. 1: If $H$ is characteristic in $N$ and $N$ is normal in $G$ then $H$ is normal in $G$. 2: If $H$ is a subgroup of $G$ and $G$ is cyclic, then $H$ is characteristic in $G$. $\endgroup$ Feb 5, 2012 at 20:19
  • $\begingroup$ Moreover, for a cyclic group $H$ of order $n$ and every $m \mid n$, there is a unique subgroup of $H$ of order $m$. $\endgroup$ Feb 5, 2012 at 20:22
  • 1
    $\begingroup$ I understand what you mean but $G$ is not necessarily a finite group. $\endgroup$
    – user23505
    Feb 5, 2012 at 21:27
  • $\begingroup$ That's a good point, but there aren't so many infinite cyclic groups! $\endgroup$ Feb 6, 2012 at 2:21

6 Answers 6

58
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Question, how'd you go from $g^{-1}(h^d)^kg=(g^{-1}hg)^{dk}$? $\endgroup$
    – anonymous
    Dec 25, 2022 at 20:44
  • $\begingroup$ @MyMathYourMath: $(h^d)^k = h^{dk}$. Then use the fact that $g^{-1} h^n g = (g^{-1}hg)^n$ for any integer $n$. $\endgroup$
    – spin
    Dec 27, 2022 at 3:27
  • $\begingroup$ and that comes from $(gH)^n=g^nH$ right? or the same idea? And im referring to the latter fact you stated in your comment. $\endgroup$
    – anonymous
    Dec 29, 2022 at 1:21
  • $\begingroup$ @MyMathYourMath: No it is not related to $(gH)^n = g^nH$. The identity $g^{-1} h^n g = (g^{-1}hg)^n$ holds in any group $G$. You can either prove it by direct computation, or use the fact that $h \mapsto g^{-1}hg$ is a homomorphism. $\endgroup$
    – spin
    Dec 29, 2022 at 8:13
13
$\begingroup$

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 $n \mathbb{Z}$.)

$\endgroup$
1
  • 1
    $\begingroup$ I see now that this answer was anticipated by @Tobias's comment. Oh, well -- I still think it is worth leaving as an actual answer. $\endgroup$ Feb 7, 2012 at 21:59
5
$\begingroup$

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 $\dots$)

$\endgroup$
2
$\begingroup$

Let $H=\langle h\rangle$. $H\unlhd G\implies\forall g\in G, ghg^{-1}=h^m$ for some $m$. All elements of $K$ looks like $h^n$. $\forall g\in G, gh^ng^{-1}=(ghg^{-1})^n=(h^m)^n=h^{mn}=(h^n)^m\in K$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
0
$\begingroup$

Another proof, suppose that $H=<h>$,$m,n\in \mathbb{N}$, $q,p\in \mathbb{Z}$ such that $m=pn+q$, $q<n$.

Generally, $x^{-1}h^{m}x=h^{n} \Leftrightarrow x^{-1}h^{pn+q}x=h^{n} \Rightarrow h^{qo(h^{n})}=e \Rightarrow q=0$.

$\endgroup$

You must log in to answer this question.