# Order of groups involving conjugates and abelian groups

• Feb 4th 2009, 08:43 PM
Janu42
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).
• Feb 4th 2009, 08:51 PM
ThePerfectHacker
Quote:

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 $b = xax^{-1}$ then $b^n = xa^n x^{-1}$.

Quote:

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 $(ab)^k = a^kb^k$ since $G$ is abelian.
If $|x|=n,|y|=m$ then $(xy)^{nm} = (x^n)^m(y^m)^n = 1$.
Thus, $|xy| \text{ divides }|x||y|$.
• Feb 4th 2009, 09:02 PM
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.
• Feb 4th 2009, 09:06 PM
ThePerfectHacker
Quote:

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 $a$ so that $G = \{ a,a^2,a^3,a^4,a^5,a^6\}$, remembmer $a^k$ means $a$ raised to $k$ reduced mod $7$.
• Feb 5th 2009, 06:00 PM
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.
• Feb 5th 2009, 09:55 PM
ThePerfectHacker
Quote:

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 $x\not = e$ there is $y$ so that $xy=e$. Thus, all elements $\not = e$ can be paired with their inverses. But then that means there are odd number of elements. (Surprised)