# Thread: Order of groups involving conjugates and abelian groups

1. ## 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).

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

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 $\displaystyle (ab)^k = a^kb^k$ since $\displaystyle G$ is abelian.
If $\displaystyle |x|=n,|y|=m$ then $\displaystyle (xy)^{nm} = (x^n)^m(y^m)^n = 1$.
Thus, $\displaystyle |xy| \text{ divides }|x||y|$.

3. 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.

4. 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 $\displaystyle a$ so that $\displaystyle G = \{ a,a^2,a^3,a^4,a^5,a^6\}$, remembmer $\displaystyle a^k$ means $\displaystyle a$ raised to $\displaystyle k$ reduced mod $\displaystyle 7$.

5. 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.

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