# 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##.

## 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##.

##\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##?

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##.

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?

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.

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##.

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##.

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.

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?

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.##

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

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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.

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.