Let G=<x, y| x^2n=e, x^n=y2, xy=yx^–1>. Show Z(G)={e, xn}. Assuming |G|=4n, show G/Z(G) is isomorphic to Dn. I really have no clue how to even start this problem, any help would be greatly appreciated.

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- May 24th 2010, 10:40 AM #1

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- May 26th 2010, 11:53 AM #2
I couldn't really understand the notation. Is this the problem?

Let $\displaystyle G=\{x, y| x^{2n}=e, x^n=y^2, xy=yx^{-1}\}$. Show $\displaystyle Z(G)=\{e, x^n\}$. Assuming $\displaystyle |G|=4n$, show $\displaystyle G/Z(G)$ is isomorphic to $\displaystyle D_n$. I really have no clue how to even start this problem, any help would be greatly appreciated.

- May 26th 2010, 06:52 PM #3

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- May 26th 2010, 08:18 PM #4

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So I tried breaking this up into cases:

Case 1: If n=1. then |x|=1 or 2. If |x|=1, then x=e and x would obviously be in the center.

If |x|=2, then xy=yx (since (y^-1)xy=x^-1 and x^-1 = x when |x|=2). Thus G is abelain, and Z(G) would be {e,x,y^2} but since y^2=x, are Z(G) would be {e,x}.

Case 2. n>1

If n>1 and the order of x>2, then xx^(n)=x^(n+1)=x^(n)x and x^(n)y=y^2y=y^3=yy^2=yx^n. Since x^n commutes with the generates of G, x^n commutes with all of G. But G is not abelain, because if it were, y^-1xy=y^-1yx=x=x^-1, which is not true when |x|>2. Thus, the only elements in Z(G) are {e,x^n}.

- May 27th 2010, 11:35 AM #5

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How about this: since x^n=y^2, then G= <x> union y<x>. So G= {y^jx^i|0<=i<= n-1, 0<=j<=1}.

Is x^k an element of Z(G) for some k? So x^ky=yx^k, thus k=2 since x^2=y (do I need to say more?). Thus x^k is contained in the Z(G).

Is yx^k an element of Z(G) for some K? The answer is no, but how would I disprove it?

- May 28th 2010, 12:42 AM #6