- #1
losiu99
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Hi, I've been vanquished by probably easy problems once again.
1. Let G be a group of order p^2 (p prime number), and H its subgroup of order p. Show that H is normal. Prove G must be abelian.
2. If a group G has exactly one subgroup H of order k, prove H is normal in G.
Lagrange theorem I think. Isomorphism theorems maybe?
1. Obviously H is cyclic. If H is not normal, G cannot be abelian, hence all the elements are of order p, except for the neutral one. G is therefore divided into p+1 cyclic, disjoint (except for e) subgroups of order p. So far I haven't succeeded deriving a contradiction.
2. Normalizer is either H or the whole group. Perhaps some property of self-normalizing groups yields a contradiction?
Thank you very much for any hints.
Homework Statement
1. Let G be a group of order p^2 (p prime number), and H its subgroup of order p. Show that H is normal. Prove G must be abelian.
2. If a group G has exactly one subgroup H of order k, prove H is normal in G.
Homework Equations
Lagrange theorem I think. Isomorphism theorems maybe?
The Attempt at a Solution
1. Obviously H is cyclic. If H is not normal, G cannot be abelian, hence all the elements are of order p, except for the neutral one. G is therefore divided into p+1 cyclic, disjoint (except for e) subgroups of order p. So far I haven't succeeded deriving a contradiction.
2. Normalizer is either H or the whole group. Perhaps some property of self-normalizing groups yields a contradiction?
Thank you very much for any hints.