**Problem**. Let and let be the dihedral group of order Find the center of

**Solution**. If or then is abelian and hence Now, suppose By definition, we have

where is an element of order 2, is an element of order and are related by the relation It then follows that and in general

for all integers Now, since and together generate an element of is in the center if and only if it commutes with both and So if and only if and The condition gives us and thus

Clearly we can have in (2) but can we have ? The answer is no because if then (2) gives and we know from (1) that But then and so which is not possible because So and thus The condition then becomes and hence, by (1), Therefore

Hence, since we get from (3) that Thus either or because If then If then is even and So we have proved

Hi, do you not mean to conclude that the centre for an even n would be {1, a^(n/2)}?

In D12 for example, we have a^3b =ba^3 and not b^3a=ab^3 as b^3a=a^5b.

No, the center of is Notice that, with my notation, in has order 2 and has order 6 (you’re probably assuming has order 6 and has order 2). So we can’t have because then would be abelian (because ).

In the very end, for even n, centre should be {1,r^n} instead of {1,r^(n/2)} which is correct if you are using D_n notation but rather you are using D_2n notation.

No, is correct; read the solution again.

Bhaskar is correct. D_8 implies n=4 and n/2=2, b^2 it’s not in the center but b^4 is.

b^2(b^3a)=b^5a

(b^3a)b^2=b^3(b^6a)=ba

Therefore, b^2 is not in the center.

No, the center of as I have proved, is

Remember that in the order of is 4. So which is obviously in the center.

Also because

This is a very useful proof.

Thanks mate, good stuff, easy to follow

really great

this is very good proof…