MHB Proof: Every Subgroup of Cyclic $H$ is Normal in $G$

[mathmari]

Hey!

I want to show that if $H$ is a cylic normal subgroup of a group $G$, then each subgroup of $H$ is a normal subgroup of $G$.

I have done the following:

Since $H$ is a normal subgroup of $G$, we have that $$ghg^{-1}=h\in H, \ \forall g \in G \text{ and } \forall h\in H \tag 1$$

Since $H$ is cyclic, each subgroup of $H$ is also cyclic.

Let $\langle a\rangle$ be a subgroup of $H$.

Then $a\in \langle a\rangle \Rightarrow a\in H$.

From $(1)$ we have that $$gag^{-1}=a\in \langle a\rangle , \ \forall g\in G \text{ and } \forall a\in \langle a\rangle$$

Since the subgroup was arbitrary, we have that each subgroup of $H$ is a normal subgroup of $G$. Is everything correct? (Wondering)

 

[Ackbach]

Hey!

I want to show that if $H$ is a cylic normal subgroup of a group $G$, then each subgroup of $H$ is a normal subgroup of $G$.

I have done the following:

Since $H$ is a normal subgroup of $G$, we have that $$ghg^{-1}=h\in H, \ \forall g \in G \text{ and } \forall h\in H \tag 1$$

Since $H$ is cyclic, each subgroup of $H$ is also cyclic.

Let $\langle a\rangle$ be a subgroup of $H$.

I don't think I would use the same symbol $a$ for the element in $\langle a\rangle$ as the element generating that subgroup. Try letting $b\in\langle a\rangle$. I think your idea is sound, though, and if you propagate this correction through your proof, I think it's fine.

Then $a\in \langle a\rangle \Rightarrow a\in H$.

From $(1)$ we have that $$gag^{-1}=a\in \langle a\rangle , \ \forall g\in G \text{ and } \forall a\in \langle a\rangle$$

Technically, the $\forall a\in\langle a\rangle$ is redundant, since $a$ is already an arbitrary element in the subgroup.

Since the subgroup was arbitrary, we have that each subgroup of $H$ is a normal subgroup of $G$. Is everything correct? (Wondering)

 

[mathmari]

So it should be as followed, shouldn't it? (Wondering) Since $H$ is a normal subgroup of $G$, we have that $$ghg^{-1}=h\in H, \ \forall g \in G \text{ and } \forall h\in H \tag 1$$

Since $H$ is cyclic, each subgroup of $H$ is also cyclic.

Let $\langle a\rangle$ be a subgroup of $H$.

Then for $b\in \langle a\rangle$ we have that $b\in H$.

From $(1)$ we have that $$gbg^{-1}=b\in \langle a\rangle , \ \forall g\in G \text{ and } \forall b\in \langle a\rangle$$

Since the subgroup was arbitrary, we have that each subgroup of $H$ is a normal subgroup of $G$. Would the statement

if $H$ is a cylic normal subgroup of a group $G$, then each subgroup of $H$ is a normal subgroup of $G$

also stand when $H$ wouldn't be cyclic? (Wondering)

 

[johng1]

Since $H$ is a normal subgroup of $G$, we have that
$$ghg^{-1}=h\in H\,\forall g\in G\text{ and }\forall h\in H.$$

This says every element of $H$ commutes with every element of $G$ or $H\subseteq Z(G)$. You are given only that $H$ is normal in $G$. So if $H=<a>$ and $g\in G,\, gag^{-1}=a^k$ for some $k$. You should now be able to finish.

I assume your last question is that for any normal subgroup, is every subgroup also normal? That is, is normality a transitive relation. The answer is definitely no; you should be able to find an example.

 

[mathmari]

So if $H=<a>$ and $g\in G,\, gag^{-1}=a^k$ for some $k$. You should now be able to finish.

Having that $H=\langle a \rangle$, is every subgroup of $H$ of the form $\langle a^i\rangle$, for some $i$ ? (Wondering)

 

[Deveno]

Hint:

Suppose $N \subseteq H$ is a subgroup of $H$.

Since $H = \langle a\rangle$ for some $a \in G$, it follows that:

$N = \langle a^i\rangle$, for some integer $i$.

Since $H \lhd G$, we have for any $g\in G$, that $ghg^{-1} \in H$, for any $h \in H$.

In particular, $gag^{-1} = a^k$, for some $k$.

Now use the fact that $ga^ig^{-1} = (gag^{-1})^i$.

(Our goal is to show that $ga^ig^{-1} \in \langle a^i\rangle$).

 

[mathmari]

Now use the fact that $ga^ig^{-1} = (gag^{-1})^i$.

