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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?

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2 Answers 2

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.

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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 at 15:15
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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 at 15:23
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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).

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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 at 15:10
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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 at 15:14
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