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jstrunk

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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.

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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