Thread: Order of groups involving conjugates and abelian groups

1. Let G be a group and let a be in G. An element b in G is called a conjugate of a if there exists an element x in G such that b = xax^-1. Show that any conjugate of a has the same order as a.

2. Let G be an abelian group and let x,y be in G. Suppose that x and y are of finite order. Show that xy is of finite order and that, in fact, o(xy) (order of xy) divides o(x)o(y).

Originally Posted by Janu42

1. Let G be a group and let a be in G. An element b in G is called a conjugate of a if there exists an element x in G such that b = xax^-1. Show that any conjugate of a has the same order as a.

Hint: If then .

2. Let G be an abelian group and let x,y be in G. Suppose that x and y are of finite order. Show that xy is of finite order and that, in fact, o(xy) (order of xy) divides o(x)o(y).

Note since is abelian.
If then .
Thus, .

Thanks again. Also, quick one:

G = {1,2,3,4,5,6}
The group (G, *), where * is multiplication mod 7. For example, 2*4 = 8 = 1, since 8 = 1 mod 7.

How do I show it's cyclic? I know it is, but which element of G can I use as the generator?

If I do <1>, doesn't that mean when I do 1^2 it is 1*1 and 1^3 is 1*1*1?
Because then I would have 1 every time.

Originally Posted by Janu42

Thanks again. Also, quick one:

G = {1,2,3,4,5,6}
The group (G, *), where * is multiplication mod 7. For example, 2*4 = 8 = 1, since 8 = 1 mod 7.

How do I show it's cyclic? I know it is, but which element of G can I use as the generator?

If I do <1>, doesn't that mean when I do 1^2 it is 1*1 and 1^3 is 1*1*1?
Because then I would have 1 every time.

Find an element so that , remembmer means raised to reduced mod .

OK, one more, I'm pretty sure this is simple. I'm pretty sure it has something to do with the order being a multiple of 2.

Show that if the order of G is an even integer, then there is an element x in G such that x is not equal to e, and x^2 = e.

Originally Posted by Janu42

OK, one more, I'm pretty sure this is simple. I'm pretty sure it has something to do with the order being a multiple of 2.

Show that if the order of G is an even integer, then there is an element x in G such that x is not equal to e, and x^2 = e.

Assume not. Then for every there is so that . Thus, all elements can be paired with their inverses. But then that means there are odd number of elements.
A contradiction.