Let G = <x, y | x2n = e, xn = y2, y-1xy = x-1>. Show that Z(G) ={e, xn}. Assuming that |G| = 4n, show that G/Z(G) is isomorphic toDn. (The group G is called the dicyclic group of order 4n.)
Let G = <x, y | x2n = e, xn = y2, y-1xy = x-1>. Show that Z(G) ={e, xn}. Assuming that |G| = 4n, show that G/Z(G) is isomorphic to
Dn. (The group G is called the dicyclic group of order 4n.)
To show that Z(G) = {e, xn}, we need to find all elements in G that commute with every other element in G.
First, note that x commutes with xn, since x(xn) = x2n = e = (xn)x. Similarly, y commutes with y-1, since y(y-1) = yy-1 = e = y-1y.
Now, suppose that z is an element in Z(G) that is not equal to e or xn. Then, z cannot be a power of x, since x does not commute with y. Similarly, z cannot be a power of y, since y does not commute with x-1. Thus, z must be a product of x and y (or their inverses) in some nontrivial way.
However, we can use the given relations to simplify any such product. For example, let z = xyx-1y-1. Then, we can use the last relation to rewrite this as:
z = xyx-1y-1 = x(yxy-1)y-1 = x(x-1yy-1)y-1 = xy-1
Thus, any element in Z(G) that is not e or xn can be written as a power of y. But this contradicts the fact that y does not commute with x-1, since (x-1)y = (y-1)x-1. Therefore, the only elements in Z(G) are e and xn.
To show that G/Z(G) is isomorphic to Dn, we need to find a surjective homomorphism from G to Dn with kernel Z(G). One way to do this is to define the following homomorphism:
φ: G -> Dn
x -> r
y -> sr
where r and s are generators of Dn satisfying r2 = s2 = (rs)n = e.
It is clear that φ is surjective, since r and s generate Dn. We just need to show that φ is well-defined and has kernel Z(G).
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