Question

Let G be defined as all formal symbols {x'yl, i, j=1), 1, 2, ..., n-1 wher
(i) x' y=x'y' if and only if i=i',j=j',
(ii) x=yh=e, n>2
(ii) xy=y''x}.
Then G is :
G={x'y',i, j=0, 1,2,..., -1,
any body tell me the answer?
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Answer 1

Step-by-step explanation:

If n=1 or n=2, then D2n is abelian and hence Z(D2n)=D2n. Suppose n≥3. By definition, we have D2n={aibj:i=0,1,j=0,1,⋯,n−1} where a is an element of order 2, b is an element of order n and a,b are related by the relation ba=ab−1. It then follows that b2a=bab−1=ab−2 and in general

bra=ab−r, (1)

for all integers r≥0. Now, since a and b together generate D2n, an element of D2n is in the center if and only if it commutes with both a and b. So x=aibj∈Z(D2n) if and only if xa=ax and xb=bx. The condition xb=bx gives us aibj+1=baibj and thus

aib=bai. (2)

Clearly we can have i=0 in (2) but can we have i=1 ? The answer is no because if i=1, then (2) gives ab=ba and we know from (1) that ba=ab−1. But then ab=ab−1 and so b2=1, which is not possible because n≥3. So i=0 and thus x=bj. The condition xa=ax then becomes bja=abj and hence, by (1), ab−j=abj. Therefore

b2j=1. (3)

Hence, since o(b)=n, we get from (3) that n∣2j. Thus either j=0 or 2j=n because 0≤j≤n−1. If j=0, then x=bj=1. If 2j=n, then n is even and x=bj=bn/2. So we have proved

Z(D2n)=⎧⎩⎨⎪⎪D2n{1}{1,bn2if n=1,2if n>2 is oddif n>2 is even