**2)(a) is generated by the transpositions: **

Proof- Let . We can write , where . Note that each . Using this we can write . Thus any permutation in can be written as a product of the given transpositions.

**(b) is generated by the transpositions **

Proof- Let . From part a we can write where . Now consider . If then leave alone. If then write . Repeat this process on each transposition of the form in the composition of for each as long as . Doing this gives written as a product of transpositions of the form so that we are done.

**12) The center of the group is if n is odd and if n is even.**

Proof- First we need to know all the normal subgroups in . If n is odd, the nontrivial subgroups are the group of rotations, and all the subgroups of the group of rotations. If n is even we have two more normal subgroups, each isomorphic to .

Let H be a normal subgroup. We have two cases, either H consists only of powers of the element a, or there is some . Suppose that for some k dividing n. Then:

because of the relation . And finally,

. Thus the subgroup is normal.

Now instead suppose there exists some . Then we have since H is normal. This gives that . This tells us that is a generator of H.

Next, . If is even let and if is odd let . Therefore either b is a generator or ab is a generator. So that or .

Now that we know all the normal subgroups we can find the center, because it must be a normal subgroup. Recall the center is the set of all elements which can commute with any other element.

First we check . We have , but n is larger than 2 so this is a contradiction. Thus this is not the center.

Next we check . We have that:

. But for n = 4, again a contradiction so this is not the center.

Lastly, assume that is the center. Then we have . Thus the center can only be which the same as but only when n is even. If n is odd then the center is just the identity.

### Like this:

Like Loading...