# Final Group Q.

1. Jan 13, 2008

### smoothman

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. Jan 13, 2008

### morphism

Just do it. Show that there's a one-one correspondence between the elements of G of order n and the elements of H of order n. Use the fact that there is an isomorphism between G and H, and that isomorphisms preserve order.

3. Jan 16, 2008

### masnevets

The best way to think of an isomorphism between two groups G and H is that it is a way to rename the elements of G as the elements of H so that you end up with exactly H. If you think of it this way, I think it should not be too hard to show (2).

4. Jan 16, 2008

### robert Ihnot

smoothman: $C_3 X C_3$

I think I can remember some of this. An example of the type of problem is the four group: $$C_2XC_2$$ What happens here?

We have (0,0), (1,0), (0,1), (1,1). You can build up a chart of this under addition, and we get for example: (1,0)+(0,1) = (0,0), which sends us back to zero. (In fact, any element added to itself gives 0.)

So this can not be the cyclic four group consisting, 0+1=1, 1+1=2, 1+2=3,1+3 = 0.

In fact, it is the Kline 4 group.

Last edited: Jan 16, 2008