Thread: Final Group Q.

ok ive managed to solve the other 2 questions.

here is my final one:

(1)
If G is a group and , define G(n) = { x E G: ord(x) = n}

(2)
If show that, for all , |G(n)| = |H(n)|.

(3)
Deduce that, is not .
Is it true that
Is it true that

What is going on here?

any help to get me started is highly appreciated. ill attempt the questions as usual once i have some idea of what to do. thnx so much

Originally Posted by smoothman

(1)
If G is a group and , define G(n) = { x E G: ord(x) = n}

This is just the set of all which have order . Note, this set can be empty.

(2)
If show that, for all , |G(n)| = |H(n)|.

Since it means there is an isomorphism . To prove that we need to define a one-to-one correspondence . First, if is empty then has to be empty, because if has order then has order also because isomorphisms preserve orders (and the inverse function is an isomorphism too). Suppose is non-empty. For define . Then is in because isomorphisms preserve orders. Now we need to prove that is one-to-one and onto. It is one-to-one because if . It is onto, because if then and . That means the sets and have equal sizes.
(3)

Deduce that, is not .
Is it true that
Is it true that

I start you off. If then . Pick an such that this is not true. Then this would mean that the isomorphism cannot exist.

Here is a useful theorem to know. If are integers such that then . That means if is factored as by fundamental theorem of arithmetic it means that .

But it turns out that above result can be strengthed. We have shown above that every finite cyclic group is a direct product of cyclic groups of prime orders. It can be shown that any finite abelian group is a direct product of cyclic groups of prime orders (but the primes can be repeated multiple times). But you will probably not learn this result because the proof is not so trivial.