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Request explain the highlighted portion of the proof (I have not pasted the full proof)

Request explain the highlighted portion of the proof (I have not pasted the full proof)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Request explain the highlighted portion of the proof (I have not pasted the full proof)

Theorem 3: Let G be a finite abelian p-group, G#{e}, and let a € G be of maximal order in G. Then G is cyclic or there is a subgroup H of G, such that G=<a>XH. Proof: Let o(a) = pk. If G=<a>, then G is cyclic. Suppose G #<a>. Consider the set A = {K≤G| <a>nK={e}}. A Ø, since {e} = A. Let H be maximal in A, that is, H belongs to A, and if K is any element of A, then K cannot properly contain H. We will prove that G=<a> x H. Since G is abelian, all its subgroups are normal. Moreover, by definition, <a>nH={e}. Therefore, in order to prove that G=<a>>H, we only need to show that G = <a > H. We will prove this by contradiction. So, suppose that G#<a > H. Then G/(<a > H) is a p-group, and hence contains an element of order p. In other words, there exists x = G/(<a > H) such that x² =ē, that is, 3x G such that x <a > H but x² €<a > H. So, x² = a¹b, for some qe Z and be H. Now, since the order of every element of G divides p*, „k-l e = x² = (x³)p²-1 = a 9p²-1 bᏢ . ⇒a € <a>nH = {e} ⇒pk|qpk-¹, since o(a) = pk ⇒p|q ⇒q=ps, for some se Z. THE PEOPLE'S
Transcribed Image Text:Theorem 3: Let G be a finite abelian p-group, G#{e}, and let a € G be of maximal order in G. Then G is cyclic or there is a subgroup H of G, such that G=<a>XH. Proof: Let o(a) = pk. If G=<a>, then G is cyclic. Suppose G #<a>. Consider the set A = {K≤G| <a>nK={e}}. A Ø, since {e} = A. Let H be maximal in A, that is, H belongs to A, and if K is any element of A, then K cannot properly contain H. We will prove that G=<a> x H. Since G is abelian, all its subgroups are normal. Moreover, by definition, <a>nH={e}. Therefore, in order to prove that G=<a>>H, we only need to show that G = <a > H. We will prove this by contradiction. So, suppose that G#<a > H. Then G/(<a > H) is a p-group, and hence contains an element of order p. In other words, there exists x = G/(<a > H) such that x² =ē, that is, 3x G such that x <a > H but x² €<a > H. So, x² = a¹b, for some qe Z and be H. Now, since the order of every element of G divides p*, „k-l e = x² = (x³)p²-1 = a 9p²-1 bᏢ . ⇒a € <a>nH = {e} ⇒pk|qpk-¹, since o(a) = pk ⇒p|q ⇒q=ps, for some se Z. THE PEOPLE'S
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