Normal subgroup generated by a subset A

[Mr Davis 97]

Homework Statement

Let $G$ be a group and let $A \subseteq G$ be a set. The normal subgroup of $G$ generated by $A$, denoted $\langle A \rangle ^N$, is the set of all products of conjugates of elements of $A$ and inverses of elements of $A$. In symbols,
$$\langle A \rangle ^N= \{ g_1a_1^{i_1}g_1^{-1} \dots g_ma_m^{i_m}g_m^{-1} \mid m \geq 0, a_j \in A, g_j \in G, i_j \in \{1,-1\}, \forall 1\leq j\leq m\}.$$

Show that $\langle A \rangle ^N$ is the intersection of all the normal subgroups of $G$ that contain $A$.

Homework Equations

The Attempt at a Solution

I amount really sure how to start here. I don't see anything that I can really grab on to to start the proof. Maybe we show that one is a subset of the other, and vice versa. But even then I still don't see how to show that an arbitrary element of $\langle A \rangle ^N$ is in the intersection of all the normal subgroups of $G$ that contain $A$.

 

[fresh_42]

$\langle A\rangle^N= \langle gAg^{-1}\,|\,g\in G \rangle$ so it is normal and contains $A$, and thus
$$
\langle A \rangle^N \supseteq \bigcap_{A\subseteq N \trianglelefteq G}N
$$
Now you have to show, that $\langle A\rangle^N \subseteq M$ for any given normal subgroup $A \subseteq M \trianglelefteq G$.

 

[Mr Davis 97]

$\langle A\rangle^N= \langle gAg^{-1}\,|\,g\in G \rangle$ so it is normal and contains $A$, and thus
$$
\langle A \rangle^N \supseteq \bigcap_{A\subseteq N \trianglelefteq G}N
$$
Now you have to show, that $\langle A\rangle^N \subseteq M$ for any given normal subgroup $A \subseteq M \trianglelefteq G$.

Just to be clear, what does the notation $\langle gAg^{-1}\,|\,g\in G \rangle$ mean? Also, don't I have to prove that $\{ g_1a_1^{i_1}g_1^{-1} \dots g_ma_m^{i_m}g_m^{-1} \mid m \geq 0, a_j \in A, g_j \in G, i_j \in \{1,-1\}, \forall 1\leq j\leq m\} = \langle gAg^{-1}\,|\,g\in G \rangle$?

 

[fresh_42]

Just to be clear, what does the notation $\langle gAg^{-1}\,|\,g\in G \rangle$ mean? Also, don't I have to prove that $\{ g_1a_1^{i_1}g_1^{-1} \dots g_ma_m^{i_m}g_m^{-1} \mid m \geq 0, a_j \in A, g_j \in G, i_j \in \{1,-1\}, \forall 1\leq j\leq m\} = \langle gAg^{-1}\,|\,g\in G \rangle$?

I think this is obvious. $\langle gAg^{-1}\,|\,g\in G\rangle$ is short for $\langle ga^{\pm 1}g^{-1}\,|\,g\in G\; , \;a\in A\rangle$ which is the same what you wrote. In any case, $A\subseteq \langle A \rangle^N$ is clear and also $g\langle A \rangle^Ng^{-1} \subseteq \langle A \rangle^N$ and so $\langle A \rangle^N \trianglelefteq G$.

 

[Mr Davis 97]

Just do be clear, I'm completely sure how to interpret $\langle ga^{\pm 1}g^{-1}\,|\,g\in G\; , \;a\in A\rangle$. I know that usually when we have something of the form $\langle ~|~ \rangle$ the left stuff consists of the generators and right stuff consists of the relations. If that's the case then how is $ga^{\pm 1}g^{-1}$ a generator and $g\in G\; , \;a\in A$ relations?

 

[fresh_42]

Just do be clear, I'm completely sure how to interpret $\langle ga^{\pm 1}g^{-1}\,|\,g\in G\; , \;a\in A\rangle$. I know that usually when we have something of the form $\langle ~|~ \rangle$ the left stuff consists of the generators and right stuff consists of the relations. If that's the case then how is $ga^{\pm 1}g^{-1}$ a generator and $g\in G\; , \;a\in A$ relations?

