Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 subgoup 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.

share|cite|improve this question
Use the fact that if $H$ is cyclic and $k|\sharp H$, then $H$ has a unique subgroup of order $k$. – user10676 Feb 5 '12 at 20:18
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 $G$. – Tobias Kildetoft Feb 5 '12 at 20:19
Moreover, for a cyclic group $H$ of order $n$ and every $m \mid n$, there is a unique subgroup of $H$ of order $m$. – Dylan Moreland Feb 5 '12 at 20:22
I understand what you mean but $G$ is not necessarily a finite group. – spohreis Feb 5 '12 at 21:27
That's a good point, but there aren't so many infinite cyclic groups! – Dylan Moreland Feb 6 '12 at 2:21
up vote 4 down vote accepted

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 $n \mathbb{Z}$.)

share|cite|improve this answer
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. – Pete L. Clark Feb 7 '12 at 21:59

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.

share|cite|improve this answer

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

share|cite|improve this answer

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.

share|cite|improve this answer
You might want to work with the element $gkg^{-1}$ directly instead of the coset $gkg^{-1}K$. That extra $K$ there doesn't really help. You need to use the fact that $ghg^{-1}=...?$ anyway. Remember that the power of a coset of $K$ does not make sense, unless the coset is inside such a group $N$ that we know has $K$ as a normal subgroup :-) – Jyrki Lahtonen Feb 7 '12 at 22:04
@JyrkiLahtonen: Thanks for that. But $gkg^{-1}\in H$ and $H/K$ is a finite abelian group, of order $d$. So $gkg^{-1}K$ has order $d$. Am I correct? – spohreis Feb 7 '12 at 22:56
Yes. For full marks you need to state that $gkg^{-1}\in H$ (which is, of course, given). But you can then as well argue that because $gkg^{-1}$ has order that is a factor of $n/d$, it has to be in $K$, because $K$ consists precisely of those elements of $H$. I'm just cutting the corner of passing over to $H/K$. +1 for you :-) – Jyrki Lahtonen Feb 8 '12 at 8:09
Arggh. For me $n=|H|$. I also forgot to add my upvote until now. Sorry :-) – Jyrki Lahtonen Feb 8 '12 at 14:56

H is a normal subgroup of G. let K is subgroup of H. since H is normal implies K is normal. since K is normal in H, H in G this implies that K is normal in G.tanveer ul haq from BKUC.

share|cite|improve this answer
It's possible to have $K$ normal in $H$ and $H$ normal in $G$ without having $K$ normal in $G$. Search for "transitivity normal subgroup". – Najib Idrissi Aug 15 '14 at 10:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.