Normal subgroup generated by a subset A

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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##.

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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##.
 

Answers and Replies

  • #2
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##\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
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##\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
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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
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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
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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
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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##?
 
  • #8
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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
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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
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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
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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
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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
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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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  • #14
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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.
 
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  • #15
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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.
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
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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.
 

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