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

Let $T$ be a cyclic subgroup of a group $G$ such that $T$ is normal in $G$. Let $S$ be a subgroup of $T$. What can we say about whether or not $S$ is normal in $G$?

My work:

Let $T \colon = \langle a \rangle$ for some $a$ in $G$. Since $S$ is a subgroup of $T$, we can write $S$ as $S = \langle a^n \rangle$, where $n$ is the smallest positive integeer such that $a^n$ is in $S$.

Now since $T$ is normal in $G$, we can say that $gtg^{-1} \in T$ for all $t \in T$ and for all $g \in G$.

Now to determine whether or not $S$ is normal in $G$, we let $s$ be an element of $S$ and $g$ an element of $G$. Then $s = a^{kn}$ for some integer $k$.

Since $S \subset T$, we must have $gsg^{-1} \in T$. Or, $$ga^{kn}g^{-1} = a^m$$ for some integer $m$, since $T$ is generated by $a$.

Now in order for $S$ to be normal in $G$, we must have $m$ to be a multiple of $n$. So we take $m = qn + r$, where $q$, $r$ are integers such that $0 \leq r < n$. Therefore, we have $$ga^{kn}g^{-1} = a^{qn+r} = a^{qn} a^r, $$ whence $$ a^r = a^{-qn}ga^{kn}g^{-1}.$$ What next?

share|cite|improve this question
up vote 2 down vote accepted

Looking at the generator $a$ of $T$, the normality of $T$ says that we must have

$$gag^{-1} = a^r$$

for some integer $r$ (that depends on $g$ of course, and $r$ must be coprime to the order of $T$ [if $T$ is infinite cyclic, we must have $r = \pm 1$], but we don't use that).

Then we use the fact that conjugation by an element of $G$ is an automorphism, in particular,

$$g\bigl(a^{kn}\bigr)g^{-1} = (gag^{-1})^{kn}.$$

The normality of $S$ follows from these observations.

share|cite|improve this answer
Daniel Fischer, thank you, but I'm afraid I didn't get you when you mention that conjugation by an element of $G$ is an automorphism; of course it is, but on the basis of the hypothesis it is an automorphism of $T$ and not of $S$. Also, please elaborate on how $r$ must be coprime to the order of $T$ if $T$ is finite and $\pm 1$ if $T$ is infinite. – Saaqib Mahmuud Apr 6 '14 at 15:15
We only need that conjugation is an automorphism of $T$. What we use is that $gx^mg^{-1} = (gxg^{-1})^m$ for all $x\in G$. This property guarantees that the special structure of the elements of $S$ is preserved, and so $gSg^{-1}\subset S$. As for the conditions on $r$, since $T$ is normal, we have $gTg^{-1} = T$, and so the image of a generator of $T$ under conjugation must again be a generator of $T$. If $a$ has finite order, the generators of $T$ are just the $a^r$ where $r$ is coprime to the order of $T$, and if $T$ is infinite cyclic the generators of $T$ are $a$ and $a^{-1}$. – Daniel Fischer Apr 6 '14 at 15:23

A shorter argument: $S$ is a subgroup of a cyclic group, so it is a characteristic subgroup. Since $T$ is normal in $G$, $S$ is normal in $G$ as it is a characteristic subgroup of a normal subgroup (conjugation is an automorphism of $T$ since $T$ is normal).

share|cite|improve this answer
Oliver Braun, what's meant by a characteristic subgroup? Furthermore, I'm afraid I'd rather you'd elaborate your argument as I'm a bit rusty on the things you've mentioned. – Saaqib Mahmuud Apr 6 '14 at 15:10
Let $G$ be a group and $H\leq G$ a subgroup. $H$ is called characteristic in $G$ if for all $\varphi \in \mathrm{Aut}(G)$ we have $\varphi(H) \subseteq H$. Now, if $T$ is cyclic of order $n$, there is precisely one subgroup of $T$ of order $k$ for all divisors $k$ of $n$. Therefore every subgroup of $T$ is characteristic. Since $T$ is normal, conjugation with an element of $G$ is an automorphism of $T$, which in conclusion must fix the subgroup $S$. So $S$ is normal. – Oliver Braun Apr 6 '14 at 15:14

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.