**Problem 1**. Let be a group with the center Given an odd integer let

Show that if is abelian, then is a normal subgroup of

**Solution**. It is clear that and implies that Also, if and then and so Thus the only thing we need to prove is that is multiplicatively closed. So let We want to show that Let

Claim 1:

*Proof*. Since is abelian, and commute, i.e. and thus

Claim 2: for all integers

*Proof*. By induction over By Claim 1, and commute and so

and that proves the claim for Now suppose the claim is true for Then

and that completes the induction.

Claim 3:

*Proof*. By Claim 2, for all integers and so, in particular, for we get

Claim 4: for all integers

*Proof*. The proof is by induction over There’s nothing to prove for Suppose now that the claim is true for By Claim 2, and thus

which completes the induction and the proof of the claim.

Now, if in Claim 4, we put then we get

because But since is odd, divides and hence because, by Claim 3, Therefore, by i.e. and that completes the solution.

**Problem 2**. Let be a non-abelian group of order where is a prime number. Let be the center of and let Show that

i)

ii) if is odd, then is a normal subgroup of and

iii) if then is not always a subgroup.

**Solution**. i) Since is a non-abelian -group, and So either or If then would be a group of order hence cyclic. But then would be abelian, which is non-sense. So

ii) By i), is a group of order hence abelian. So, by Problem 1, is a normal subgroup of It is clear that because, by i), You may also argue that because, by Cauchy, has an element of order Thus since divides we must have

iii) Consider the dihedral group of order

Then but , i.e. in this case, is not closed under multiplication hence not a subgroup.

**Exercise**. Let be a commutative ring with unity and let be the Heisenberg group over Show that for any odd integer the set is a normal subgroup of