22.Prove that a finite Abelian group of square-free order is cyclic.?

Question

1.Show that the endomorphism of a semigroup S constitute, under composition, a semigroup with identity. What are the units of this semigroup?
2.Determine all the endomorphisms of the additive semigroup Z. Which of these are automorphisms?
9. Show that any group of 200 is soluble?
14.Show that a finite group is never the union of the conjugate of a proper subgroup. Equivalently, if a subgroup H of the finite group G meets every conjugacy class of G, then H=G.
16.Prove that every subgroup of the quaternion group of order 8 is normal in it(even though this group is not Abelian).
17.If the cyclic subgroup A of G is normal in G, show that every subgroup of A is again normal in G.
18. If H, K are normal subgroups of G having intersections E={1}, show that every element of H commutes with every element of K.
20.If all subgroups of the group G are normal in G, show that the commutator [a,b] commutes with each of a,b.
21.Prove that a every quotient of a cyclic group is cyclic.

Answer 1

Wow, 10 questions in one! I pretty much hate when people do that. To answer 22, use the Fundamental Theorem of Finitely Generated Abelian Groups. Some hints on the others:

1 & 2: I don't really care about semigroups.

9: Use Sylow's third theorem on the number of groups of order 25, and the fact that any group of order 25 is abelian.

16: All the subgroups either have index 2 or are Z(G).

17: Conjugate by any element in G.

18. Note [h,k] is in the intersection.

21. Plenty of options here: use the Fundamental Theorem mentioned above, use that Z and Z/nZ are all the cyclic groups up to isomorphism, or deal directly with the cosets in the factor group.



Steve

Answer 2

I'm a bit rusty on this, so better answers are welcome. In particular, I'm a little worried that my proof doesn't even use the abelian property... Consider the order of an element g, i.e. the lowest positive n such that g^n = e, where e is the identity. If n is the order of the G, then G is cyclic. If n is not the order of G, then we have a cyclic subgroup, let's call it H. The question is, can we always find a cyclic subgroup of prime order? If n is prime, then obviously we have. If n is not prime, then we can find a cyclic subgroup whose order is one of the prime factors of H, by considering the order of g^m, where n/m is prime. And any cyclic group order p (where p is prime) is isomorphic to Z_p_ (+) Z_p_. EDIT: I assume the TD was from the other answerer. I don't care about getting a TD but I'm curious... what's wrong with my proof?

Answer 3

I'm just doing the one in the title, number 22.



Let n be the order of the group G. Then n is the product of distinct primes, say p1,... pk. So G is isomorphic by the fundamental theorem of abelian groups to



Zp1 x Zp2 x ... Zpk.



Then the element of G corresponding to (1,1,1,...,1) is of order n.