If G is an abelian group of order (p^t)m, and (p,m)=1, show that G(p) has order p^t and G(p) = {a e G| |a|=p^m where m is a natural number} any suggestions?
What do you have to work with? This problem is trivial if you've developed enough tools. You can also do this with your "bare hands" if you want to. What have you tried so far?
Geez...I have no idea where to start... Ok, I was looking at a lemma... "Let G be an abelian group and a [tex]\in[/tex] G and is an element of finite order. Then a = a_{1}+a_{2}+...+a_{k}, with a_{i} an element of G(p_{i}), where p_{1},....,p_{k} are the distinct positive primes that divide the order of a" .... now G(p) is only the set of all elements whose order is some power of p... I was thinking you could use that lemma to show that G(p) has power p^{t} ...seems like that could help somehow.. but it seems I need to use the fact that (p,m)=1 ...this surely means that (p^{t},m)=1 which means you can write 1=p^{t}*u + m*v...for some numbers u,v... and I have no idea where to go...
Well...Actually, I could use Cauchy's theorem for Abelian Groups... "If G is a finite abelian group and p is a prime that divides |G|, prove that G contains an element of order p.... we know that p divides |G|...and p^{2} divides the order of G...and so on and so on until we have p^{t} divides the order of G...so all these elements exist in G(p)..then use the (p,m)=1 to show that each one of these elements is uniquely the product of m with some interger...I think that is it...
Does Cauchy's theorem imply that if p^m divides |G|, then G contains an element of order p^m? The fact that G is abelian is vital here. First note that G(p) is a subgroup of G. Next note that G(p) has order a power of p, by Cauchy. If |G(p)| isn't p^t, then we can consider the group G/G(p). G/G(p) is still abelian, and p divides |G/G(p)|. Apply Cauchy's theorem again to get an element g+G(p) of order p. Now what?