You are right. This was a different use of notation and thus confusing. The relations are hidden in $a,g \in G$ so the relations of $G$ hold. It should only note the group generated by all conjugates of elements of $A$ and their inverses. One could also write $\langle G\langle A \rangle G\cup G\langle A^{-1} \rangle G \rangle$ or describe it in words. It is simply the smallest subgroup of $G$ which contains $A$ and all conjugates to $A$ and $A^{-1}$ and shorter than what you wrote. Point is, it is automatically a normal subgroup which contains $A$ and thus contains the intersection of even more of those.

 

[Mr Davis 97]

You are right. This was a different use of notation and thus confusing. The relations are hidden in $a,g \in G$ so the relations of $G$ hold. It should only note the group generated by all conjugates of elements of $A$ and their inverses. One could also write $\langle G\langle A \rangle G\cup G\langle A^{-1} \rangle G \rangle$ or describe it in words. It is simply the smallest subgroup of $G$ which contains $A$ and all conjugates to $A$ and $A^{-1}$ and shorter than what you wrote. Point is, it is automatically a normal subgroup which contains $A$ and thus contains the intersection of even more of those.

Last question before I attempt to write a proof. Why in $\langle ga^{\pm 1}g^{-1}\,|\,g\in G\; , \;a\in A\rangle$ do we have $\pm 1$ and not just $1$? If we were taking the normal closure of a subgroup would we just need $+1$ instead of $\pm 1$?

 

[fresh_42]

Yes, here we have only a set $A$, so the inverse elements have to be added somehow. Of course one could as well first define a subgroup $U= \langle A \rangle$ generated by $A$ and then consider $U^N=\{\,gUg^{-1}\,|\,g\in G\,\}$, but this is more to write than $\pm 1$ as a power.

Before you start: Remember that you only have to show $gUg^{-1} \subseteq M$ for a normal $A \subseteq M \trianglelefteq G$.

 

[Mr Davis 97]

Let's show that $\langle A \rangle ^N \trianglelefteq G$. Let $x \in \langle A \rangle ^N$ and let $g \in G$. We need to show that $gxg^{-1} \in \langle A \rangle ^N$. By the definition, every element $x$ in $\langle A \rangle ^N$ can be written as $$x=g_1a_1^{i_1}g_1^{-1}g_2a_2^{i_2}g_2^{-1}\dots g_na_n^{i_n}g_n^{-1}$$ for some positive integer $n$, where each $g_i \in G$ and each $a_i \in A$, and where $i_j \in \{1,-1\}$. Consequently:
$\begin{align*} gxg^{-1} &= g(g_1a_1^{i_1}g_1^{-1}g_2a_2^{i_2}g_2^{-1}\dots g_na_n^{i_n}g_n^{-1})g^{-1}\\ &= (gg_1a_1^{i_1}g_1^{-1}g^{-1})(gg_2a_2^{i_2}g_2^{-1}g^{-1})\dots (gg_na_n^{i_n}g_n^{-1}g^{-1})\\ &= ((gg_1)a_1^{i_1}(gg_1)^{-1})((gg_2)a_2^{i_2}(gg_2)^{-1})\dots (gg_na_n^{i_n}(gg_n)^{-1}) \end{align*}$
i.e. $gxg^{-1}$ is a product of elements of the form $ja^{i}j^{-1}$ with $a \in A$, $j \in G$, and $i \in \{1,-1\}$. So $gxg^{-1} \in \langle A \rangle^N$. So $\langle A \rangle^N \trianglelefteq G$.


Now we show that $\displaystyle \langle A \rangle^N = \bigcap_{A\subseteq N \trianglelefteq G}N$
As a lemma, note that $\displaystyle \bigcap_{A\subseteq N \trianglelefteq G}N$ is the smallest normal subgroup containing $A$. First we note that since $A \subseteq \langle A \rangle^N$, we have that $\displaystyle \bigcap_{A\subseteq N \trianglelefteq G}N \subseteq \langle A \rangle^N$.

Now we show that $\displaystyle \langle A \rangle^N \subseteq \bigcap_{A\subseteq N \trianglelefteq G}N$. Suppose that $M$ is a normal subgroup of $H$ containing $A$. Then $N$ must contain $ga^ig^{-1}$ for all $g\in G$, $a\in A$ and $i\in \{1,-1\}$. in order to be normal and contain $A$, and $M$ must contain all products of such elements in order to be a subgroup of $G$. So $M$ must contain $\langle A \rangle^N$. So $\langle A \rangle^N$ is the smallest normal subgroup of $G$ containing the set $A$. So $\displaystyle \langle A \rangle^N \subseteq \bigcap_{A\subseteq N \trianglelefteq G}N$.

