# Thread: Final Group Q.

1. ## Final Group Q.

ok ive managed to solve the other 2 questions.

here is my final one:

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

(2)
If $G \cong H$ show that, for all $n \geq 1$, |G(n)| = |H(n)|.

(3)
Deduce that, $C_3 X C_3$ is not $\cong C_9$.
Is it true that $C_3 X C_5 \cong C_15$
Is it true that $C_2 X C_6 \cong C_12$

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

2. Originally Posted by smoothman

(1)
If G is a group and $n \geq 1$, define G(n) = { x E G: ord(x) = n}
This is just the set of all $x\in G$ which have order $n$. Note, this set can be empty.

(2)
If $G \cong H$ show that, for all $n \geq 1$, |G(n)| = |H(n)|.
Since $G\simeq H$ it means there is an isomorphism $\phi : G\mapsto H$. To prove that $|G(n)| = |H(n)|$ we need to define a one-to-one correspondence $\theta : G(n)\mapsto H(n)$. First, if $G(n)$ is empty then $H(n)$ has to be empty, because if $b \in H(n)$ has order $n$ then $a= \phi^{-1} (b)\in G(n)$ has order $n$ also because isomorphisms preserve orders (and the inverse function is an isomorphism too). Suppose $G(n)$ is non-empty. For $x\in G(n)$ define $\theta (x) = \phi(x)$. Then $\phi(x)$ is in $H(n)$ because isomorphisms preserve orders. Now we need to prove that $\theta : G(n)\mapsto H(n)$ is one-to-one and onto. It is one-to-one because if $\theta (x) = \theta (y) \implies \phi (x) = \phi (y) \implies x = y$. It is onto, because if $y\in H(n)$ then $\phi^{-1} (y) \in G(n)$ and $\theta (\phi^{-1} (y)) = y$. That means the sets $G(n)$ and $H(n)$ have equal sizes.
(3)

Deduce that, $C_3 X C_3$ is not $\cong C_9$.
Is it true that $C_3 X C_5 \cong C_15$
Is it true that $C_2 X C_6 \cong C_12$
I start you off. If $C_3\times C_3 \simeq C_9$ then $C_3\times C_3 (n) = C_9(n)$. Pick an $n$ such that this is not true. Then this would mean that the isomorphism cannot exist.

3. Here is a useful theorem to know. If $n,m\geq 1$ are integers such that $\gcd (n,m)=1$ then $\mathbb{Z}_{nm}\simeq \mathbb{Z}_n \times \mathbb{Z}_m$. That means if $n>1$ is factored as $n=p_1^{a_1}\cdot ... \cdot p_k^{a_k}$ by fundamental theorem of arithmetic it means that $\mathbb{Z}_n \simeq \mathbb{Z}_{p_1^{a_1}}\times ... \times \mathbb{Z}_{p_k^{a_k}}$.

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.