We have the following:

$$ga^ig^{-1} = g\underset{i}{\underbrace{aa\cdots a a}}g^{-1}=g\underset{i}{\underbrace{ag^{-1}ga g^{-1}g\cdots g^{-1}ga g^{-1}ga}}g^{-1}=(gag^{-1})^i$$

right? (Wondering)

Hint:

Suppose $N \subseteq H$ is a subgroup of $H$.

Since $H = \langle a\rangle$ for some $a \in G$, it follows that:

$N = \langle a^i\rangle$, for some integer $i$.

Since $H \lhd G$, we have for any $g\in G$, that $ghg^{-1} \in H$, for any $h \in H$.

In particular, $gag^{-1} = a^k$, for some $k$.

Now use the fact that $ga^ig^{-1} = (gag^{-1})^i$.

(Our goal is to show that $ga^ig^{-1} \in \langle a^i\rangle$).

Since $H \lhd G$, we have for any $g\in G$, that $ghg^{-1} \in H$, for any $h \in H$.

So since $a\in H=\langle a\rangle$ we have that $gag^{-1} = a^k$, for some $k$.

We have the following: $$gag^{-1} = a^k \Rightarrow (gag^{-1})^i = (a^k)^i \Rightarrow ga^ig^{-1}=a^{ki} \Rightarrow ga^ig^{-1}=(a^i)^k \in \langle a^i\rangle$$

Is everything correct? (Wondering)

 

[Deveno]

We have the following:

$$ga^ig^{-1} = g\underset{i}{\underbrace{aa\cdots a a}}g^{-1}=g\underset{i}{\underbrace{ag^{-1}ga g^{-1}g\cdots g^{-1}ga g^{-1}ga}}g^{-1}=(gag^{-1})^i$$

right? (Wondering)


Since $H \lhd G$, we have for any $g\in G$, that $ghg^{-1} \in H$, for any $h \in H$.

So since $a\in H=\langle a\rangle$ we have that $gag^{-1} = a^k$, for some $k$.

We have the following: $$gag^{-1} = a^k \Rightarrow (gag^{-1})^i = (a^k)^i \Rightarrow ga^ig^{-1}=a^{ki} \Rightarrow ga^ig^{-1}=(a^i)^k \in \langle a^i\rangle$$

Is everything correct? (Wondering)

Looks good.

 

[caffeinemachine]

Having that $H=\langle a \rangle$, is every subgroup of $H$ of the form $\langle a^i\rangle$, for some $i$ ? (Wondering)

Yes. This is same as saying that every subgroup of a cyclic group is a cyclic group.

 

[mathmari]

Thank you so much! (Sun)

 

[mathmari]

Hint:

Suppose $N \subseteq H$ is a subgroup of $H$.

Since $H = \langle a\rangle$ for some $a \in G$, it follows that:

$N = \langle a^i\rangle$, for some integer $i$.

Since $H \lhd G$, we have for any $g\in G$, that $ghg^{-1} \in H$, for any $h \in H$.

In particular, $gag^{-1} = a^k$, for some $k$.

Now use the fact that $ga^ig^{-1} = (gag^{-1})^i$.

(Our goal is to show that $ga^ig^{-1} \in \langle a^i\rangle$).

We have that $a^i$ is the generator of the subgroup $N=\langle a^i\rangle$.
Do we have to take the generator to show that $ga^ig^{-1} \in \langle a^i\rangle$ or could we have taken also an element of that cyclic group, for example $a^{im}$, so $ga^{im}g^{-1} \in \langle a^i\rangle$ ? (Wondering)

 

[Deveno]

We have that $a^i$ is the generator of the subgroup $N=\langle a^i\rangle$.
Do we have to take the generator to show that $ga^ig^{-1} \in \langle a^i\rangle$ or could we have taken also an element of that cyclic group, for example $a^{im}$, so $ga^{im}g^{-1} \in \langle a^i\rangle$ ? (Wondering)

We could do either one.

 

[mathmari]

We could do either one.

So, would we do then the following? (Wondering)

Since $H \lhd G$, we have for any $g\in G$, that $ghg^{-1} \in H$, for any $h \in H$.

So since $a\in H=\langle a\rangle$ we have that $gag^{-1} = a^k$, for some $k$.

We have the following: $$gag^{-1} = a^k \Rightarrow (gag^{-1})^{im} = (a^k)^{im} \Rightarrow ga^{im}g^{-1}=a^{kim} \Rightarrow ga^{im}g^{-1}=(a^i)^{km} \in \langle a^i\rangle$$

Is everything correct? (Wondering)