Normal subgroup generated by a subset A

In summary: 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
Mr Davis 97
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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##.
 
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  • #2
##\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##.
 
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  • #3
fresh_42 said:
##\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##?
 
  • #4
Mr Davis 97 said:
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##.
 
  • #5
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?
 
  • #6
Mr Davis 97 said:
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.
 
  • #7
fresh_42 said:
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##?
 
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  • #8
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##.
 
  • #9
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##.
 
  • #10
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.
 
  • #11
fresh_42 said:
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?
 
  • #12
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.##
 
  • #13
fresh_42 said:
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
 
Last edited:
  • #14
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.
 
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  • #15
fresh_42 said:
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.
 
  • #16
Mr Davis 97 said:
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.
 

Related to Normal subgroup generated by a subset A

What is a normal subgroup generated by a subset A?

A normal subgroup generated by a subset A is a subgroup of a group G that is closed under conjugation by elements of the group. This means that for any element a in A and any element g in G, the conjugate of a by g (i.e. g^-1ag) is also in the subgroup.

How is a normal subgroup generated by a subset A different from a regular subgroup?

A regular subgroup is simply a subset of a group that is closed under the group's operation. A normal subgroup, on the other hand, is also closed under conjugation by elements of the group. This means that normal subgroups are a more restrictive type of subgroup compared to regular subgroups.

What is the significance of a normal subgroup generated by a subset A?

Normal subgroups are important in the study of group theory because they help us to understand the structure of a group. They also play a key role in the definition of quotient groups, which are important in many areas of mathematics.

How can one determine if a subgroup is a normal subgroup generated by a subset A?

To determine if a subgroup is a normal subgroup generated by a subset A, one can use the definition of normality. This means checking if the subgroup is closed under conjugation by all elements of the group. Alternatively, one can use the subgroup criterion, which states that a subgroup is normal if and only if it is the kernel of a homomorphism.

What are some examples of normal subgroups generated by a subset A?

Some examples of normal subgroups generated by a subset A include the center of a group, the commutator subgroup, and the subgroup generated by all elements of the form aba^-1b^-1. These subgroups are all normal because they are closed under conjugation by all elements of the group.

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