Question

Suppose that the order of some finite Abelian group is divisible by 10. Prove that the group has a cyclic subgroup of order 10.

My professor says I cannot used the corollary : "If m divides the order of a finite Abelian group G, then G has a subgroup of order m." So I don't know what to do know

Answer 1

you know cauchy theorem?

Answer 2

oh i see, you cannot use it

Answer 3

oh, yes you can
cauchy says if \(p|o(G)\) then there is an element of order \(p\)

Answer 4

he never said I couldn't use cauchy's theorem just that corollary above

my book says cauchy's theorem:
Let G be a finite abelian group and let p be a prime that divides the order of G. Then G has an element of order p.

Answer 5

therefore your abelian group must have an element \(x\) of order \(2\) and also an element \(y\) of order \(5\)
now what is left for you to show is the subgroup generated by \(xy\) has order \(10\) and you are done

Answer 6

so woud it be the cyclic group <10> or would it be 2<5> ? I'm reading the proofs and examples that go along with Cauchy's theorem and they arent making sense.

Answer 7

or perhaps Z10 would be the group? I can't seem to find agreeing statements within the book