1. ## Groups-Permutation groups

I have the following questions but I don't totally understand them, so im not sure about it all:

Let H be the subgroup of $S_8$ generated by the permutations

$\alpha = (1 2 3 4)(5 6 7 8), \beta = (1 5 3 7)(2 8 4 6)
$

(1) Find the order of $\alpha, \alpha\beta$. What's the order of $\beta^{-1}\alpha\beta$ ?

(2) Find the order of H.

(3) Find the conjugacy classes of H.

(4) Show that $Z=\{e, \alpha^2 \}$ is a normal subgroup of H

(5) To which well-known group is the quotient H/Z isomorphic?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
For (1) I know the orders are all 4 I think.

For (2) Is the fastest way of doing this to actually go through and compute every new element? Or is there an easier way of finding the order?

For (3) I think the conjugacy classes are the sets in which the permutions all have the same cycle structure?

For (4) Z is normal iff Z is a union of conjugacy classes?

For (5) I think this is isomorphic to the Klein Four Group. Find the orders of elements in H/Z ?

Thanks for any help.

2. i am nebie....., but i enjoy grup theory, can i follow this thread?

i am nebie....., but i enjoy grup theory, can i follow this thread?
Yes you can follow this thread and feel free to join in.

4. Originally Posted by Jason Bourne
I have the following questions but I don't totally understand them, so im not sure about it all:

Let H be the subgroup of $S_8$ generated by the permutations

$\alpha = (1 2 3 4)(5 6 7 8), \beta = (1 5 3 7)(2 8 4 6)
$

(1) Find the order of $\alpha, \alpha\beta$. What's the order of $\beta^{-1}\alpha\beta$ ?

(2) Find the order of H.

(3) Find the conjugacy classes of H.

(4) Show that $Z=\{e, \alpha^2 \}$ is a normal subgroup of H

(5) To which well-known group is the quotient H/Z isomorphic?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
For (1) I know the orders are all 4 I think.

For (2) Is the fastest way of doing this to actually go through and compute every new element? Or is there an easier way of finding the order?

For (3) I think the conjugacy classes are the sets in which the permutions all have the same cycle structure?

For (4) Z is normal iff Z is a union of conjugacy classes?

For (5) I think this is isomorphic to the Klein Four Group. Find the orders of elements in H/Z ?

Thanks for any help.
this is a long exercise ... here are some hints:

1. In any group, the order of an element and any of its conjugates are equal because $(y^{-1}xy)^k=y^{-1}x^ky.$ so $o(\beta^{-1} \alpha \beta)=o(\alpha)=4.$

2. show that $\alpha^2=\beta^2$ and $\alpha \beta \alpha = \beta$ and therefore $H \cong Q_8,$ the quaternion group of order 8.

3. $H$ has 5 conjugacy classes: $\{1 \}, \ \{\alpha, \alpha^3 \}, \ \{\alpha^2 \}, \ \{\beta, \beta^3 \}, \ \{\alpha \beta, \alpha \beta^3 \}.$

5. 2. show that $\alpha^2=\beta^2$ and $\alpha \beta \alpha = \beta$ and therefore $H \cong Q_8,$ the quaternion group of order 8.
I don't understand how this implies H is isomorphic to Q8

3. $H$ has 4 conjugacy classes: $\{1 \}, \ \{\alpha, \alpha^3 \}, \ \{\alpha^2 \}, \ \{\beta, \beta^3 \}, \ \{\alpha \beta, \alpha \beta^3 \}.$
I would have though that the conjugacy classes are:

$\{1 \},\ \{\alpha^2 \}, \ \{\alpha, \alpha^3 , \beta, \beta^3, \alpha \beta, \alpha \beta^3 \}$

because the elements in each set have the same disjoint cycle structure?

6. Originally Posted by Jason Bourne

I don't understand how this implies H is isomorphic to Q8
because they have the same presentations.

I would have though that the conjugacy classes are:

$\{1 \},\ \{\alpha^2 \}, \ \{\alpha, \alpha^3 , \beta, \beta^3, \alpha \beta, \alpha \beta^3 \}$

because the elements in each set have the same disjoint cycle structure?
those are conjugates in $S_8$ not in $H.$ don't forget that we want to find the conjugacy classes of $H$.

7. because they have the same presentations.
Presentations????

those are conjugates in $S_8$ not in $H.$ don't forget that we want to find the conjugacy classes of $H$.
Now im confused. So how do I find the conjugacy classes of H? I'm clearly not understanding this stuff.