I Trying to get the point of some Group Theory Lemmas

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There are two related Lemmas in Schaum's Outline of Group Theory, Chapter 4 that seem excessively convoluted. Either I am missing something or they can be made much simpler and clearer.
There are two related Lemmas in Schaum's Outline of Group Theory, Chapter 4 that seem excessively convoluted.
Either I am missing something or they can be made much simpler and clearer.

Lemma 4.2:
If H is a subgroup of G and [itex]{\rm{X}} \subseteq {\rm{H}}[/itex] then [itex]{\rm{H}} \supseteq \left\{ {{\rm{x}}_1^{{\varepsilon _1}}...{\rm{x}}_n^{{\varepsilon _n}}|{x_i} \in X,{\varepsilon _i} = \pm 1,n{\text{ a positive integer}}} \right\}[/itex].

Lemma 4.3
Let G be a group and let X be a non-empty subset of G.
Let [itex]{\rm{S = }}\left\{ {{\rm{x}}_1^{{\varepsilon _1}}...{\rm{x}}_n^{{\varepsilon _n}}|{x_i} \in X,{\varepsilon _i} = \pm 1,n{\text{ a positive integer}}} \right\}[/itex]. Then S is a subgroup of G. If H is any subgroup containing X, then [itex]{\rm{H}} \supseteq {\rm{S}}[/itex].

Question1: In Lemma 4.2, does H being a subgroup of G add anything? I think we could replace "If H is a subgroup of G" with "If H is group".
Question2: In Lemma 4.3, the isn't statement
If H is any subgroup containing X, then [itex]{\rm{H}} \supseteq {\rm{S}}[/itex] exactly equivalent to Lemma 4.2?

So the two Lemmas seem to amount to this (if we replace that expression in curly braces with gp(X)):
If H is a group and [itex]{\rm{X}} \subseteq {\rm{H}}[/itex] then [itex]{\rm{gp}}\left( X \right) \subseteq H[/itex].
If H is a group and [itex]\left( {{\rm{X}} \ne \emptyset } \right) \subseteq {\rm{H}}[/itex] then [itex]{\rm{gp}}\left( {\rm{X}} \right)[/itex] is a subgroup of H.
 
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fresh_42

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Summary: There are two related Lemmas in Schaum's Outline of Group Theory, Chapter 4 that seem excessively convoluted. Either I am missing something or they can be made much simpler and clearer.

There are two related Lemmas in Schaum's Outline of Group Theory, Chapter 4 that seem excessively convoluted.
Either I am missing something or they can be made much simpler and clearer.

Lemma 4.2:
If H is a subgroup of G and [itex]{\rm{X}} \subseteq {\rm{H}}[/itex] then [itex]{\rm{H}} \supseteq \left\{ {{\rm{x}}_1^{{\varepsilon _1}}...{\rm{x}}_n^{{\varepsilon _n}}|{x_i} \in X,{\varepsilon _i} = \pm 1,n{\text{ a positive integer}}} \right\}[/itex].

Lemma 4.3
Let G be a group and let X be a non-empty subset of G.
Let [itex]{\rm{S = }}\left\{ {{\rm{x}}_1^{{\varepsilon _1}}...{\rm{x}}_n^{{\varepsilon _n}}|{x_i} \in X,{\varepsilon _i} = \pm 1,n{\text{ a positive integer}}} \right\}[/itex]. Then S is a subgroup of G. If H is any subgroup containing X, then [itex]{\rm{H}} \supseteq {\rm{S}}[/itex].

Question1: In Lemma 4.2, does H being a subgroup of G add anything? I think we could replace "If H is a subgroup of G" with "If H is group".
Why should we restrict the statement? With respect of Lemma 4.3. it might be useful that it holds for any subgroup, ##G=H## included.
Question2: In Lemma 4.3, the isn't statement
If H is any subgroup containing X, then [itex]{\rm{H}} \supseteq {\rm{S}}[/itex] exactly equivalent to Lemma 4.2?
Lemma 4.3. says that ##S## is the smallest possible group which contains ##X##. Lemma 4.2. does not contain the statement, that the words over ##X## are actually a group, which is the first part. Only the second part is the same as Lemma 4.2.
So the two Lemmas seem to amount to this (if we replace that expression in curly braces with gp(X)):
If H is a group and [itex]{\rm{X}} \subseteq {\rm{H}}[/itex] then [itex]{\rm{gp}}\left( X \right) \subseteq H[/itex].
If H is a group and [itex]\left( {{\rm{X}} \ne \emptyset } \right) \subseteq {\rm{H}}[/itex] then [itex]{\rm{gp}}\left( {\rm{X}} \right)[/itex] is a subgroup of H.
This is not quite correct. We need two sets ##F(X)=\left\{ {{\rm{x}}_1^{{\varepsilon _1}}...{\rm{x}}_n^{{\varepsilon _n}}|{x_i} \in X,{\varepsilon _i} = \pm 1,n{\text{ a positive integer}}} \right\}## and ##gp(X)=\bigcap \{\,H \leq G\,|\,X \subseteq H \,\}##. ##F(X)## is a set, the set of all words over ##X##. ##gp(X)## is the smallest group which contains ##X##.

Lemma 4.2.: ##X \subseteq H \leq G \Longrightarrow F(X) \subseteq H##.
Lemma 4.3.: ##F(X) \leq G \,\wedge \,F(X)=gp(X)##.
 
Thank you. This helps.
A couple of points:

1) You say "Why should we restrict the statement? With respect of Lemma 4.3. it might be useful that it holds for any subgroup, G=HG=H included". My only answer is that find this material difficult and it helps if there are no
extraneous distractions. I even do things like use either all
⊇ or all
⊆.
Expressing everything in a parallel way makes it clearer to me.

2) I am not familiar with the term "
the words over
X".
I guess that means
{
x
ε1
1
...
x
εn
n
|
x
i
X,
ε
i
=±1,n
a positive integer
}?
 

fresh_42

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Not sure what you mean by the subset signs.

If it is about meaning, then ##\subseteq## refers to sets, ##\leq## to subgroups, ##\trianglelefteq## to normal subgroups.

If you asked about my comment on Lemma 4.2. then I meant, that choosing subgroups ##H## instead of ##G## makes the statement easier for applications. If we have a situation (*) ##X \subseteq H \leq G## and want to apply it to ##H##, it is immediately clear. If Schaum had written it as ##X \subseteq G## and wanted to apply it for a situation (*), he would had been forced to complicatedly write: "Now we apply Lemma 4.2. with ##G=H##, but here at (*) we still have possibly ##H \subsetneq G## ..." which is a total mess. Generality in Lemma 4.2. as it stands is as good and easier to apply. And Lemmata are meant to apply on various situations.

Words in group theory are elements written as products and integer powers of certain group elements, called the alphabet. Here we have ##X=\{\,x_1,\ldots,x_n\,\}## as alphabet and words are any products or inverses of these.
 

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