 

[fresh_42]

Too long, especially the first part, but o.k. An $H$ should be a $G$ and an $N$ an $M$, and I would say normal subgroup instead of just normal in the line before last, since we need the subgroup property as well. And I wouldn't write "as a Lemma" since this is what we actually have to show. Lemma sounds as something separated. However, these aren't crucial, noted just for completion.

 

[Mr Davis 97]

Too long, especially the first part, but o.k. An $H$ should be a $G$ and an $N$ an $M$, and I would say normal subgroup instead of just normal in the line before last, since we need the subgroup property as well. And I wouldn't write "as a Lemma" since this is what we actually have to show. Lemma sounds as something separated. However, these aren't crucial, noted just for completion.

So I'm a little concerned about the second part. You said that I need to show that $\displaystyle \bigcap_{A\subseteq N \trianglelefteq G}N$ is the smallest nromal subgroup containing $A$, but in supposing that this is true I am allowed to show the two directions of inclusion for $\displaystyle \bigcap_{A\subseteq N \trianglelefteq G}N$ and $\langle A \rangle^N$. So do I need to prove that $\displaystyle \bigcap_{A\subseteq N \trianglelefteq G}N$ is the smallest nromal subgroup containing $A$ before I do anything else, or is it obvious?

 

[fresh_42]

If there was a smaller one, say $M$, then $\langle A\rangle^N \supsetneq \bigcap_{A\subseteq N \trianglelefteq G} N \cap M$ but you have shown that $M$ contains $\langle A\rangle^N.$

 

[Mr Davis 97]

If there was a smaller one, say $M$, then $\langle A\rangle^N \supsetneq \bigcap_{A\subseteq N \trianglelefteq G} N \cap M$ but you have shown that $M$ contains $\langle A\rangle^N.$

I don't see quite what you mean by this.

Here is my argument. I suppose that $\displaystyle \bigcap_{A\subseteq N \trianglelefteq G}N$ is the smallest normal subgroup containing $A$. Then since $A \subseteq \langle A \rangle^N$ it's immediately apparent that $\displaystyle \bigcap_{A\subseteq N \trianglelefteq G}N \subseteq \langle A \rangle^N$ by the minimality of $\displaystyle \bigcap_{A\subseteq N \trianglelefteq G}N$. Next, I show that $\langle A \rangle^N$ is minimal, and since $\displaystyle \bigcap_{A\subseteq N \trianglelefteq G}N$ is a normal subgroup containing $A$, we conclude that $\displaystyle \langle A \rangle^N \subseteq \bigcap_{A\subseteq N \trianglelefteq G}N$. So $\displaystyle \langle A \rangle^N = \bigcap_{A\subseteq N \trianglelefteq G}N$

So all of this hinges on the assumption that $\displaystyle \bigcap_{A\subseteq N \trianglelefteq G}N$ is minimal. So do I need to show why this is the case explicitly before the proof?

EDITED

 

[fresh_42]

What you want to show depends on you and your readers. My argument was, that the intersection is already minimal, as every other normal subgroup with $A$ in it is part of the sets we take the intersection over. So there is no smaller possible, because of the property of the intersection. As $\langle A \rangle^N$ is in a subgroup of the intersection, it has to be the smallest. So what you already have shown implies minimality without further need to prove anything.

 

[Mr Davis 97]

What you want to show depends on you and your readers. My argument was, that the intersection is already minimal, as every other normal subgroup with $A$ in it is part of the sets we take the intersection over. So there is no smaller possible, because of the property of the intersection. As $\langle A \rangle^N$ is in a subgroup of the intersection, it has to be the smallest. So what you already have shown implies minimality without further need to prove anything.

Makes sense. One more thing. Should I also show that $\displaystyle\bigcap_{A\subseteq N \trianglelefteq G}N$ is actually even a normal subgroup of $G$ at all? I feel like I might be taking this for granted.

 

[fresh_42]

Makes sense. One more thing. Should I also show that $\displaystyle\bigcap_{A\subseteq N \trianglelefteq G}N$ is actually even a normal subgroup of $G$ at all? I feel like I might be taking this for granted.

i had the same thought. But as it equals a normal subgroup, namely $\langle A \rangle^N$, it has to be a group itself. And one can practically see that it's a normal subgroup, because the all quantifier of an intersection is preserved.