G denotes a non-cyclic group with p^2 elements, where p is a prime number. a) Le
ID: 3106743 • Letter: G
Question
G denotes a non-cyclic group with p^2 elements, where p is a prime number.a) Let A be the set of all subgroups of G which have p elements. Show that |A| = p + 1.
b) Show that G has a normal subgroup N with p elements
c) Show that G is abelian. (You may need to use Fermat’s Little Theorem; recall that this says a^p = a (mod p) for any integer a.)
Explanation / Answer
a))) Since G is a p-group, we must have |Z(G)| > 1. As |G| = p^2, we must have |Z(G)| = p or p^2 by Lagrange's Theorem. (i) If |Z(G)| = p^2 = |G|, then Z(G) = G, and thus G is abelian. (ii) If |Z(G)| = p, then |G/Z(G)| = p. So, G/Z(G) is cyclic ==> G is abelian. (This last claim is straightforward to establish.) b)) Let g be a nontrivial element of G. By Lagrange's Theorem, we know that the order of g (call it |g|) must divide |G| = p^2. So, |g| = 1, p, or p^2. Since g is nontrivial, |g| > 1. Since G is not cyclic, |g| All nontrivial elements have order p. Let g be one such element. [G being abelian, combined with G/ ? Z_p, we can conclude that G ? Z_p x Z_p.] Since G has (p^2 - 1) elements of order p (ignoring the identity), and each subgroup of order p has (p - 1) elements of order p in it, we conclude that |A| = (p^2 - 1)/(p - 1) = p + 1.Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.